Taikydami Gryno formulę, apskaičiuokime kreivinį integralą
∫
L
x
y
d
x
+
(
x
2
+
y
2
)
d
y
,
{\displaystyle \int _{L}xy\mathbf {d} x+(x^{2}+y^{2})\mathbf {d} y,}
L - apskritimas
x
2
+
y
2
=
a
x
{\displaystyle x^{2}+y^{2}=ax}
Kadangi skritulyje
x
2
+
y
2
≤
a
x
{\displaystyle x^{2}+y^{2}\leq ax}
P
(
x
,
y
)
=
x
y
{\displaystyle P(x,y)=xy}
Q
(
x
,
y
)
=
x
2
+
y
2
{\displaystyle Q(x,y)=x^{2}+y^{2}}
∂
P
∂
y
=
x
{\displaystyle {\partial P \over \partial y}=x}
∂
Q
∂
x
=
2
x
{\displaystyle {\partial Q \over \partial x}=2x}
Turime:
∫
L
x
y
d
x
+
(
x
2
+
y
2
)
d
y
=
∬
D
(
2
x
−
x
)
d
x
d
y
=
∬
D
x
d
x
d
y
.
{\displaystyle \int _{L}xy\;dx+(x^{2}+y^{2})dy=\iint _{D}(2x-x)\;dx\;dy=\iint _{D}x\;dx\;dy.}
Randame apskritimo y išraišką:
x
2
+
y
2
=
a
x
,
{\displaystyle x^{2}+y^{2}=ax,}
y
2
=
a
x
−
x
2
,
{\displaystyle y^{2}=ax-x^{2},}
y
=
a
x
−
x
2
.
{\displaystyle y={\sqrt {ax-x^{2}}}.}
Randame y išvestinę, o paskui ir dy :
y
′
=
d
y
d
x
=
(
a
x
−
x
2
)
′
=
(
a
x
−
x
2
)
′
2
a
x
−
x
2
=
a
−
2
x
2
a
x
−
x
2
;
{\displaystyle y'={\frac {dy}{dx}}=({\sqrt {ax-x^{2}}})'={\frac {(ax-x^{2})'}{2{\sqrt {ax-x^{2}}}}}={\frac {a-2x}{2{\sqrt {ax-x^{2}}}}};}
d
y
=
a
−
2
x
2
a
x
−
x
2
d
x
.
{\displaystyle dy={\frac {a-2x}{2{\sqrt {ax-x^{2}}}}}\;dx.}
Apskritimo spindulys
R
=
a
2
,
{\displaystyle R={\frac {a}{2}},}
a
=
3
,
{\displaystyle a=3,}
x
2
+
y
2
=
3
x
.
{\displaystyle x^{2}+y^{2}=3x.}
O (0; 0) ir kad ašis Ox dalina apskritimą pusiau. Vadinasi, kai
x
=
0
{\displaystyle x=0}
y
=
0
,
{\displaystyle y=0,}
0
2
+
0
2
=
3
⋅
0.
{\displaystyle 0^{2}+0^{2}=3\cdot 0.}
x
2
+
y
2
=
3
x
.
{\displaystyle x^{2}+y^{2}=3x.}
Ox ašies. Įstačius taško (3; 0) koordinates į apskritimo lygtį
x
2
+
y
2
=
3
x
{\displaystyle x^{2}+y^{2}=3x}
3
2
+
0
2
=
3
⋅
3.
{\displaystyle 3^{2}+0^{2}=3\cdot 3.}
Ox ašies. Todėl apskritimo
x
2
+
y
2
=
3
x
{\displaystyle x^{2}+y^{2}=3x}
R
=
3
2
=
a
2
.
{\displaystyle R={\frac {3}{2}}={\frac {a}{2}}.}
Kadangi apskritimo
x
2
+
y
2
=
a
x
{\displaystyle x^{2}+y^{2}=ax}
R
=
a
2
{\displaystyle R={\frac {a}{2}}}
Ox ašis dalina apskritimą per pusę, tai didžiausia y reikšmė gali būti
R
=
a
2
.
{\displaystyle R={\frac {a}{2}}.}
x
2
+
y
2
≤
a
x
{\displaystyle x^{2}+y^{2}\leq ax}
x
2
+
y
2
≤
a
x
{\displaystyle x^{2}+y^{2}\leq ax}
x kinta nuo 0 iki a, o y kinta nuo 0 iki
R
=
a
2
.
{\displaystyle R={\frac {a}{2}}.}
Taikydami Gryno formulę, integruojame (pasinaudodami internetiniu integratoriumi):
∬
D
x
d
x
d
y
=
2
∫
0
a
x
(
∫
0
a
x
−
x
2
d
y
)
d
x
=
2
∫
0
a
x
(
y
|
0
a
x
−
x
2
)
d
x
=
2
∫
0
a
x
a
x
−
x
2
d
x
=
{\displaystyle \iint _{D}xdxdy=2\int _{0}^{a}x(\int _{0}^{\sqrt {ax-x^{2}}}dy)dx=2\int _{0}^{a}x(y|_{0}^{\sqrt {ax-x^{2}}})dx=2\int _{0}^{a}x{\sqrt {ax-x^{2}}}dx=}
=
2
⋅
−
x
(
x
−
a
)
(
3
a
3
arctan
(
x
a
−
x
)
+
x
a
−
x
(
−
3
a
2
−
2
a
x
+
8
x
2
)
)
24
x
a
−
x
|
0
a
=
{\displaystyle =2\cdot {\frac {{\sqrt {-x(x-a)}}\left(3a^{3}\arctan \left({\frac {\sqrt {x}}{\sqrt {a-x}}}\right)+{\sqrt {x}}{\sqrt {a-x}}(-3a^{2}-2ax+8x^{2})\right)}{24{\sqrt {x}}{\sqrt {a-x}}}}|_{0}^{a}=}
=
3
a
3
arctan
(
x
a
−
x
)
+
x
a
−
x
(
−
3
a
2
−
2
a
x
+
8
x
2
)
12
|
0
a
=
{\displaystyle ={\frac {3a^{3}\arctan \left({\frac {\sqrt {x}}{\sqrt {a-x}}}\right)+{\sqrt {x}}{\sqrt {a-x}}(-3a^{2}-2ax+8x^{2})}{12}}|_{0}^{a}=}
=
3
a
3
arctan
(
a
a
−
a
)
+
a
a
−
a
(
−
3
a
2
−
2
a
⋅
a
+
8
a
2
)
12
−
3
a
3
arctan
(
0
a
−
0
)
+
0
a
−
0
(
−
3
a
2
−
2
a
⋅
0
+
8
⋅
0
2
)
12
=
{\displaystyle ={\frac {3a^{3}\arctan \left({\frac {\sqrt {a}}{\sqrt {a-a}}}\right)+{\sqrt {a}}{\sqrt {a-a}}(-3a^{2}-2a\cdot a+8a^{2})}{12}}-{\frac {3a^{3}\arctan \left({\frac {\sqrt {0}}{\sqrt {a-0}}}\right)+{\sqrt {0}}{\sqrt {a-0}}(-3a^{2}-2a\cdot 0+8\cdot 0^{2})}{12}}=}
=
3
a
3
arctan
a
0
12
−
3
a
3
arctan
0
a
12
=
3
a
3
arctan
(
error
)
12
−
3
a
3
arctan
(
0
)
12
=
{\displaystyle ={\frac {3a^{3}\arctan {\frac {\sqrt {a}}{0}}}{12}}-{\frac {3a^{3}\arctan {\frac {0}{\sqrt {a}}}}{12}}={\frac {3a^{3}\arctan({\text{error}})}{12}}-{\frac {3a^{3}\arctan(0)}{12}}=}
=
3
a
3
arctan
(
∞
)
12
−
3
a
3
⋅
0
12
=
3
a
3
⋅
π
2
12
−
0
=
3
a
3
⋅
π
24
=
a
3
π
8
.
{\displaystyle ={\frac {3a^{3}\arctan(\infty )}{12}}-{\frac {3a^{3}\cdot 0}{12}}={\frac {3a^{3}\cdot {\frac {\pi }{2}}}{12}}-0={\frac {3a^{3}\cdot \pi }{24}}={\frac {a^{3}\pi }{8}}.}
Pasitikriname (įstatydami y ir dy ir pasinaudodami internetiniu integratoriumi ):
∫
L
x
y
d
x
+
(
x
2
+
y
2
)
d
y
=
∫
L
x
a
x
−
x
2
d
x
+
(
x
2
+
(
a
x
−
x
2
)
2
)
a
−
2
x
2
a
x
−
x
2
d
x
=
{\displaystyle \int _{L}xy\;dx+(x^{2}+y^{2})dy=\int _{L}x{\sqrt {ax-x^{2}}}\;dx+(x^{2}+({\sqrt {ax-x^{2}}})^{2}){\frac {a-2x}{2{\sqrt {ax-x^{2}}}}}\;dx=}
=
2
∫
0
a
(
x
a
x
−
x
2
+
(
x
2
+
(
a
x
−
x
2
)
2
)
a
−
2
x
2
a
x
−
x
2
)
d
x
=
{\displaystyle =2\int _{0}^{a}\left(x{\sqrt {ax-x^{2}}}+(x^{2}+({\sqrt {ax-x^{2}}})^{2}){\frac {a-2x}{2{\sqrt {ax-x^{2}}}}}\right)dx=}
=
2
∫
0
a
(
x
a
x
−
x
2
+
(
x
2
+
a
x
−
x
2
)
a
−
2
x
2
a
x
−
x
2
)
d
x
=
{\displaystyle =2\int _{0}^{a}\left(x{\sqrt {ax-x^{2}}}+(x^{2}+ax-x^{2}){\frac {a-2x}{2{\sqrt {ax-x^{2}}}}}\right)dx=}
=
2
∫
0
a
(
x
a
x
−
x
2
+
a
x
(
a
−
2
x
)
2
a
x
−
x
2
)
d
x
=
{\displaystyle =2\int _{0}^{a}\left(x{\sqrt {ax-x^{2}}}+{\frac {ax(a-2x)}{2{\sqrt {ax-x^{2}}}}}\right)dx=}
=
2
∫
0
a
(
x
a
x
−
x
2
+
a
2
x
−
2
a
x
2
2
a
x
−
x
2
)
d
x
=
{\displaystyle =2\int _{0}^{a}\left(x{\sqrt {ax-x^{2}}}+{\frac {a^{2}x-2ax^{2}}{2{\sqrt {ax-x^{2}}}}}\right)dx=}
=
2
∫
0
a
2
x
(
a
x
−
x
2
)
+
a
2
x
−
2
a
x
2
2
a
x
−
x
2
d
x
=
{\displaystyle =2\int _{0}^{a}{\frac {2x(ax-x^{2})+a^{2}x-2ax^{2}}{2{\sqrt {ax-x^{2}}}}}dx=}
=
2
∫
0
a
2
a
x
2
−
2
x
3
+
a
2
x
−
2
a
x
2
2
a
x
−
x
2
d
x
=
{\displaystyle =2\int _{0}^{a}{\frac {2ax^{2}-2x^{3}+a^{2}x-2ax^{2}}{2{\sqrt {ax-x^{2}}}}}dx=}
=
∫
0
a
−
2
x
3
+
a
2
x
a
x
−
x
2
d
x
=
∫
0
a
x
(
a
2
−
2
x
2
)
a
x
−
x
2
d
x
=
{\displaystyle =\int _{0}^{a}{\frac {-2x^{3}+a^{2}x}{\sqrt {ax-x^{2}}}}dx=\int _{0}^{a}{\frac {x(a^{2}-2x^{2})}{\sqrt {ax-x^{2}}}}dx=}
=
1
3
a
x
−
x
2
(
−
3
a
2
+
4
a
x
+
2
x
2
)
|
0
a
=
{\displaystyle ={\frac {1}{3}}{\sqrt {ax-x^{2}}}(-3a^{2}+4ax+2x^{2})|_{0}^{a}=}
=
1
3
a
⋅
a
−
a
2
(
−
3
a
2
+
4
a
⋅
a
+
2
a
2
)
−
1
3
a
⋅
0
−
0
2
(
−
3
a
2
+
4
a
⋅
0
+
2
⋅
0
2
)
=
{\displaystyle ={\frac {1}{3}}{\sqrt {a\cdot a-a^{2}}}(-3a^{2}+4a\cdot a+2a^{2})-{\frac {1}{3}}{\sqrt {a\cdot 0-0^{2}}}(-3a^{2}+4a\cdot 0+2\cdot 0^{2})=}
=
1
3
0
(
−
3
a
2
+
4
a
2
+
2
a
2
)
−
1
3
0
(
−
3
a
2
+
0
+
0
)
=
0
−
0
=
0.
{\displaystyle ={\frac {1}{3}}{\sqrt {0}}(-3a^{2}+4a^{2}+2a^{2})-{\frac {1}{3}}{\sqrt {0}}(-3a^{2}+0+0)=0-0=0.}
Pastaba, kad taip gauname dalyba iš nulio ir neįmanoma vietomis išintegruoti įstatant a arba 0. Bet gauname kažką panašesnio į teisingą atsakymą:
=
∫
0
a
−
2
x
3
+
a
2
x
a
x
−
x
2
d
x
=
{\displaystyle =\int _{0}^{a}{\frac {-2x^{3}+a^{2}x}{\sqrt {ax-x^{2}}}}dx=}
=
x
(
3
a
3
+
7
a
2
x
−
2
a
x
2
−
8
x
3
)
−
3
a
3
x
a
−
x
arctan
x
a
−
x
12
−
x
(
x
−
a
)
|
0
a
=
{\displaystyle ={\frac {x(3a^{3}+7a^{2}x-2ax^{2}-8x^{3})-3a^{3}{\sqrt {x}}{\sqrt {a-x}}\arctan {\frac {\sqrt {x}}{\sqrt {a-x}}}}{12{\sqrt {-x(x-a)}}}}|_{0}^{a}=}
=
(
x
(
3
a
3
+
7
a
2
x
−
2
a
x
2
−
8
x
3
)
12
x
(
a
−
x
)
−
3
a
3
x
(
a
−
x
)
arctan
x
a
−
x
12
x
(
a
−
x
)
)
|
0
a
=
{\displaystyle =\left({\frac {x(3a^{3}+7a^{2}x-2ax^{2}-8x^{3})}{12{\sqrt {x(a-x)}}}}-{\frac {3a^{3}{\sqrt {x(a-x)}}\arctan {\frac {\sqrt {x}}{\sqrt {a-x}}}}{12{\sqrt {x(a-x)}}}}\right)|_{0}^{a}=}
=
(
x
(
3
a
3
+
7
a
2
x
−
2
a
x
2
−
8
x
3
)
12
x
(
a
−
x
)
−
3
a
3
arctan
x
a
−
x
12
)
|
0
a
=
{\displaystyle =\left({\frac {x(3a^{3}+7a^{2}x-2ax^{2}-8x^{3})}{12{\sqrt {x(a-x)}}}}-{\frac {3a^{3}\arctan {\frac {\sqrt {x}}{\sqrt {a-x}}}}{12}}\right)|_{0}^{a}=}
=
(
x
(
3
a
3
+
7
a
2
x
−
2
a
x
2
−
8
x
3
)
12
a
−
x
−
a
3
arctan
x
a
−
x
4
)
|
0
a
=
{\displaystyle =\left({\frac {{\sqrt {x}}(3a^{3}+7a^{2}x-2ax^{2}-8x^{3})}{12{\sqrt {a-x}}}}-{\frac {a^{3}\arctan {\frac {\sqrt {x}}{\sqrt {a-x}}}}{4}}\right)|_{0}^{a}=}
=
(
x
(
7
a
2
x
−
2
a
x
2
−
5
x
3
)
a
−
x
12
(
a
−
x
)
−
a
3
arctan
x
a
−
x
4
)
|
0
a
.
{\displaystyle =\left({\frac {{\sqrt {x}}(7a^{2}x-2ax^{2}-5x^{3}){\sqrt {a-x}}}{12(a-x)}}-{\frac {a^{3}\arctan {\frac {\sqrt {x}}{\sqrt {a-x}}}}{4}}\right)|_{0}^{a}.}
Toliau pasinaudojame internetiniu polinomų dalikliu parinkę
a
=
3
{\displaystyle a=3}
(
7
a
2
x
−
2
a
x
2
−
5
x
3
)
/
(
a
−
x
)
=
(
−
5
x
3
−
6
x
2
+
63
x
)
/
(
−
x
+
3
)
=
5
x
2
+
21
x
.
{\displaystyle (7a^{2}x-2ax^{2}-5x^{3})/(a-x)=(-5x^{3}-6x^{2}+63x)/(-x+3)=5x^{2}+21x.}
(
x
(
7
a
2
x
−
2
a
x
2
−
5
x
3
)
a
−
x
12
(
a
−
x
)
−
a
3
arctan
x
a
−
x
4
)
|
0
a
=
{\displaystyle \left({\frac {{\sqrt {x}}(7a^{2}x-2ax^{2}-5x^{3}){\sqrt {a-x}}}{12(a-x)}}-{\frac {a^{3}\arctan {\frac {\sqrt {x}}{\sqrt {a-x}}}}{4}}\right)|_{0}^{a}=}
=
(
x
(
7
⋅
3
2
x
−
2
⋅
3
x
2
−
5
x
3
)
3
−
x
12
(
3
−
x
)
−
3
3
arctan
x
3
−
x
4
)
|
0
3
=
{\displaystyle =\left({\frac {{\sqrt {x}}(7\cdot 3^{2}x-2\cdot 3x^{2}-5x^{3}){\sqrt {3-x}}}{12(3-x)}}-{\frac {3^{3}\arctan {\frac {\sqrt {x}}{\sqrt {3-x}}}}{4}}\right)|_{0}^{3}=}
=
(
(
−
5
x
3
−
6
x
2
+
63
x
)
x
(
3
−
x
)
12
(
−
x
+
3
)
−
27
arctan
x
3
−
x
4
)
|
0
3
=
{\displaystyle =\left({\frac {(-5x^{3}-6x^{2}+63x){\sqrt {x(3-x)}}}{12(-x+3)}}-{\frac {27\arctan {\frac {\sqrt {x}}{\sqrt {3-x}}}}{4}}\right)|_{0}^{3}=}
=
(
(
5
x
2
+
21
x
)
x
(
3
−
x
)
12
−
27
arctan
x
3
−
x
4
)
|
0
3
=
{\displaystyle =\left({\frac {(5x^{2}+21x){\sqrt {x(3-x)}}}{12}}-{\frac {27\arctan {\frac {\sqrt {x}}{\sqrt {3-x}}}}{4}}\right)|_{0}^{3}=}
=
(
(
5
⋅
3
2
+
21
⋅
3
)
3
(
3
−
3
)
12
−
27
arctan
3
3
−
3
4
)
−
(
(
5
⋅
0
2
+
21
⋅
0
)
0
(
3
−
0
)
12
−
27
arctan
0
3
−
0
4
)
=
{\displaystyle =\left({\frac {(5\cdot 3^{2}+21\cdot 3){\sqrt {3(3-3)}}}{12}}-{\frac {27\arctan {\frac {\sqrt {3}}{\sqrt {3-3}}}}{4}}\right)-\left({\frac {(5\cdot 0^{2}+21\cdot 0){\sqrt {0(3-0)}}}{12}}-{\frac {27\arctan {\frac {\sqrt {0}}{\sqrt {3-0}}}}{4}}\right)=}
=
(
(
45
+
63
)
3
⋅
0
12
−
27
arctan
3
0
4
)
−
(
0
−
27
arctan
(
0
)
4
)
=
{\displaystyle =\left({\frac {(45+63){\sqrt {3\cdot 0}}}{12}}-{\frac {27\arctan {\frac {\sqrt {3}}{0}}}{4}}\right)-\left(0-{\frac {27\arctan(0)}{4}}\right)=}
=
(
0
−
27
arctan
(
∞
)
4
)
−
(
−
27
⋅
0
4
)
=
{\displaystyle =\left(0-{\frac {27\arctan(\infty )}{4}}\right)-\left(-{\frac {27\cdot 0}{4}}\right)=}
=
(
−
27
⋅
π
2
4
)
−
0
=
−
27
π
8
.
{\displaystyle =\left(-{\frac {27\cdot {\frac {\pi }{2}}}{4}}\right)-0=-{\frac {27\pi }{8}}.}
Kadangi
a
3
=
3
3
=
27
,
{\displaystyle a^{3}=3^{3}=27,}
a
3
π
8
{\displaystyle {\frac {a^{3}\pi }{8}}}
a
=
3
{\displaystyle a=3}
Apskaičiuosime darbą jėgos
F
(
x
,
y
)
{\displaystyle F(x,y)}
Pagal sąlyga,
|
F
(
x
,
y
)
|
=
x
2
+
y
2
;
{\displaystyle |F(x,y)|={\sqrt {x^{2}+y^{2}}};}
P
=
−
x
,
{\displaystyle P=-x,}
Q
=
−
y
{\displaystyle Q=-y}
−
{\displaystyle -}
A
=
−
∮
L
x
d
x
+
y
d
y
,
{\displaystyle A=-\oint _{L}xdx+ydy,}
L - elipsė
x
=
a
cos
t
,
y
=
b
sin
t
,
{\displaystyle x=a\cos t,\;y=b\sin t,}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
A
=
−
∫
0
2
π
a
cos
t
(
−
a
sin
t
)
d
t
+
b
sin
(
t
)
b
cos
(
t
)
d
t
=
−
∫
0
2
π
(
b
2
−
a
2
)
sin
(
t
)
cos
t
d
t
=
{\displaystyle A=-\int _{0}^{2\pi }a\cos t(-a\sin t)dt+b\sin(t)b\cos(t)dt=-\int _{0}^{2\pi }(b^{2}-a^{2})\sin(t)\cos t\;dt=}
=
a
2
−
b
2
2
∫
0
2
π
sin
(
2
t
)
d
t
=
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
2
π
=
0.
{\displaystyle ={a^{2}-b^{2} \over 2}\int _{0}^{2\pi }\sin(2t)dt={a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{2\pi }=0.}
Jei t keistusi nuo 0 iki
π
2
,
{\displaystyle {\pi \over 2},}
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
π
2
=
a
2
−
b
2
2
.
{\displaystyle {a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{\pi \over 2}={a^{2}-b^{2} \over 2}.}
a
=
5
,
b
=
3
,
{\displaystyle a=5,\;b=3,}
A
=
a
2
−
b
2
2
=
5
2
−
3
2
2
=
25
−
9
2
=
16
2
=
8.
{\displaystyle A={a^{2}-b^{2} \over 2}={5^{2}-3^{2} \over 2}={\frac {25-9}{2}}={\frac {16}{2}}=8.}
Kad patikrinti ar apskaičiuota teisingai reikia sudėti visas x reikšmes ant elipsės linijos pirmame ketvirtyje. Taip pat reikia sudėti visas y reikšmes ant elipsės linijos pirmame ketvirtyje. Galiausiai reikia sudėti sumas x ir y reikšmių, kad gauti darbą A pirmame ketvirtyje.
Arba tiesiog darbas yra lygus
A
=
∑
n
=
1
100
x
n
2
+
y
n
2
;
{\displaystyle A=\sum _{n=1}^{100}{\sqrt {x_{n}^{2}+y_{n}^{2}}};}
x
n
{\displaystyle x_{n}}
x reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki a ; analogiškai
y
n
{\displaystyle y_{n}}
y reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki b .
Kuris iš šių variantų yra darbas A , paaiškės pasumavus ir paskaičiavus. Sutapimas ar ne, bet
A
=
3
+
5
=
8
{\displaystyle A=3+5=8}
A
=
∑
n
=
1
10
(
n
(
3
/
10
)
)
2
+
(
(
10
−
n
)
(
5
/
10
)
)
2
=
{\displaystyle A=\sum _{n=1}^{10}{\sqrt {(n(3/10))^{2}+((10-n)(5/10))^{2}}}=}
A
=
0.3
2
+
5
2
+
0.6
2
+
4.5
2
+
0.9
2
+
4
2
+
1.2
2
+
3.5
2
+
1.5
2
+
3
2
+
{\displaystyle A={\sqrt {0.3^{2}+5^{2}}}+{\sqrt {0.6^{2}+4.5^{2}}}+{\sqrt {0.9^{2}+4^{2}}}+{\sqrt {1.2^{2}+3.5^{2}}}+{\sqrt {1.5^{2}+3^{2}}}+}
+
1.8
2
+
2.5
2
+
2.1
2
+
2
2
+
2.4
2
+
1.5
2
+
2.7
2
+
1
2
+
3
2
+
0.5
2
=
{\displaystyle +{\sqrt {1.8^{2}+2.5^{2}}}+{\sqrt {2.1^{2}+2^{2}}}+{\sqrt {2.4^{2}+1.5^{2}}}+{\sqrt {2.7^{2}+1^{2}}}+{\sqrt {3^{2}+0.5^{2}}}=}
=
5.0089919
+
4.539823785
+
4.1
+
3.7
+
3.354101966
+
{\displaystyle =5.0089919+4.539823785+4.1+3.7+3.354101966+}
+
3.08058436
+
2.9
+
2.83019434
+
2.879236
+
3.082207
=
{\displaystyle +3.08058436+2.9+2.83019434+2.879236+3.082207=}
=
35.475139351.
{\displaystyle =35.475139351.}
Na, pagal pirmą variantą tikrai ne sutapimas, nes, jei
a
=
8
,
b
=
6
,
{\displaystyle a=8,\;b=6,}
A
=
8
+
6
=
14
{\displaystyle A=8+6=14}
A
=
a
2
−
b
2
2
=
8
2
−
6
2
2
=
64
−
36
2
=
28
2
=
14.
{\displaystyle A={a^{2}-b^{2} \over 2}={8^{2}-6^{2} \over 2}={64-36 \over 2}={\frac {28}{2}}=14.}
Bet ne, vis dėl to tai sutapimas, nes jei
a
=
8
,
b
=
7
,
{\displaystyle a=8,\;b=7,}
A
=
8
+
7
=
15
{\displaystyle A=8+7=15}
A
=
a
2
−
b
2
2
=
8
2
−
7
2
2
=
64
−
49
2
=
15
2
=
7.5.
{\displaystyle A={a^{2}-b^{2} \over 2}={8^{2}-7^{2} \over 2}={64-49 \over 2}={\frac {15}{2}}=7.5.}
Integruojant gauname:
A
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
x
=
{\displaystyle A=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}dx=}
=
1
340
(
34
x
−
45
)
34
x
2
−
90
x
+
225
−
1125
arcsinh
(
3
5
−
34
x
75
)
68
34
|
0
5
=
{\displaystyle ={\frac {1}{340}}(34x-45){\sqrt {34x^{2}-90x+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34x}{75}})}{68{\sqrt {34}}}}|_{0}^{5}=}
=
34
⋅
5
−
45
340
34
⋅
5
2
−
90
⋅
5
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
5
75
)
68
34
−
(
34
⋅
0
−
45
340
34
⋅
0
2
−
90
⋅
0
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
0
75
)
68
34
)
=
{\displaystyle ={\frac {34\cdot 5-45}{340}}{\sqrt {34\cdot 5^{2}-90\cdot 5+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 5}{75}})}{68{\sqrt {34}}}}-\left({\frac {34\cdot 0-45}{340}}{\sqrt {34\cdot 0^{2}-90\cdot 0+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 0}{75}})}{68{\sqrt {34}}}}\right)=}
=
170
−
45
340
850
−
450
+
225
−
1125
arcsinh
(
3
5
−
170
75
)
68
34
−
(
−
45
340
225
−
1125
arcsinh
(
3
5
)
68
34
)
=
{\displaystyle ={\frac {170-45}{340}}{\sqrt {850-450+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {170}{75}})}{68{\sqrt {34}}}}-\left({\frac {-45}{340}}{\sqrt {225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}})}{68{\sqrt {34}}}}\right)=}
=
125
340
625
−
1125
arcsinh
(
3
⋅
75
−
170
⋅
5
5
⋅
75
)
68
34
−
(
−
9
68
⋅
15
−
1125
arcsinh
(
0.6
)
396.50472884948
)
=
{\displaystyle ={\frac {125}{340}}{\sqrt {625}}-{\frac {1125\operatorname {arcsinh} ({\frac {3\cdot 75-170\cdot 5}{5\cdot 75}})}{68{\sqrt {34}}}}-\left({\frac {-9}{68}}\cdot 15-{\frac {1125\operatorname {arcsinh} (0.6)}{396.50472884948}}\right)=}
=
25
68
⋅
25
−
1125
arcsinh
(
225
−
850
375
)
396.50472884948
−
(
−
135
68
−
1125
⋅
0.5688248987
396.50472884948
)
=
{\displaystyle ={\frac {25}{68}}\cdot 25-{\frac {1125\operatorname {arcsinh} ({\frac {225-850}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {1125\cdot 0.5688248987}{396.50472884948}}\right)=}
=
625
68
−
1125
arcsinh
(
−
625
375
)
396.50472884948
−
(
−
135
68
−
639.928011
396.50472884948
)
=
{\displaystyle ={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} ({\frac {-625}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {639.928011}{396.50472884948}}\right)=}
=
9.19117647
−
1125
arcsinh
(
−
5
3
)
396.50472884948
−
(
−
1.98529412
−
1.613922772
)
=
{\displaystyle =9.19117647-{\frac {1125\operatorname {arcsinh} (-{\frac {5}{3}})}{396.50472884948}}-(-1.98529412-1.613922772)=}
=
9.19117647
−
1125
⋅
(
−
1.283795663
)
396.50472884948
−
(
−
3.599216892
)
=
{\displaystyle =9.19117647-{\frac {1125\cdot (-1.283795663)}{396.50472884948}}-(-3.599216892)=}
=
9.19117647
+
3.642504151
+
3.599216892
=
16.43289751.
{\displaystyle =9.19117647+3.642504151+3.599216892=16.43289751.}
Pastebime, kad
2
A
/
10
=
2
⋅
16.43289751
/
10
=
3.286579503.
{\displaystyle 2A/10=2\cdot 16.43289751/10=3.286579503.}
Apskaičiuosime darbą jėgos
F
(
x
,
y
)
{\displaystyle F(x,y)}
Pagal sąlyga,
|
F
(
x
,
y
)
|
=
x
2
+
y
2
;
{\displaystyle |F(x,y)|={\sqrt {x^{2}+y^{2}}};}
P
=
−
x
,
{\displaystyle P=-x,}
Q
=
−
y
{\displaystyle Q=-y}
−
{\displaystyle -}
A
=
−
∮
L
x
d
x
+
y
d
y
,
{\displaystyle A=-\oint _{L}xdx+ydy,}
L - elipsė
x
=
a
cos
t
,
y
=
b
sin
t
,
{\displaystyle x=a\cos t,\;y=b\sin t,}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
A
=
−
∫
0
2
π
a
cos
t
(
−
a
sin
t
)
d
t
+
b
sin
(
t
)
b
cos
(
t
)
d
t
=
−
∫
0
2
π
(
b
2
−
a
2
)
sin
(
t
)
cos
t
d
t
=
{\displaystyle A=-\int _{0}^{2\pi }a\cos t(-a\sin t)dt+b\sin(t)b\cos(t)dt=-\int _{0}^{2\pi }(b^{2}-a^{2})\sin(t)\cos t\;dt=}
=
a
2
−
b
2
2
∫
0
2
π
sin
(
2
t
)
d
t
=
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
2
π
=
0.
{\displaystyle ={a^{2}-b^{2} \over 2}\int _{0}^{2\pi }\sin(2t)dt={a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{2\pi }=0.}
Jei t keistusi nuo 0 iki
π
2
,
{\displaystyle {\pi \over 2},}
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
π
2
=
a
2
−
b
2
4
(
−
cos
(
2
⋅
π
2
)
)
−
a
2
−
b
2
4
(
−
cos
(
2
⋅
0
)
)
=
{\displaystyle {a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{\pi \over 2}={a^{2}-b^{2} \over 4}(-\cos(2\cdot {\pi \over 2}))-{a^{2}-b^{2} \over 4}(-\cos(2\cdot 0))=}
=
a
2
−
b
2
4
(
−
cos
π
+
cos
(
0
)
)
=
a
2
−
b
2
4
(
−
(
−
1
)
+
1
)
=
a
2
−
b
2
2
.
{\displaystyle ={a^{2}-b^{2} \over 4}(-\cos \pi +\cos(0))={a^{2}-b^{2} \over 4}(-(-1)+1)={a^{2}-b^{2} \over 2}.}
Tarkime, jei
a
=
5
,
b
=
3
,
{\displaystyle a=5,\;b=3,}
A
=
a
2
−
b
2
2
=
5
2
−
3
2
2
=
25
−
9
2
=
16
2
=
8.
{\displaystyle A={a^{2}-b^{2} \over 2}={5^{2}-3^{2} \over 2}={\frac {25-9}{2}}={\frac {16}{2}}=8.}
Kad patikrinti ar apskaičiuota teisingai reikia sudėti visas x reikšmes ant elipsės linijos pirmame ketvirtyje. Taip pat reikia sudėti visas y reikšmes ant elipsės linijos pirmame ketvirtyje. Galiausiai reikia sudėti sumas x ir y reikšmių, kad gauti darbą A pirmame ketvirtyje.
Arba tiesiog darbas yra lygus
A
=
1
100
∑
n
=
1
100
x
n
2
+
y
n
2
;
{\displaystyle A={\frac {1}{100}}\sum _{n=1}^{100}{\sqrt {x_{n}^{2}+y_{n}^{2}}};}
x
n
{\displaystyle x_{n}}
x reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki a ; analogiškai
y
n
{\displaystyle y_{n}}
y reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki b .
Kuris iš šių variantų yra darbas A , paaiškės pasumavus ir paskaičiavus. Sutapimas ar ne, bet
A
=
3
+
5
=
8
{\displaystyle A=3+5=8}
A
=
1
10
∑
n
=
1
10
(
n
(
3
/
10
)
)
2
+
(
(
10
−
n
)
(
5
/
10
)
)
2
=
{\displaystyle A={\frac {1}{10}}\sum _{n=1}^{10}{\sqrt {(n(3/10))^{2}+((10-n)(5/10))^{2}}}=}
A
=
(
0.3
2
+
5
2
+
0.6
2
+
4.5
2
+
0.9
2
+
4
2
+
1.2
2
+
3.5
2
+
1.5
2
+
3
2
+
{\displaystyle A=({\sqrt {0.3^{2}+5^{2}}}+{\sqrt {0.6^{2}+4.5^{2}}}+{\sqrt {0.9^{2}+4^{2}}}+{\sqrt {1.2^{2}+3.5^{2}}}+{\sqrt {1.5^{2}+3^{2}}}+}
+
1.8
2
+
2.5
2
+
2.1
2
+
2
2
+
2.4
2
+
1.5
2
+
2.7
2
+
1
2
+
3
2
+
0.5
2
)
/
10
=
{\displaystyle +{\sqrt {1.8^{2}+2.5^{2}}}+{\sqrt {2.1^{2}+2^{2}}}+{\sqrt {2.4^{2}+1.5^{2}}}+{\sqrt {2.7^{2}+1^{2}}}+{\sqrt {3^{2}+0.5^{2}}})/10=}
=
(
5.0089919
+
4.539823785
+
4.1
+
3.7
+
3.354101966
+
{\displaystyle =(5.0089919+4.539823785+4.1+3.7+3.354101966+}
+
3.08058436
+
2.9
+
2.83019434
+
2.879236
+
3.041381265
)
/
10
=
{\displaystyle +3.08058436+2.9+2.83019434+2.879236+3.041381265)/10=}
=
35.43431361
/
10
=
3.543431361.
{\displaystyle =35.43431361/10=3.543431361.}
Iš tikro, ko gero, mes apskaičiuojame tokiu budu ne darbą atlikta apeinant elipsės liniją pirmame ketvirtyje, o darbą atlikta apeinant tiesę pirmame ketvirtyje. Štai kodas programos "Free Pascal" (FreePascal IDE for Win32 for i386; Target CPU: i386; Version 1.0.12 2011/04/23; <Compiler Version 2.4.4>; <Debugger GDB 7.2>; Copyright <C> 1998-2009): var
a:longint;
b,d,c:real;
begin
c:=0;
for a:=1 to 10
do
c:=c+0.1*sqrt(sqr(a*0.3)+sqr((10-a)*0.5));
writeln(c);
readln;
end.
Gauname atsakymą
A
=
3.19325040614418.
{\displaystyle A=3.19325040614418.}
Arba darome taip (pradedame ne nuo 1, o nuo 0, bet tada reikėtų dalinti iš 11, o ne iš 10 kol a iki mažo skaičiaus skaičiuojamas): var
a:longint;
b,d,c:real;
begin
c:=0;
for a:=0 to 10
do
c:=c+0.1*sqrt(sqr(a*0.3)+sqr((10-a)*0.5));
writeln(c);
readln;
end.
Ir gauname atsakymą
A
=
3.69325040614418.
{\displaystyle A=3.69325040614418.}
Panaudojus šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 10
do
c:=c+0.1*sqrt(sqr(a*0.3)+sqr((11-a)*0.5));
writeln(c);
readln;
end.
gauname
A
=
3.54343136406170.
{\displaystyle A=3.54343136406170.}
Kad gauti tikslesnį atsakymą naudojame šį kodą: var
a:longint;
b,d,c:real;
begin
c:=0;
for a:=1 to 100
do
c:=c+0.01*sqrt(sqr(a*0.03)+sqr((100-a)*0.05));
writeln(c);
readln;
end.
Tokiu budu gauname gana tikslų atsakymą
A
=
3.27664616925457.
{\displaystyle A=3.27664616925457.}
Kad gauti labai tikslų atsakymą naudojame tokį kodą: var
a:longint;
c:real;
begin
for a:=1 to 1000000
do
c:=c+0.000001*sqrt(sqr(a*0.000003)+sqr((1000000-a)*0.000005));
writeln(c);
readln;
end.
Gauname atsakymą
A
=
3.28657850216630.
{\displaystyle A=3.28657850216630.}
Kad gauti labai labai tikslų atsakymą naudojame šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+0.000000001*sqrt(sqr(a*0.000000003)+sqr((1000000000-a)*0.000000005));
writeln(c);
readln;
end.
Gauname atsakymą
A
=
3.28657950114378
{\displaystyle A=3.28657950114378}
Jei naudojame šį kodą: var
a:longint;
c:real;
begin
for a:=0 to 1000000000 do
c:=c+0.000000001*sqrt(sqr(a*0.000000003)+sqr((1000000000-a)*0.000000005));
writeln(c);
readln;
end.
tai gauname atsakymą
A
=
3.28657950614378.
{\displaystyle A=3.28657950614378.}
Panaudojus šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+0.000000001*sqrt(sqr(a*0.000000003)+sqr((1000000001-a)*0.000000005));
writeln(c);
readln;
end.
gauname atsakymą
A
=
3.28657950471730.
{\displaystyle A=3.28657950471730.}
Integruojant gauname:
A
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
x
=
{\displaystyle A=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}dx=}
=
1
340
(
34
x
−
45
)
34
x
2
−
90
x
+
225
−
1125
arcsinh
(
3
5
−
34
x
75
)
68
34
|
0
5
=
{\displaystyle ={\frac {1}{340}}(34x-45){\sqrt {34x^{2}-90x+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34x}{75}})}{68{\sqrt {34}}}}|_{0}^{5}=}
=
34
⋅
5
−
45
340
34
⋅
5
2
−
90
⋅
5
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
5
75
)
68
34
−
(
34
⋅
0
−
45
340
34
⋅
0
2
−
90
⋅
0
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
0
75
)
68
34
)
=
{\displaystyle ={\frac {34\cdot 5-45}{340}}{\sqrt {34\cdot 5^{2}-90\cdot 5+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 5}{75}})}{68{\sqrt {34}}}}-\left({\frac {34\cdot 0-45}{340}}{\sqrt {34\cdot 0^{2}-90\cdot 0+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 0}{75}})}{68{\sqrt {34}}}}\right)=}
=
170
−
45
340
850
−
450
+
225
−
1125
arcsinh
(
3
5
−
170
75
)
68
34
−
(
−
45
340
225
−
1125
arcsinh
(
3
5
)
68
34
)
=
{\displaystyle ={\frac {170-45}{340}}{\sqrt {850-450+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {170}{75}})}{68{\sqrt {34}}}}-\left({\frac {-45}{340}}{\sqrt {225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}})}{68{\sqrt {34}}}}\right)=}
=
125
340
625
−
1125
arcsinh
(
3
⋅
75
−
170
⋅
5
5
⋅
75
)
68
34
−
(
−
9
68
⋅
15
−
1125
arcsinh
(
0.6
)
396.50472884948
)
=
{\displaystyle ={\frac {125}{340}}{\sqrt {625}}-{\frac {1125\operatorname {arcsinh} ({\frac {3\cdot 75-170\cdot 5}{5\cdot 75}})}{68{\sqrt {34}}}}-\left({\frac {-9}{68}}\cdot 15-{\frac {1125\operatorname {arcsinh} (0.6)}{396.50472884948}}\right)=}
=
25
68
⋅
25
−
1125
arcsinh
(
225
−
850
375
)
396.50472884948
−
(
−
135
68
−
1125
⋅
0.5688248987
396.50472884948
)
=
{\displaystyle ={\frac {25}{68}}\cdot 25-{\frac {1125\operatorname {arcsinh} ({\frac {225-850}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {1125\cdot 0.5688248987}{396.50472884948}}\right)=}
=
625
68
−
1125
arcsinh
(
−
625
375
)
396.50472884948
−
(
−
135
68
−
639.928011
396.50472884948
)
=
{\displaystyle ={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} ({\frac {-625}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {639.928011}{396.50472884948}}\right)=}
=
9.19117647
−
1125
arcsinh
(
−
5
3
)
396.50472884948
−
(
−
1.98529412
−
1.613922772
)
=
{\displaystyle =9.19117647-{\frac {1125\operatorname {arcsinh} (-{\frac {5}{3}})}{396.50472884948}}-(-1.98529412-1.613922772)=}
=
9.19117647
−
1125
⋅
(
−
1.283795663
)
396.50472884948
−
(
−
3.599216892
)
=
{\displaystyle =9.19117647-{\frac {1125\cdot (-1.283795663)}{396.50472884948}}-(-3.599216892)=}
=
9.19117647
+
3.642504151
+
3.599216892
=
16.43289751.
{\displaystyle =9.19117647+3.642504151+3.599216892=16.43289751.}
Pastebime, kad
2
A
/
10
=
2
⋅
16.43289751
/
10
=
3.286579503.
{\displaystyle 2A/10=2\cdot 16.43289751/10=3.286579503.}
A
=
625
68
−
1125
arcsinh
(
−
5
/
3
)
68
34
−
(
−
135
68
−
1125
arcsinh
(
3
/
5
)
68
34
)
=
{\displaystyle A={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} (-5/3)}{68{\sqrt {34}}}}-\left({\frac {-135}{68}}-{\frac {1125\operatorname {arcsinh} (3/5)}{68{\sqrt {34}}}}\right)=}
=
12.833680621236990323769396193322
−
(
−
3.5992168895913637545005254840611
)
=
16.432897510828354078269921677383.
{\displaystyle =12.833680621236990323769396193322-(-3.5992168895913637545005254840611)=16.432897510828354078269921677383.}
Randame, kad
A
/
5
=
16.432897510828354078269921677383
/
5
=
3.2865795021656708156539843354766.
{\displaystyle A/5=16.432897510828354078269921677383/5=3.2865795021656708156539843354766.}
Toliau bandome rasti atlikta darbą apeita tiese
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
A
=
−
∮
L
x
d
x
+
y
d
y
.
{\displaystyle A=-\oint _{L}xdx+ydy.}
d
y
=
−
3
5
d
x
.
{\displaystyle dy=-{\frac {3}{5}}\;dx.}
x integravimo ribos yra 0 iki 5, o y integravimo ribos yra nuo 0 iki 3. Gauname:
A
=
−
∮
L
x
d
x
+
y
d
y
=
−
(
∫
0
5
x
d
x
−
3
5
(
3
−
3
x
5
)
d
x
)
=
−
∫
0
5
(
x
−
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=-\oint _{L}xdx+ydy=-(\int _{0}^{5}x\;dx-{\frac {3}{5}}(3-{\frac {3x}{5}})dx)=-\int _{0}^{5}(x-{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
−
∫
0
5
(
x
−
9
5
+
9
x
25
)
d
x
=
−
∫
0
5
(
−
9
5
+
25
x
+
9
x
25
)
d
x
=
−
∫
0
5
(
34
x
25
−
9
5
)
d
x
=
{\displaystyle =-\int _{0}^{5}(x-{\frac {9}{5}}+{\frac {9x}{25}})dx=-\int _{0}^{5}(-{\frac {9}{5}}+{\frac {25x+9x}{25}})dx=-\int _{0}^{5}({\frac {34x}{25}}-{\frac {9}{5}})dx=}
=
−
(
34
x
2
25
⋅
2
−
9
5
x
)
|
0
5
=
−
[
(
17
⋅
5
2
25
−
9
5
⋅
5
)
−
(
17
⋅
0
2
25
−
9
5
⋅
0
)
]
=
−
(
17
−
9
)
=
−
8.
{\displaystyle =-({\frac {34x^{2}}{25\cdot 2}}-{\frac {9}{5}}x)|_{0}^{5}=-[({\frac {17\cdot 5^{2}}{25}}-{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}-{\frac {9}{5}}\cdot 0)]=-(17-9)=-8.}
Akivaizdu, kad atsakymas
A
=
−
8
{\displaystyle A=-8}
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
Pastebime, kad jeigu elipsė būtų apskritimas tai pirmame ketvirtyje atliktas darbas A=0. Taip yra todėl, kad kai x didėja, tada y reikšmės mažėja. Todėl skaičiuoti darbą, ko gero, galėtų būti teisingiau
A
=
∮
L
x
d
x
−
y
d
y
{\displaystyle A=\oint _{L}xdx-ydy}
A
=
∮
L
−
x
d
x
+
y
d
y
.
{\displaystyle A=\oint _{L}-xdx+ydy.}
A
=
a
2
+
b
2
2
=
5
2
+
3
2
2
=
17.
{\displaystyle A={\frac {a^{2}+b^{2}}{2}}={\frac {5^{2}+3^{2}}{2}}=17.}
Tą patį darba gauname ir apeinant tiese:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
−
−
3
5
(
3
−
3
x
5
)
d
x
=
∫
0
5
(
x
+
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx-{\frac {-3}{5}}(3-{\frac {3x}{5}})dx=\int _{0}^{5}(x+{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
∫
0
5
(
x
+
9
5
−
9
x
25
)
d
x
=
∫
0
5
(
9
5
+
25
x
−
9
x
25
)
d
x
=
∫
0
5
(
16
x
25
+
9
5
)
d
x
=
{\displaystyle =\int _{0}^{5}(x+{\frac {9}{5}}-{\frac {9x}{25}})dx=\int _{0}^{5}({\frac {9}{5}}+{\frac {25x-9x}{25}})dx=\int _{0}^{5}({\frac {16x}{25}}+{\frac {9}{5}})dx=}
=
(
16
x
2
25
⋅
2
+
9
5
x
)
|
0
5
=
(
8
⋅
5
2
25
+
9
5
⋅
5
)
−
(
17
⋅
0
2
25
+
9
5
⋅
0
)
=
8
+
9
=
17.
{\displaystyle =({\frac {16x^{2}}{25\cdot 2}}+{\frac {9}{5}}x)|_{0}^{5}=({\frac {8\cdot 5^{2}}{25}}+{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}+{\frac {9}{5}}\cdot 0)=8+9=17.}
Lygiai tą patį darbą gausime ir integruojant taip:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
+
∫
0
3
y
d
y
=
x
2
2
|
0
5
+
y
2
2
|
0
3
=
25
2
+
9
2
=
17.
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx+\int _{0}^{3}y\;dy={\frac {x^{2}}{2}}|_{0}^{5}+{\frac {y^{2}}{2}}|_{0}^{3}={\frac {25}{2}}+{\frac {9}{2}}=17.}
Tą patį gausime ir taip integruojant nuo 0 iki 5:
A
=
∫
F
d
F
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
(
x
2
+
(
3
−
3
x
5
)
2
)
=
{\displaystyle A=\int F\;\mathbf {d} F=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\;\mathbf {d} \left({\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\right)=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
0.5
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
2
x
+
2
(
3
−
3
x
/
5
)
(
−
0.6
)
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}0.5(x^{2}+(3-3x/5)^{2})^{-0.5}(2x+2(3-3x/5)(-0.6))\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
x
−
0.6
(
3
−
3
x
/
5
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x^{2}+(3-3x/5)^{2})^{-0.5}(x-0.6(3-3x/5)\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
−
1.8
+
0.36
x
)
x
2
+
(
3
−
3
x
/
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\frac {{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x-1.8+0.36x)}{\sqrt {x^{2}+(3-3x/5)^{2}}}}\;dx=}
=
∫
0
5
(
1.36
x
−
1.8
)
d
x
=
(
1.36
x
2
2
−
1.8
x
)
|
0
5
=
{\displaystyle =\int _{0}^{5}(1.36x-1.8)\;dx=(1.36{\frac {x^{2}}{2}}-1.8x)|_{0}^{5}=}
=
[
(
0.68
x
−
1.8
)
x
]
|
0
5
=
(
0.68
⋅
5
−
1.8
)
⋅
5
=
(
3.4
−
1.8
)
⋅
5
=
1.6
⋅
5
=
8.
{\displaystyle =[(0.68x-1.8)x]|_{0}^{5}=(0.68\cdot 5-1.8)\cdot 5=(3.4-1.8)\cdot 5=1.6\cdot 5=8.}
Jėgos darbas padarytas judant kreive plokštumoje apskaičiuojamas pagal formulę
A
=
∫
B
C
P
d
x
+
Q
d
y
.
{\displaystyle A=\int _{BC}Pdx+Qdy.}
A
=
∫
B
C
P
d
x
+
Q
d
y
+
R
d
z
.
{\displaystyle A=\int _{BC}Pdx+Qdy+Rdz.}
Apskaičiuosime darbą jėgos
F
(
x
,
y
)
{\displaystyle F(x,y)}
Pagal sąlyga,
|
F
(
x
,
y
)
|
=
x
2
+
y
2
;
{\displaystyle |F(x,y)|={\sqrt {x^{2}+y^{2}}};}
P
=
−
x
,
{\displaystyle P=-x,}
Q
=
−
y
{\displaystyle Q=-y}
−
{\displaystyle -}
A
=
−
∮
L
x
d
x
+
y
d
y
,
{\displaystyle A=-\oint _{L}xdx+ydy,}
L - elipsė
x
=
a
cos
t
,
y
=
b
sin
t
,
{\displaystyle x=a\cos t,\;y=b\sin t,}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
A
=
−
∫
0
2
π
a
cos
t
(
−
a
sin
t
)
d
t
+
b
sin
(
t
)
b
cos
(
t
)
d
t
=
−
∫
0
2
π
(
b
2
−
a
2
)
sin
(
t
)
cos
t
d
t
=
{\displaystyle A=-\int _{0}^{2\pi }a\cos t(-a\sin t)dt+b\sin(t)b\cos(t)dt=-\int _{0}^{2\pi }(b^{2}-a^{2})\sin(t)\cos t\;dt=}
=
a
2
−
b
2
2
∫
0
2
π
sin
(
2
t
)
d
t
=
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
2
π
=
0.
{\displaystyle ={a^{2}-b^{2} \over 2}\int _{0}^{2\pi }\sin(2t)dt={a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{2\pi }=0.}
Jei t keistusi nuo 0 iki
π
2
,
{\displaystyle {\pi \over 2},}
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
π
2
=
a
2
−
b
2
4
(
−
cos
(
2
⋅
π
2
)
)
−
a
2
−
b
2
4
(
−
cos
(
2
⋅
0
)
)
=
{\displaystyle {a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{\pi \over 2}={a^{2}-b^{2} \over 4}(-\cos(2\cdot {\pi \over 2}))-{a^{2}-b^{2} \over 4}(-\cos(2\cdot 0))=}
=
a
2
−
b
2
4
(
−
cos
π
+
cos
(
0
)
)
=
a
2
−
b
2
4
(
−
(
−
1
)
+
1
)
=
a
2
−
b
2
2
.
{\displaystyle ={a^{2}-b^{2} \over 4}(-\cos \pi +\cos(0))={a^{2}-b^{2} \over 4}(-(-1)+1)={a^{2}-b^{2} \over 2}.}
Tarkime, jei
a
=
5
,
b
=
3
,
{\displaystyle a=5,\;b=3,}
A
=
a
2
−
b
2
2
=
5
2
−
3
2
2
=
25
−
9
2
=
16
2
=
8.
{\displaystyle A={a^{2}-b^{2} \over 2}={5^{2}-3^{2} \over 2}={\frac {25-9}{2}}={\frac {16}{2}}=8.}
Kad patikrinti ar apskaičiuota teisingai reikia sudėti visas x reikšmes ant elipsės linijos pirmame ketvirtyje. Taip pat reikia sudėti visas y reikšmes ant elipsės linijos pirmame ketvirtyje. Galiausiai reikia sudėti sumas x ir y reikšmių, kad gauti darbą A pirmame ketvirtyje.
Arba tiesiog darbas yra lygus
A
=
1
100
∑
n
=
1
100
x
n
2
+
y
n
2
;
{\displaystyle A={\frac {1}{100}}\sum _{n=1}^{100}{\sqrt {x_{n}^{2}+y_{n}^{2}}};}
x
n
{\displaystyle x_{n}}
x reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki a ; analogiškai
y
n
{\displaystyle y_{n}}
y reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki b .
Kuris iš šių variantų yra darbas A , paaiškės pasumavus ir paskaičiavus. Sutapimas ar ne, bet
A
=
3
+
5
=
8
{\displaystyle A=3+5=8}
A
=
1
10
∑
n
=
1
10
(
n
(
3
/
10
)
)
2
+
(
(
10
−
n
)
(
5
/
10
)
)
2
=
{\displaystyle A={\frac {1}{10}}\sum _{n=1}^{10}{\sqrt {(n(3/10))^{2}+((10-n)(5/10))^{2}}}=}
A
=
(
0.3
2
+
5
2
+
0.6
2
+
4.5
2
+
0.9
2
+
4
2
+
1.2
2
+
3.5
2
+
1.5
2
+
3
2
+
{\displaystyle A=({\sqrt {0.3^{2}+5^{2}}}+{\sqrt {0.6^{2}+4.5^{2}}}+{\sqrt {0.9^{2}+4^{2}}}+{\sqrt {1.2^{2}+3.5^{2}}}+{\sqrt {1.5^{2}+3^{2}}}+}
+
1.8
2
+
2.5
2
+
2.1
2
+
2
2
+
2.4
2
+
1.5
2
+
2.7
2
+
1
2
+
3
2
+
0.5
2
)
/
10
=
{\displaystyle +{\sqrt {1.8^{2}+2.5^{2}}}+{\sqrt {2.1^{2}+2^{2}}}+{\sqrt {2.4^{2}+1.5^{2}}}+{\sqrt {2.7^{2}+1^{2}}}+{\sqrt {3^{2}+0.5^{2}}})/10=}
=
(
5.0089919
+
4.539823785
+
4.1
+
3.7
+
3.354101966
+
{\displaystyle =(5.0089919+4.539823785+4.1+3.7+3.354101966+}
+
3.08058436
+
2.9
+
2.83019434
+
2.879236
+
3.041381265
)
/
10
=
{\displaystyle +3.08058436+2.9+2.83019434+2.879236+3.041381265)/10=}
=
35.43431361
/
10
=
3.543431361.
{\displaystyle =35.43431361/10=3.543431361.}
Iš tikro, ko gero, mes apskaičiuojame tokiu budu ne darbą atlikta apeinant elipsės liniją pirmame ketvirtyje, o darbą atlikta apeinant tiesę pirmame ketvirtyje. Štai kodas programos "Free Pascal" (FreePascal IDE for Win32 for i386; Target CPU: i386; Version 1.0.12 2011/04/23; <Compiler Version 2.4.4>; <Debugger GDB 7.2>; Copyright <C> 1998-2009): var
a:longint;
c:real;
begin
for a:=1 to 10
do
c:=c+0.1*sqrt(sqr(a*0.3)+sqr((11-a)*0.5));
writeln(c);
readln;
end.
gauname
A
=
3.54343136406170.
{\displaystyle A=3.54343136406170.}
Panaudojus šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+0.000000001*sqrt(sqr(a*0.000000003)+sqr((1000000001-a)*0.000000005));
writeln(c);
readln;
end.
gauname atsakymą
A
=
3.28657950471730
{\displaystyle A=3.28657950471730}
Panaudojus šį (teisingesnį) kodą: var
a:longint;
c:real;
begin
for a:=0 to 1000000000 do
c:=c+sqrt(sqr(a*0.000000003)+sqr((1000000000-a)*0.000000005))/1000000001;
writeln(c);
readln;
end.
gauname atsakymą
A
=
3.28657950285701
{\displaystyle A=3.28657950285701}
Integruojant gauname:
A
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
x
=
{\displaystyle A=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}dx=}
=
1
340
(
34
x
−
45
)
34
x
2
−
90
x
+
225
−
1125
arcsinh
(
3
5
−
34
x
75
)
68
34
|
0
5
=
{\displaystyle ={\frac {1}{340}}(34x-45){\sqrt {34x^{2}-90x+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34x}{75}})}{68{\sqrt {34}}}}|_{0}^{5}=}
=
34
⋅
5
−
45
340
34
⋅
5
2
−
90
⋅
5
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
5
75
)
68
34
−
(
34
⋅
0
−
45
340
34
⋅
0
2
−
90
⋅
0
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
0
75
)
68
34
)
=
{\displaystyle ={\frac {34\cdot 5-45}{340}}{\sqrt {34\cdot 5^{2}-90\cdot 5+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 5}{75}})}{68{\sqrt {34}}}}-\left({\frac {34\cdot 0-45}{340}}{\sqrt {34\cdot 0^{2}-90\cdot 0+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 0}{75}})}{68{\sqrt {34}}}}\right)=}
=
170
−
45
340
850
−
450
+
225
−
1125
arcsinh
(
3
5
−
170
75
)
68
34
−
(
−
45
340
225
−
1125
arcsinh
(
3
5
)
68
34
)
=
{\displaystyle ={\frac {170-45}{340}}{\sqrt {850-450+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {170}{75}})}{68{\sqrt {34}}}}-\left({\frac {-45}{340}}{\sqrt {225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}})}{68{\sqrt {34}}}}\right)=}
=
125
340
625
−
1125
arcsinh
(
3
⋅
75
−
170
⋅
5
5
⋅
75
)
68
34
−
(
−
9
68
⋅
15
−
1125
arcsinh
(
0.6
)
396.50472884948
)
=
{\displaystyle ={\frac {125}{340}}{\sqrt {625}}-{\frac {1125\operatorname {arcsinh} ({\frac {3\cdot 75-170\cdot 5}{5\cdot 75}})}{68{\sqrt {34}}}}-\left({\frac {-9}{68}}\cdot 15-{\frac {1125\operatorname {arcsinh} (0.6)}{396.50472884948}}\right)=}
=
25
68
⋅
25
−
1125
arcsinh
(
225
−
850
375
)
396.50472884948
−
(
−
135
68
−
1125
⋅
0.5688248987
396.50472884948
)
=
{\displaystyle ={\frac {25}{68}}\cdot 25-{\frac {1125\operatorname {arcsinh} ({\frac {225-850}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {1125\cdot 0.5688248987}{396.50472884948}}\right)=}
=
625
68
−
1125
arcsinh
(
−
625
375
)
396.50472884948
−
(
−
135
68
−
639.928011
396.50472884948
)
=
{\displaystyle ={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} ({\frac {-625}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {639.928011}{396.50472884948}}\right)=}
=
9.19117647
−
1125
arcsinh
(
−
5
3
)
396.50472884948
−
(
−
1.98529412
−
1.613922772
)
=
{\displaystyle =9.19117647-{\frac {1125\operatorname {arcsinh} (-{\frac {5}{3}})}{396.50472884948}}-(-1.98529412-1.613922772)=}
=
9.19117647
−
1125
⋅
(
−
1.283795663
)
396.50472884948
−
(
−
3.599216892
)
=
{\displaystyle =9.19117647-{\frac {1125\cdot (-1.283795663)}{396.50472884948}}-(-3.599216892)=}
=
9.19117647
+
3.642504151
+
3.599216892
=
16.43289751.
{\displaystyle =9.19117647+3.642504151+3.599216892=16.43289751.}
(Update : Perskaičiavus gaunasi:
(9,1911764705882352941176470588235+3,6425041506487550296517491344981)-(-3,3088235294117647058823529411765-1,6139227719443049309711137193553)=
=12,833680621236990323769396193322-(-4,9227463013560696368534666605317)=17,756426922593059960622862853854.
Update 2 : Ne -3,3088, o -1,9852941176470588235294117647059.)Pastebime, kad
2
A
/
10
=
2
⋅
16.43289751
/
10
=
3.286579503.
{\displaystyle 2A/10=2\cdot 16.43289751/10=3.286579503.}
A
=
625
68
−
1125
arcsinh
(
−
5
/
3
)
68
34
−
(
−
135
68
−
1125
arcsinh
(
3
/
5
)
68
34
)
=
{\displaystyle A={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} (-5/3)}{68{\sqrt {34}}}}-\left({\frac {-135}{68}}-{\frac {1125\operatorname {arcsinh} (3/5)}{68{\sqrt {34}}}}\right)=}
=
12.833680621236990323769396193322
−
(
−
3.5992168895913637545005254840611
)
=
16.432897510828354078269921677383.
{\displaystyle =12.833680621236990323769396193322-(-3.5992168895913637545005254840611)=16.432897510828354078269921677383.}
Randame, kad
A
/
5
=
16.432897510828354078269921677383
/
5
=
3.2865795021656708156539843354766.
{\displaystyle A/5=16.432897510828354078269921677383/5=3.2865795021656708156539843354766.}
Toliau bandome rasti atlikta darbą apeita tiese
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
A
=
−
∮
L
x
d
x
+
y
d
y
.
{\displaystyle A=-\oint _{L}xdx+ydy.}
d
y
=
−
3
5
d
x
.
{\displaystyle dy=-{\frac {3}{5}}\;dx.}
x integravimo ribos yra 0 iki 5, o y integravimo ribos yra nuo 0 iki 3. Gauname:
A
=
−
∮
L
x
d
x
+
y
d
y
=
−
(
∫
0
5
x
d
x
−
3
5
(
3
−
3
x
5
)
d
x
)
=
−
∫
0
5
(
x
−
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=-\oint _{L}xdx+ydy=-(\int _{0}^{5}x\;dx-{\frac {3}{5}}(3-{\frac {3x}{5}})dx)=-\int _{0}^{5}(x-{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
−
∫
0
5
(
x
−
9
5
+
9
x
25
)
d
x
=
−
∫
0
5
(
−
9
5
+
25
x
+
9
x
25
)
d
x
=
−
∫
0
5
(
34
x
25
−
9
5
)
d
x
=
{\displaystyle =-\int _{0}^{5}(x-{\frac {9}{5}}+{\frac {9x}{25}})dx=-\int _{0}^{5}(-{\frac {9}{5}}+{\frac {25x+9x}{25}})dx=-\int _{0}^{5}({\frac {34x}{25}}-{\frac {9}{5}})dx=}
=
−
(
34
x
2
25
⋅
2
−
9
5
x
)
|
0
5
=
−
[
(
17
⋅
5
2
25
−
9
5
⋅
5
)
−
(
17
⋅
0
2
25
−
9
5
⋅
0
)
]
=
−
(
17
−
9
)
=
−
8.
{\displaystyle =-({\frac {34x^{2}}{25\cdot 2}}-{\frac {9}{5}}x)|_{0}^{5}=-[({\frac {17\cdot 5^{2}}{25}}-{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}-{\frac {9}{5}}\cdot 0)]=-(17-9)=-8.}
Akivaizdu, kad atsakymas
A
=
−
8
{\displaystyle A=-8}
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
Pastebime, kad jeigu elipsė būtų apskritimas tai pirmame ketvirtyje atliktas darbas A=0. Taip yra todėl, kad kai x didėja, tada y reikšmės mažėja. Todėl skaičiuoti darbą, ko gero, galėtų būti teisingiau
A
=
∮
L
x
d
x
−
y
d
y
{\displaystyle A=\oint _{L}xdx-ydy}
A
=
∮
L
−
x
d
x
+
y
d
y
.
{\displaystyle A=\oint _{L}-xdx+ydy.}
A
=
a
2
+
b
2
2
=
5
2
+
3
2
2
=
17.
{\displaystyle A={\frac {a^{2}+b^{2}}{2}}={\frac {5^{2}+3^{2}}{2}}=17.}
Tą patį darba gauname ir apeinant tiese:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
−
−
3
5
(
3
−
3
x
5
)
d
x
=
∫
0
5
(
x
+
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx-{\frac {-3}{5}}(3-{\frac {3x}{5}})dx=\int _{0}^{5}(x+{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
∫
0
5
(
x
+
9
5
−
9
x
25
)
d
x
=
∫
0
5
(
9
5
+
25
x
−
9
x
25
)
d
x
=
∫
0
5
(
16
x
25
+
9
5
)
d
x
=
{\displaystyle =\int _{0}^{5}(x+{\frac {9}{5}}-{\frac {9x}{25}})dx=\int _{0}^{5}({\frac {9}{5}}+{\frac {25x-9x}{25}})dx=\int _{0}^{5}({\frac {16x}{25}}+{\frac {9}{5}})dx=}
=
(
16
x
2
25
⋅
2
+
9
5
x
)
|
0
5
=
(
8
⋅
5
2
25
+
9
5
⋅
5
)
−
(
17
⋅
0
2
25
+
9
5
⋅
0
)
=
8
+
9
=
17.
{\displaystyle =({\frac {16x^{2}}{25\cdot 2}}+{\frac {9}{5}}x)|_{0}^{5}=({\frac {8\cdot 5^{2}}{25}}+{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}+{\frac {9}{5}}\cdot 0)=8+9=17.}
Lygiai tą patį darbą gausime ir integruojant taip:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
+
∫
0
3
y
d
y
=
x
2
2
|
0
5
+
y
2
2
|
0
3
=
25
2
+
9
2
=
17.
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx+\int _{0}^{3}y\;dy={\frac {x^{2}}{2}}|_{0}^{5}+{\frac {y^{2}}{2}}|_{0}^{3}={\frac {25}{2}}+{\frac {9}{2}}=17.}
Tą patį gausime ir taip integruojant nuo 0 iki 5:
A
=
∫
F
d
F
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
(
x
2
+
(
3
−
3
x
5
)
2
)
=
{\displaystyle A=\int F\;\mathbf {d} F=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\;\mathbf {d} \left({\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\right)=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
0.5
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
2
x
+
2
(
3
−
3
x
/
5
)
(
−
0.6
)
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}0.5(x^{2}+(3-3x/5)^{2})^{-0.5}(2x+2(3-3x/5)(-0.6))\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
x
−
0.6
(
3
−
3
x
/
5
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x^{2}+(3-3x/5)^{2})^{-0.5}(x-0.6(3-3x/5)\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
−
1.8
+
0.36
x
)
x
2
+
(
3
−
3
x
/
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\frac {{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x-1.8+0.36x)}{\sqrt {x^{2}+(3-3x/5)^{2}}}}\;dx=}
=
∫
0
5
(
1.36
x
−
1.8
)
d
x
=
(
1.36
x
2
2
−
1.8
x
)
|
0
5
=
{\displaystyle =\int _{0}^{5}(1.36x-1.8)\;dx=(1.36{\frac {x^{2}}{2}}-1.8x)|_{0}^{5}=}
=
[
(
0.68
x
−
1.8
)
x
]
|
0
5
=
(
0.68
⋅
5
−
1.8
)
⋅
5
=
(
3.4
−
1.8
)
⋅
5
=
1.6
⋅
5
=
8.
{\displaystyle =[(0.68x-1.8)x]|_{0}^{5}=(0.68\cdot 5-1.8)\cdot 5=(3.4-1.8)\cdot 5=1.6\cdot 5=8.}
Nustatyti tiesės
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Pasinaudosime masės skaičiavimo formule
m
=
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
6
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {6x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
6
x
5
+
34
x
2
25
34
25
d
x
.
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {6x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx.}
Toliau integruodami gauname:
m
=
(
34
25
34
x
2
−
30
x
+
225
(
x
2
2
−
15
68
)
+
7425
arcsinh
34
x
−
15
15
33
68
34
)
|
0
5
=
{\displaystyle m=\left({\frac {\sqrt {34}}{25}}{\sqrt {34x^{2}-30x+225}}\left({\frac {x^{2}}{2}}-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {34x-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)|_{0}^{5}=}
=
(
34
25
34
⋅
25
−
30
⋅
5
+
225
(
25
2
−
15
68
)
+
7425
arcsinh
34
⋅
5
−
15
15
33
68
34
)
−
(
34
25
34
⋅
0
−
30
⋅
0
+
225
(
0
2
2
−
15
68
)
+
7425
arcsinh
34
⋅
0
−
15
15
33
68
34
)
=
{\displaystyle =\left({\frac {\sqrt {34}}{25}}{\sqrt {34\cdot 25-30\cdot 5+225}}\left({\frac {25}{2}}-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {34\cdot 5-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)-\left({\frac {\sqrt {34}}{25}}{\sqrt {34\cdot 0-30\cdot 0+225}}\left({\frac {0^{2}}{2}}-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {34\cdot 0-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)=}
=
(
34
25
850
−
150
+
225
25
⋅
68
−
30
2
⋅
68
+
7425
arcsinh
170
−
15
15
33
68
34
)
−
(
34
25
225
(
−
15
68
)
+
7425
arcsinh
−
15
15
33
68
34
)
=
{\displaystyle =\left({\frac {\sqrt {34}}{25}}{\sqrt {850-150+225}}{\frac {25\cdot 68-30}{2\cdot 68}}+{\frac {7425\operatorname {arcsinh} {\frac {170-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)-\left({\frac {\sqrt {34}}{25}}{\sqrt {225}}\left(-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)=}
=
(
5
34
25
37
1670
136
+
7425
arcsinh
31
3
33
68
34
)
−
(
3
34
5
(
−
15
68
)
+
7425
arcsinh
(
−
1
33
)
68
34
)
=
{\displaystyle =\left({\frac {5{\sqrt {34}}}{25}}{\sqrt {37}}{\frac {1670}{136}}+{\frac {7425\operatorname {arcsinh} {\frac {31}{3{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)-\left({\frac {3{\sqrt {34}}}{5}}\left(-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} (-{\frac {1}{\sqrt {33}}})}{68{\sqrt {34}}}}\right)=}
=
(
1258
5
1670
136
+
7425
arcsinh
(
1.798802445
)
396.5047288
)
−
(
−
9
34
68
+
7425
arcsinh
(
−
0.174077656
)
396.5047288
)
=
{\displaystyle =\left({\frac {\sqrt {1258}}{5}}{\frac {1670}{136}}+{\frac {7425\operatorname {arcsinh} (1.798802445)}{396.5047288}}\right)-\left(-{\frac {9{\sqrt {34}}}{68}}+{\frac {7425\operatorname {arcsinh} (-0.174077656)}{396.5047288}}\right)=}
=
(
334
1258
136
+
7425
⋅
1.349859008
396.5047288
)
−
(
−
9
34
68
+
7425
⋅
(
−
0.173210253
)
396.5047288
)
=
{\displaystyle =\left({\frac {334{\sqrt {1258}}}{136}}+{\frac {7425\cdot 1.349859008}{396.5047288}}\right)-\left(-{\frac {9{\sqrt {34}}}{68}}+{\frac {7425\cdot (-0.173210253)}{396.5047288}}\right)=}
=
(
167
1258
68
+
10022.7031344
396.5047288
)
−
(
−
9
34
68
+
−
1286.086127
396.5047288
)
=
{\displaystyle =\left({\frac {167{\sqrt {1258}}}{68}}+{\frac {10022.7031344}{396.5047288}}\right)-\left(-{\frac {9{\sqrt {34}}}{68}}+{\frac {-1286.086127}{396.5047288}}\right)=}
=
(
87.1059615
+
25.27763834
)
−
(
−
0.771743633
−
3.243558105
)
=
{\displaystyle =(87.1059615+25.27763834)-(-0.771743633-3.243558105)=}
=
112.3835998
−
(
−
4.015301739
)
=
116.3989015.
{\displaystyle =112.3835998-(-4.015301739)=116.3989015.}
Perskaičiavus gaunas m=41.4470084394957645512263519847-(-4.015301738572567547333986879663)=45.462310178068332098560338864363.
Wolframo integratoriui varžteliai atsisuko, nes integruojant tą patį taip ir taip nuo 0 iki 5 gauname skirtingus atsakymus atitinkamai:
16.43289751
{\displaystyle 16.43289751}
45.46231018
34
25
=
38.98360936.
{\displaystyle {\frac {45.46231018}{\sqrt {\frac {34}{25}}}}=38.98360936.}
Beje, integruojant taip nuo 0 iki 5 gauname:
(13,865120473473883114631633205975+5,0555276680167165753752595311952)-(-0,66176470588235294117647058823529-0,64871162108625555582601097623286)=
=18,92064814149059969000689273717-(-1,3104763269686084970024815644681)=20,231124468459208187009374301638.
Nustatyti tiesės
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Pasinaudosime masės skaičiavimo formule
m
=
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
6
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {6x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
6
x
5
+
34
x
2
25
34
25
d
x
.
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {6x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx.}
Toliau integruodami gauname:
m
=
34
25
(
34
x
2
−
30
x
+
225
(
x
2
2
−
15
68
)
+
7425
arcsinh
34
x
−
15
15
33
68
34
)
|
0
5
=
{\displaystyle m={\frac {\sqrt {34}}{25}}\left({\sqrt {34x^{2}-30x+225}}\left({\frac {x^{2}}{2}}-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {34x-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)|_{0}^{5}=}
=
34
25
(
34
⋅
25
−
30
⋅
5
+
225
(
25
2
−
15
68
)
+
7425
arcsinh
34
⋅
5
−
15
15
33
68
34
)
−
34
25
(
34
⋅
0
−
30
⋅
0
+
225
(
0
2
2
−
15
68
)
+
7425
arcsinh
34
⋅
0
−
15
15
33
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 25-30\cdot 5+225}}\left({\frac {25}{2}}-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {34\cdot 5-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 0-30\cdot 0+225}}\left({\frac {0^{2}}{2}}-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {34\cdot 0-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
850
−
150
+
225
25
⋅
68
−
30
2
⋅
68
+
7425
arcsinh
170
−
15
15
33
68
34
)
−
34
25
(
225
(
−
15
68
)
+
7425
arcsinh
−
15
15
33
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {850-150+225}}{\frac {25\cdot 68-30}{2\cdot 68}}+{\frac {7425\operatorname {arcsinh} {\frac {170-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {225}}\left(-{\frac {15}{68}}\right)+{\frac {7425\operatorname {arcsinh} {\frac {-15}{15{\sqrt {33}}}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
373.46372888228
+
25.27763834
)
−
34
25
(
−
3.3088235294
−
3.243558105
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}(373.46372888228+25.27763834)-{\frac {\sqrt {34}}{25}}(-3.3088235294-3.243558105)=}
=
34
25
⋅
398.74136722228
−
34
25
(
−
6.5523816344
)
=
405.293749
34
25
=
94.52993411.
{\displaystyle ={\frac {\sqrt {34}}{25}}\cdot 398.74136722228-{\frac {\sqrt {34}}{25}}(-6.5523816344)={\frac {405.293749{\sqrt {34}}}{25}}=94.52993411.}
Wolframo integratoriui varžteliai atsisuko, nes integruojant tą patį taip ir taip nuo 0 iki 5 gauname skirtingus atsakymus atitinkamai:
16.43289751
{\displaystyle 16.43289751}
94.52993411
34
25
=
81.05874977.
{\displaystyle {\frac {94.52993411}{\sqrt {\frac {34}{25}}}}=81.05874977.}
Beje, integruojant taip nuo 0 iki 5 gauname:
(13,865120473473883114631633205975+5,0555276680167165753752595311952)-(-0,66176470588235294117647058823529-0,64871162108625555582601097623286)=
=18,92064814149059969000689273717-(-1,3104763269686084970024815644681)=20,231124468459208187009374301638. Na, bent jau apytiksliai gauname kažką panašaus į
3.286579502
{\displaystyle 3.286579502}
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
20.23112447
34
25
34
=
23.59334271
34
=
20.23112447
5
=
4.046224894.
{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {20.23112447{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {23.59334271}{\sqrt {34}}}={\frac {20.23112447}{5}}=4.046224894.}
Nors šita formulė ir skirta kreivės svorio centro koordinačių nustatymui (
x
C
=
4.046224894
{\displaystyle x_{C}=4.046224894}
Tačiau, ko gero,
x
C
=
m
=
16.43289751
34
25
34
=
19.16388697
34
=
16.43289751
5
=
3.286579502.
{\displaystyle x_{C}=m={\frac {16.43289751{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.16388697}{\sqrt {34}}}={\frac {16.43289751}{5}}=3.286579502.}
(Taip 1.16619*sqrt(9-18*x/5 +(34*x^2)/25) integruojant gauname 19,163865128930380119022361244004; čia 1,16619=(34/25)^(1/2).) Nustatyti tiesės
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Pasinaudosime masės skaičiavimo formule
m
=
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
18
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {18x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
18
x
5
+
34
x
2
25
34
25
d
x
.
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {18x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx.}
Toliau integruodami gauname:
m
=
34
25
(
34
x
2
−
90
x
+
225
(
x
2
2
−
45
68
)
+
5625
arcsinh
34
x
−
45
75
68
34
)
|
0
5
=
{\displaystyle m={\frac {\sqrt {34}}{25}}\left({\sqrt {34x^{2}-90x+225}}\left({\frac {x^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34x-45}{75}}}{68{\sqrt {34}}}}\right)|_{0}^{5}=}
=
34
25
(
34
⋅
25
−
90
⋅
5
+
225
(
25
2
−
45
68
)
+
5625
arcsinh
34
⋅
5
−
45
75
68
34
)
−
34
25
(
34
⋅
0
−
90
⋅
0
+
225
(
0
2
2
−
45
68
)
+
5625
arcsinh
34
⋅
0
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 25-90\cdot 5+225}}\left({\frac {25}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 5-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 0-90\cdot 0+225}}\left({\frac {0^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 0-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
850
−
450
+
225
25
⋅
68
−
90
2
⋅
68
+
5625
arcsinh
170
−
45
75
68
34
)
−
34
25
(
225
(
−
45
68
)
+
5625
arcsinh
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {850-450+225}}{\frac {25\cdot 68-90}{2\cdot 68}}+{\frac {5625\operatorname {arcsinh} {\frac {170-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {225}}\left(-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
295.95588235294
+
18.21252075
)
−
34
25
(
−
9.926470588
−
8.0696138597
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}(295.95588235294+18.21252075)-{\frac {\sqrt {34}}{25}}(-9.926470588-8.0696138597)=}
=
34
25
⋅
314.1684031
−
34
25
(
−
17.99608445
)
=
332.1644875
34
25
=
77.47340592.
{\displaystyle ={\frac {\sqrt {34}}{25}}\cdot 314.1684031-{\frac {\sqrt {34}}{25}}(-17.99608445)={\frac {332.1644875{\sqrt {34}}}{25}}=77.47340592.}
Wolframo integratoriui varžteliai atsisuko, nes integruojant tą patį taip ir taip nuo 0 iki 5 gauname skirtingus atsakymus atitinkamai:
16.43289751
{\displaystyle 16.43289751}
77.47340592
34
25
=
66.4328975.
{\displaystyle {\frac {77.47340592}{\sqrt {\frac {34}{25}}}}=66.4328975.}
Beje, integruojant taip nuo 0 iki 5 gauname:
[34^0.5]*[(45,955882352941176470588235294118+18,21252075324377514825874567249)-(-9,9264705882352941176470588235294-8,0696138597215246548555685967763)]/25=
=[34^0.5]*[64,168403106184951618846980966608-(-17,996084447956818772502627420306)]/25=
=[34^0.5]*82,164487554141770391349608386914/25=19,163886975712642545064105297031. Na, bent jau apytiksliai gauname kažką panašaus į
3.286579502
{\displaystyle 3.286579502}
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
19.1638869757
34
25
34
=
19.1638869757
5
=
3.832777395.
{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.1638869757{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.1638869757}{5}}=3.832777395.}
Nors šita formulė ir skirta kreivės svorio centro koordinačių nustatymui (
x
C
=
3.832777395
{\displaystyle x_{C}=3.832777395}
Tačiau, ko gero,
x
C
=
m
=
16.43289751
34
25
34
=
19.16388697
34
=
16.43289751
5
=
3.286579502.
{\displaystyle x_{C}=m={\frac {16.43289751{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.16388697}{\sqrt {34}}}={\frac {16.43289751}{5}}=3.286579502.}
Nustatyti tiesės
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Pasinaudosime masės skaičiavimo formule
m
=
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
18
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {18x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
18
x
5
+
34
x
2
25
34
25
d
x
.
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {18x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx.}
Toliau integruodami gauname:
m
=
34
25
(
34
x
2
−
90
x
+
225
(
x
2
2
−
45
68
)
+
5625
arcsinh
34
x
−
45
75
68
34
)
|
0
5
=
{\displaystyle m={\frac {\sqrt {34}}{25}}\left({\sqrt {34x^{2}-90x+225}}\left({\frac {x^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34x-45}{75}}}{68{\sqrt {34}}}}\right)|_{0}^{5}=}
=
34
25
(
34
⋅
25
−
90
⋅
5
+
225
(
25
2
−
45
68
)
+
5625
arcsinh
34
⋅
5
−
45
75
68
34
)
−
34
25
(
34
⋅
0
−
90
⋅
0
+
225
(
0
2
2
−
45
68
)
+
5625
arcsinh
34
⋅
0
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 25-90\cdot 5+225}}\left({\frac {25}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 5-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 0-90\cdot 0+225}}\left({\frac {0^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 0-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
850
−
450
+
225
25
⋅
68
−
90
2
⋅
68
+
5625
arcsinh
170
−
45
75
68
34
)
−
34
25
(
225
(
−
45
68
)
+
5625
arcsinh
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {850-450+225}}{\frac {25\cdot 68-90}{2\cdot 68}}+{\frac {5625\operatorname {arcsinh} {\frac {170-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {225}}\left(-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
295.95588235294
+
18.21252075
)
−
34
25
(
−
9.926470588
−
8.0696138597
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}(295.95588235294+18.21252075)-{\frac {\sqrt {34}}{25}}(-9.926470588-8.0696138597)=}
=
34
25
⋅
314.1684031
−
34
25
(
−
17.99608445
)
=
332.1644875
34
25
=
77.47340592.
{\displaystyle ={\frac {\sqrt {34}}{25}}\cdot 314.1684031-{\frac {\sqrt {34}}{25}}(-17.99608445)={\frac {332.1644875{\sqrt {34}}}{25}}=77.47340592.}
Wolframo integratoriui varžteliai atsisuko, nes integruojant tą patį taip ir taip nuo 0 iki 5 gauname skirtingus atsakymus atitinkamai:
16.43289751
{\displaystyle 16.43289751}
77.47340592
34
25
=
66.4328975.
{\displaystyle {\frac {77.47340592}{\sqrt {\frac {34}{25}}}}=66.4328975.}
Beje, integruojant taip nuo 0 iki 5 gauname:
[34^0.5]*[(45,955882352941176470588235294118+18,21252075324377514825874567249)-(-9,9264705882352941176470588235294-8,0696138597215246548555685967763)]/25=
=[34^0.5]*[64,168403106184951618846980966608-(-17,996084447956818772502627420306)]/25=
=[34^0.5]*82,164487554141770391349608386914/25=19,163886975712642545064105297031.
Wolframo integratorius duoda keistus sutapimus:
m
=
∫
0
5
γ
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
19.1638869757
34
=
3.286579502.
{\displaystyle m={\frac {\int _{0}^{5}\gamma dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.1638869757}{\sqrt {34}}}=3.286579502.}
Nors formulė (
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
{\displaystyle {\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}}
x
C
{\displaystyle x_{C}}
Tačiau, ko gero,
x
C
=
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
16.43289751
34
25
34
=
19.16388697
34
=
16.43289751
5
=
3.286579502.
{\displaystyle x_{C}=m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {16.43289751{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.16388697}{\sqrt {34}}}={\frac {16.43289751}{5}}=3.286579502.}
Nustatyti tiesės
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Pasinaudosime masės skaičiavimo formule
m
=
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
18
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {18x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
18
x
5
+
34
x
2
25
34
25
d
x
.
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {18x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx.}
Toliau integruodami gauname:
m
=
34
25
(
34
x
2
−
90
x
+
225
(
x
2
2
−
45
68
)
+
5625
arcsinh
34
x
−
45
75
68
34
)
|
0
5
=
{\displaystyle m={\frac {\sqrt {34}}{25}}\left({\sqrt {34x^{2}-90x+225}}\left({\frac {x^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34x-45}{75}}}{68{\sqrt {34}}}}\right)|_{0}^{5}=}
=
34
25
(
34
⋅
25
−
90
⋅
5
+
225
(
25
2
−
45
68
)
+
5625
arcsinh
34
⋅
5
−
45
75
68
34
)
−
34
25
(
34
⋅
0
−
90
⋅
0
+
225
(
0
2
2
−
45
68
)
+
5625
arcsinh
34
⋅
0
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 25-90\cdot 5+225}}\left({\frac {25}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 5-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 0-90\cdot 0+225}}\left({\frac {0^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 0-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
850
−
450
+
225
25
⋅
68
−
90
2
⋅
68
+
5625
arcsinh
170
−
45
75
68
34
)
−
34
25
(
225
(
−
45
68
)
+
5625
arcsinh
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {850-450+225}}{\frac {25\cdot 68-90}{2\cdot 68}}+{\frac {5625\operatorname {arcsinh} {\frac {170-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {225}}\left(-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
295.95588235294
+
18.21252075
)
−
34
25
(
−
9.926470588
−
8.0696138597
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}(295.95588235294+18.21252075)-{\frac {\sqrt {34}}{25}}(-9.926470588-8.0696138597)=}
=
34
25
⋅
314.1684031
−
34
25
(
−
17.99608445
)
=
332.1644875
34
25
=
77.47340592.
{\displaystyle ={\frac {\sqrt {34}}{25}}\cdot 314.1684031-{\frac {\sqrt {34}}{25}}(-17.99608445)={\frac {332.1644875{\sqrt {34}}}{25}}=77.47340592.}
Wolframo integratoriui varžteliai atsisuko, nes integruojant tą patį taip ir taip nuo 0 iki 5 gauname skirtingus atsakymus atitinkamai:
16.43289751
{\displaystyle 16.43289751}
77.47340592
34
25
=
66.4328975.
{\displaystyle {\frac {77.47340592}{\sqrt {\frac {34}{25}}}}=66.4328975.}
Beje, integruojant taip nuo 0 iki 5 gauname:
[34^0.5]*[(45,955882352941176470588235294118+18,21252075324377514825874567249)-(-9,9264705882352941176470588235294-8,0696138597215246548555685967763)]/25=
=[34^0.5]*[64,168403106184951618846980966608-(-17,996084447956818772502627420306)]/25=
=[34^0.5]*82,164487554141770391349608386914/25=19,163886975712642545064105297031.
Wolframo integratorius duoda keistus sutapimus:
m
=
∫
0
5
γ
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
19.1638869757
34
=
3.286579502.
{\displaystyle m={\frac {\int _{0}^{5}\gamma dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.1638869757}{\sqrt {34}}}=3.286579502.}
Nors formulė (
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
{\displaystyle {\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}}
x
C
{\displaystyle x_{C}}
Tačiau, ko gero,
x
C
=
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
16.43289751
34
25
34
=
19.16388697
34
=
16.43289751
5
=
3.286579502.
{\displaystyle x_{C}=m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {16.43289751{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.16388697}{\sqrt {34}}}={\frac {16.43289751}{5}}=3.286579502.}
Kad išsiaiškinti teisybę pasinaudosime integralų lentele :
∫
a
x
2
+
b
x
+
c
d
x
=
b
+
2
a
x
4
a
a
x
2
+
b
x
+
c
+
4
a
c
−
b
2
8
a
3
/
2
ln
|
2
a
x
+
b
+
2
a
(
a
x
2
+
b
x
+
c
)
|
.
{\displaystyle \int {\sqrt {ax^{2}+bx+c}}\;dx={\frac {b+2ax}{4a}}{\sqrt {ax^{2}+bx+c}}+{\frac {4ac-b^{2}}{8a^{3/2}}}\ln |2ax+b+2{\sqrt {a(ax^{2}+bx+c)}}|.}
Tada integruojame:
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle \int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
18
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {18x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
18
x
5
+
34
x
2
25
34
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {18x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx=}
=
34
5
∫
0
5
34
25
x
2
−
18
5
x
+
9
d
x
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\int _{0}^{5}{\sqrt {{\frac {34}{25}}x^{2}-{\frac {18}{5}}x+9}}\;dx=}
=
34
5
∫
0
5
1.36
x
2
−
3.6
x
+
9
d
x
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\int _{0}^{5}{\sqrt {1.36x^{2}-3.6x+9}}\;dx=}
=
34
5
(
−
3.6
+
2
⋅
1.36
x
4
⋅
1.36
1.36
x
2
−
3.6
x
+
9
+
4
⋅
1.36
⋅
9
−
(
−
3.6
)
2
8
⋅
1.36
3
/
2
ln
|
2
⋅
1.36
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2\cdot 1.36x}{4\cdot 1.36}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {4\cdot 1.36\cdot 9-(-3.6)^{2}}{8\cdot 1.36^{3/2}}}\ln |2\cdot 1.36x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
x
5.44
1.36
x
2
−
3.6
x
+
9
+
48.96
−
12.96
8
⋅
2.515456
1
/
2
ln
|
2.72
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72x}{5.44}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {48.96-12.96}{8\cdot 2.515456^{1/2}}}\ln |2.72x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
x
5.44
1.36
x
2
−
3.6
x
+
9
+
36
12.68815132
ln
|
2.72
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72x}{5.44}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {36}{12.68815132}}\ln |2.72x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
⋅
5
5.44
1.36
⋅
25
−
3.6
⋅
5
+
9
+
2.8372927689
ln
|
2.72
⋅
5
−
3.6
+
2
1.36
(
1.36
⋅
25
−
3.6
⋅
5
+
9
)
|
)
−
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72\cdot 5}{5.44}}{\sqrt {1.36\cdot 25-3.6\cdot 5+9}}+2.8372927689\ln |2.72\cdot 5-3.6+2{\sqrt {1.36(1.36\cdot 25-3.6\cdot 5+9)}}|\right)-}
−
34
5
(
−
3.6
+
2.72
⋅
0
5.44
1.36
⋅
0
2
−
3.6
⋅
0
+
9
+
2.8372927689
ln
|
2.72
⋅
0
−
3.6
+
2
1.36
(
1.36
⋅
0
2
−
3.6
⋅
0
+
9
)
|
)
=
{\displaystyle -{\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72\cdot 0}{5.44}}{\sqrt {1.36\cdot 0^{2}-3.6\cdot 0+9}}+2.8372927689\ln |2.72\cdot 0-3.6+2{\sqrt {1.36(1.36\cdot 0^{2}-3.6\cdot 0+9)}}|\right)=}
=
34
5
(
10
5.44
34
−
18
+
9
+
2.8372927689
ln
|
13.6
−
3.6
+
2
1.36
(
34
−
18
+
9
)
|
)
−
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {10}{5.44}}{\sqrt {34-18+9}}+2.8372927689\ln |13.6-3.6+2{\sqrt {1.36(34-18+9)}}|\right)-}
−
34
5
(
−
3.6
5.44
9
+
2.8372927689
ln
|
−
3.6
+
2
1.36
⋅
9
|
)
=
{\displaystyle -{\frac {\sqrt {34}}{5}}\left({\frac {-3.6}{5.44}}{\sqrt {9}}+2.8372927689\ln |-3.6+2{\sqrt {1.36\cdot 9}}|\right)=}
=
34
5
(
10
5.44
25
+
2.8372927689
ln
|
10
+
2
1.36
⋅
25
|
)
−
34
5
(
−
3.6
5.44
⋅
3
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {10}{5.44}}{\sqrt {25}}+2.8372927689\ln |10+2{\sqrt {1.36\cdot 25}}|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-3.6}{5.44}}\cdot 3+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
50
5.44
+
2.8372927689
ln
|
10
+
2
34
|
)
−
34
5
(
−
10.8
5.44
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {50}{5.44}}+2.8372927689\ln |10+2{\sqrt {34}}|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-10.8}{5.44}}+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
9.191176471
+
2.8372927689
ln
|
21.66190379
|
)
−
34
5
(
−
10.8
5.44
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left(9.191176471+2.8372927689\ln |21.66190379|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-10.8}{5.44}}+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
9.191176471
+
8.726250336
)
−
34
5
(
−
1.985294118
+
3.469823414
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}(9.191176471+8.726250336)-{\frac {\sqrt {34}}{5}}(-1.985294118+3.469823414)=}
=
34
5
⋅
17.91742681
−
34
5
⋅
1.484529296
=
16.43289751
34
5
=
19.16388698.
{\displaystyle ={\frac {\sqrt {34}}{5}}\cdot 17.91742681-{\frac {\sqrt {34}}{5}}\cdot 1.484529296={\frac {16.43289751{\sqrt {34}}}{5}}=19.16388698.}
Vadinasi, randame tiesės masę kaip prašyta sąlygoje:
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
19.16388698
∫
0
5
1
+
(
−
3
5
)
2
d
x
=
19.16388698
∫
0
5
1
+
9
25
d
x
=
19.16388698
∫
0
5
34
25
d
x
=
{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {1+{\frac {9}{25}}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {\frac {34}{25}}}dx}}=}
=
19.16388698
34
5
x
|
0
5
=
19.16388698
34
5
⋅
5
−
34
5
⋅
0
=
19.16388698
34
=
3.286579502.
{\displaystyle ={\frac {19.16388698}{{\frac {\sqrt {34}}{5}}x|_{0}^{5}}}={\frac {19.16388698}{{\frac {\sqrt {34}}{5}}\cdot 5-{\frac {\sqrt {34}}{5}}\cdot 0}}={\frac {19.16388698}{\sqrt {34}}}=3.286579502.}
Jėgos darbas padarytas judant kreive plokštumoje apskaičiuojamas pagal formulę
A
=
∫
B
C
P
d
x
+
Q
d
y
.
{\displaystyle A=\int _{BC}Pdx+Qdy.}
A
=
∫
B
C
P
d
x
+
Q
d
y
+
R
d
z
.
{\displaystyle A=\int _{BC}Pdx+Qdy+Rdz.}
Apskaičiuosime darbą jėgos
F
(
x
,
y
)
{\displaystyle F(x,y)}
Pagal sąlyga,
|
F
(
x
,
y
)
|
=
x
2
+
y
2
;
{\displaystyle |F(x,y)|={\sqrt {x^{2}+y^{2}}};}
P
=
−
x
,
{\displaystyle P=-x,}
Q
=
−
y
{\displaystyle Q=-y}
−
{\displaystyle -}
A
=
−
∮
L
x
d
x
+
y
d
y
,
{\displaystyle A=-\oint _{L}xdx+ydy,}
L - elipsė
x
=
a
cos
t
,
y
=
b
sin
t
,
{\displaystyle x=a\cos t,\;y=b\sin t,}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
A
=
−
∫
0
2
π
a
cos
t
(
−
a
sin
t
)
d
t
+
b
sin
(
t
)
b
cos
(
t
)
d
t
=
−
∫
0
2
π
(
b
2
−
a
2
)
sin
(
t
)
cos
t
d
t
=
{\displaystyle A=-\int _{0}^{2\pi }a\cos t(-a\sin t)dt+b\sin(t)b\cos(t)dt=-\int _{0}^{2\pi }(b^{2}-a^{2})\sin(t)\cos t\;dt=}
=
a
2
−
b
2
2
∫
0
2
π
sin
(
2
t
)
d
t
=
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
2
π
=
0.
{\displaystyle ={a^{2}-b^{2} \over 2}\int _{0}^{2\pi }\sin(2t)dt={a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{2\pi }=0.}
Jei t keistusi nuo 0 iki
π
2
,
{\displaystyle {\pi \over 2},}
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
π
2
=
a
2
−
b
2
4
(
−
cos
(
2
⋅
π
2
)
)
−
a
2
−
b
2
4
(
−
cos
(
2
⋅
0
)
)
=
{\displaystyle {a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{\pi \over 2}={a^{2}-b^{2} \over 4}(-\cos(2\cdot {\pi \over 2}))-{a^{2}-b^{2} \over 4}(-\cos(2\cdot 0))=}
=
a
2
−
b
2
4
(
−
cos
π
+
cos
(
0
)
)
=
a
2
−
b
2
4
(
−
(
−
1
)
+
1
)
=
a
2
−
b
2
2
.
{\displaystyle ={a^{2}-b^{2} \over 4}(-\cos \pi +\cos(0))={a^{2}-b^{2} \over 4}(-(-1)+1)={a^{2}-b^{2} \over 2}.}
Tarkime, jei
a
=
5
,
b
=
3
,
{\displaystyle a=5,\;b=3,}
A
=
a
2
−
b
2
2
=
5
2
−
3
2
2
=
25
−
9
2
=
16
2
=
8.
{\displaystyle A={a^{2}-b^{2} \over 2}={5^{2}-3^{2} \over 2}={\frac {25-9}{2}}={\frac {16}{2}}=8.}
Kad patikrinti ar apskaičiuota teisingai reikia sudėti visas x reikšmes ant elipsės linijos pirmame ketvirtyje. Taip pat reikia sudėti visas y reikšmes ant elipsės linijos pirmame ketvirtyje. Galiausiai reikia sudėti sumas x ir y reikšmių, kad gauti darbą A pirmame ketvirtyje.
Arba tiesiog darbas yra lygus
A
=
1
100
∑
n
=
1
100
x
n
2
+
y
n
2
;
{\displaystyle A={\frac {1}{100}}\sum _{n=1}^{100}{\sqrt {x_{n}^{2}+y_{n}^{2}}};}
x
n
{\displaystyle x_{n}}
x reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki a ; analogiškai
y
n
{\displaystyle y_{n}}
y reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki b .
Kuris iš šių variantų yra darbas A , paaiškės pasumavus ir paskaičiavus. Sutapimas ar ne, bet
A
=
3
+
5
=
8
{\displaystyle A=3+5=8}
A
=
1
10
∑
n
=
1
10
(
n
(
3
/
10
)
)
2
+
(
(
10
−
n
)
(
5
/
10
)
)
2
=
{\displaystyle A={\frac {1}{10}}\sum _{n=1}^{10}{\sqrt {(n(3/10))^{2}+((10-n)(5/10))^{2}}}=}
A
=
(
0.3
2
+
5
2
+
0.6
2
+
4.5
2
+
0.9
2
+
4
2
+
1.2
2
+
3.5
2
+
1.5
2
+
3
2
+
{\displaystyle A=({\sqrt {0.3^{2}+5^{2}}}+{\sqrt {0.6^{2}+4.5^{2}}}+{\sqrt {0.9^{2}+4^{2}}}+{\sqrt {1.2^{2}+3.5^{2}}}+{\sqrt {1.5^{2}+3^{2}}}+}
+
1.8
2
+
2.5
2
+
2.1
2
+
2
2
+
2.4
2
+
1.5
2
+
2.7
2
+
1
2
+
3
2
+
0.5
2
)
/
10
=
{\displaystyle +{\sqrt {1.8^{2}+2.5^{2}}}+{\sqrt {2.1^{2}+2^{2}}}+{\sqrt {2.4^{2}+1.5^{2}}}+{\sqrt {2.7^{2}+1^{2}}}+{\sqrt {3^{2}+0.5^{2}}})/10=}
=
(
5.0089919
+
4.539823785
+
4.1
+
3.7
+
3.354101966
+
{\displaystyle =(5.0089919+4.539823785+4.1+3.7+3.354101966+}
+
3.08058436
+
2.9
+
2.83019434
+
2.879236
+
3.041381265
)
/
10
=
{\displaystyle +3.08058436+2.9+2.83019434+2.879236+3.041381265)/10=}
=
35.43431361
/
10
=
3.543431361.
{\displaystyle =35.43431361/10=3.543431361.}
Iš tikro, ko gero, mes apskaičiuojame tokiu budu ne darbą atlikta apeinant elipsės liniją pirmame ketvirtyje, o darbą atlikta apeinant tiesę pirmame ketvirtyje. Štai kodas programos "Free Pascal" (FreePascal IDE for Win32 for i386; Target CPU: i386; Version 1.0.12 2011/04/23; <Compiler Version 2.4.4>; <Debugger GDB 7.2>; Copyright <C> 1998-2009): var
a:longint;
c:real;
begin
for a:=1 to 10
do
c:=c+0.1*sqrt(sqr(a*0.3)+sqr((11-a)*0.5));
writeln(c);
readln;
end.
gauname
A
=
3.54343136406170.
{\displaystyle A=3.54343136406170.}
Panaudojus šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+0.000000001*sqrt(sqr(a*0.000000003)+sqr((1000000001-a)*0.000000005));
writeln(c);
readln;
end.
gauname atsakymą
A
=
3.28657950471730
{\displaystyle A=3.28657950471730}
Panaudojus šį (teisingesnį) kodą: var
a:longint;
c:real;
begin
for a:=0 to 1000000000 do
c:=c+sqrt(sqr(a*0.000000003)+sqr((1000000000-a)*0.000000005))/1000000001;
writeln(c);
readln;
end.
gauname atsakymą
A
=
3.28657950285701
{\displaystyle A=3.28657950285701}
Integruojant gauname:
A
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
x
=
{\displaystyle A=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}dx=}
=
1
340
(
34
x
−
45
)
34
x
2
−
90
x
+
225
−
1125
arcsinh
(
3
5
−
34
x
75
)
68
34
|
0
5
=
{\displaystyle ={\frac {1}{340}}(34x-45){\sqrt {34x^{2}-90x+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34x}{75}})}{68{\sqrt {34}}}}|_{0}^{5}=}
=
34
⋅
5
−
45
340
34
⋅
5
2
−
90
⋅
5
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
5
75
)
68
34
−
(
34
⋅
0
−
45
340
34
⋅
0
2
−
90
⋅
0
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
0
75
)
68
34
)
=
{\displaystyle ={\frac {34\cdot 5-45}{340}}{\sqrt {34\cdot 5^{2}-90\cdot 5+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 5}{75}})}{68{\sqrt {34}}}}-\left({\frac {34\cdot 0-45}{340}}{\sqrt {34\cdot 0^{2}-90\cdot 0+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 0}{75}})}{68{\sqrt {34}}}}\right)=}
=
170
−
45
340
850
−
450
+
225
−
1125
arcsinh
(
3
5
−
170
75
)
68
34
−
(
−
45
340
225
−
1125
arcsinh
(
3
5
)
68
34
)
=
{\displaystyle ={\frac {170-45}{340}}{\sqrt {850-450+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {170}{75}})}{68{\sqrt {34}}}}-\left({\frac {-45}{340}}{\sqrt {225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}})}{68{\sqrt {34}}}}\right)=}
=
125
340
625
−
1125
arcsinh
(
3
⋅
75
−
170
⋅
5
5
⋅
75
)
68
34
−
(
−
9
68
⋅
15
−
1125
arcsinh
(
0.6
)
396.50472884948
)
=
{\displaystyle ={\frac {125}{340}}{\sqrt {625}}-{\frac {1125\operatorname {arcsinh} ({\frac {3\cdot 75-170\cdot 5}{5\cdot 75}})}{68{\sqrt {34}}}}-\left({\frac {-9}{68}}\cdot 15-{\frac {1125\operatorname {arcsinh} (0.6)}{396.50472884948}}\right)=}
=
25
68
⋅
25
−
1125
arcsinh
(
225
−
850
375
)
396.50472884948
−
(
−
135
68
−
1125
⋅
0.5688248987
396.50472884948
)
=
{\displaystyle ={\frac {25}{68}}\cdot 25-{\frac {1125\operatorname {arcsinh} ({\frac {225-850}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {1125\cdot 0.5688248987}{396.50472884948}}\right)=}
=
625
68
−
1125
arcsinh
(
−
625
375
)
396.50472884948
−
(
−
135
68
−
639.928011
396.50472884948
)
=
{\displaystyle ={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} ({\frac {-625}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {639.928011}{396.50472884948}}\right)=}
=
9.19117647
−
1125
arcsinh
(
−
5
3
)
396.50472884948
−
(
−
1.98529412
−
1.613922772
)
=
{\displaystyle =9.19117647-{\frac {1125\operatorname {arcsinh} (-{\frac {5}{3}})}{396.50472884948}}-(-1.98529412-1.613922772)=}
=
9.19117647
−
1125
⋅
(
−
1.283795663
)
396.50472884948
−
(
−
3.599216892
)
=
{\displaystyle =9.19117647-{\frac {1125\cdot (-1.283795663)}{396.50472884948}}-(-3.599216892)=}
=
9.19117647
+
3.642504151
+
3.599216892
=
16.43289751.
{\displaystyle =9.19117647+3.642504151+3.599216892=16.43289751.}
Pastebime, kad
2
A
/
10
=
2
⋅
16.43289751
/
10
=
3.286579503.
{\displaystyle 2A/10=2\cdot 16.43289751/10=3.286579503.}
A
=
625
68
−
1125
arcsinh
(
−
5
/
3
)
68
34
−
(
−
135
68
−
1125
arcsinh
(
3
/
5
)
68
34
)
=
{\displaystyle A={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} (-5/3)}{68{\sqrt {34}}}}-\left({\frac {-135}{68}}-{\frac {1125\operatorname {arcsinh} (3/5)}{68{\sqrt {34}}}}\right)=}
=
12.833680621236990323769396193322
−
(
−
3.5992168895913637545005254840611
)
=
16.432897510828354078269921677383.
{\displaystyle =12.833680621236990323769396193322-(-3.5992168895913637545005254840611)=16.432897510828354078269921677383.}
Randame, kad
A
/
5
=
16.432897510828354078269921677383
/
5
=
3.2865795021656708156539843354766.
{\displaystyle A/5=16.432897510828354078269921677383/5=3.2865795021656708156539843354766.}
Toliau bandome rasti atlikta darbą apeita tiese
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
A
=
−
∮
L
x
d
x
+
y
d
y
.
{\displaystyle A=-\oint _{L}xdx+ydy.}
d
y
=
−
3
5
d
x
.
{\displaystyle dy=-{\frac {3}{5}}\;dx.}
x integravimo ribos yra 0 iki 5, o y integravimo ribos yra nuo 0 iki 3. Gauname:
A
=
−
∮
L
x
d
x
+
y
d
y
=
−
(
∫
0
5
x
d
x
−
3
5
(
3
−
3
x
5
)
d
x
)
=
−
∫
0
5
(
x
−
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=-\oint _{L}xdx+ydy=-(\int _{0}^{5}x\;dx-{\frac {3}{5}}(3-{\frac {3x}{5}})dx)=-\int _{0}^{5}(x-{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
−
∫
0
5
(
x
−
9
5
+
9
x
25
)
d
x
=
−
∫
0
5
(
−
9
5
+
25
x
+
9
x
25
)
d
x
=
−
∫
0
5
(
34
x
25
−
9
5
)
d
x
=
{\displaystyle =-\int _{0}^{5}(x-{\frac {9}{5}}+{\frac {9x}{25}})dx=-\int _{0}^{5}(-{\frac {9}{5}}+{\frac {25x+9x}{25}})dx=-\int _{0}^{5}({\frac {34x}{25}}-{\frac {9}{5}})dx=}
=
−
(
34
x
2
25
⋅
2
−
9
5
x
)
|
0
5
=
−
[
(
17
⋅
5
2
25
−
9
5
⋅
5
)
−
(
17
⋅
0
2
25
−
9
5
⋅
0
)
]
=
−
(
17
−
9
)
=
−
8.
{\displaystyle =-({\frac {34x^{2}}{25\cdot 2}}-{\frac {9}{5}}x)|_{0}^{5}=-[({\frac {17\cdot 5^{2}}{25}}-{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}-{\frac {9}{5}}\cdot 0)]=-(17-9)=-8.}
Akivaizdu, kad atsakymas
A
=
−
8
{\displaystyle A=-8}
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
Pastebime, kad jeigu elipsė būtų apskritimas tai pirmame ketvirtyje atliktas darbas A=0. Taip yra todėl, kad kai x didėja, tada y reikšmės mažėja. Todėl skaičiuoti darbą, ko gero, galėtų būti teisingiau
A
=
∮
L
x
d
x
−
y
d
y
{\displaystyle A=\oint _{L}xdx-ydy}
A
=
∮
L
−
x
d
x
+
y
d
y
.
{\displaystyle A=\oint _{L}-xdx+ydy.}
A
=
a
2
+
b
2
2
=
5
2
+
3
2
2
=
17.
{\displaystyle A={\frac {a^{2}+b^{2}}{2}}={\frac {5^{2}+3^{2}}{2}}=17.}
Tą patį darba gauname ir apeinant tiese:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
−
−
3
5
(
3
−
3
x
5
)
d
x
=
∫
0
5
(
x
+
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx-{\frac {-3}{5}}(3-{\frac {3x}{5}})dx=\int _{0}^{5}(x+{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
∫
0
5
(
x
+
9
5
−
9
x
25
)
d
x
=
∫
0
5
(
9
5
+
25
x
−
9
x
25
)
d
x
=
∫
0
5
(
16
x
25
+
9
5
)
d
x
=
{\displaystyle =\int _{0}^{5}(x+{\frac {9}{5}}-{\frac {9x}{25}})dx=\int _{0}^{5}({\frac {9}{5}}+{\frac {25x-9x}{25}})dx=\int _{0}^{5}({\frac {16x}{25}}+{\frac {9}{5}})dx=}
=
(
16
x
2
25
⋅
2
+
9
5
x
)
|
0
5
=
(
8
⋅
5
2
25
+
9
5
⋅
5
)
−
(
17
⋅
0
2
25
+
9
5
⋅
0
)
=
8
+
9
=
17.
{\displaystyle =({\frac {16x^{2}}{25\cdot 2}}+{\frac {9}{5}}x)|_{0}^{5}=({\frac {8\cdot 5^{2}}{25}}+{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}+{\frac {9}{5}}\cdot 0)=8+9=17.}
Lygiai tą patį darbą gausime ir integruojant taip:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
+
∫
0
3
y
d
y
=
x
2
2
|
0
5
+
y
2
2
|
0
3
=
25
2
+
9
2
=
17.
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx+\int _{0}^{3}y\;dy={\frac {x^{2}}{2}}|_{0}^{5}+{\frac {y^{2}}{2}}|_{0}^{3}={\frac {25}{2}}+{\frac {9}{2}}=17.}
Tą patį gausime ir taip integruojant nuo 0 iki 5:
A
=
∫
F
d
F
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
(
x
2
+
(
3
−
3
x
5
)
2
)
=
{\displaystyle A=\int F\;\mathbf {d} F=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\;\mathbf {d} \left({\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\right)=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
0.5
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
2
x
+
2
(
3
−
3
x
/
5
)
(
−
0.6
)
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}0.5(x^{2}+(3-3x/5)^{2})^{-0.5}(2x+2(3-3x/5)(-0.6))\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
x
−
0.6
(
3
−
3
x
/
5
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x^{2}+(3-3x/5)^{2})^{-0.5}(x-0.6(3-3x/5)\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
−
1.8
+
0.36
x
)
x
2
+
(
3
−
3
x
/
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\frac {{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x-1.8+0.36x)}{\sqrt {x^{2}+(3-3x/5)^{2}}}}\;dx=}
=
∫
0
5
(
1.36
x
−
1.8
)
d
x
=
(
1.36
x
2
2
−
1.8
x
)
|
0
5
=
{\displaystyle =\int _{0}^{5}(1.36x-1.8)\;dx=(1.36{\frac {x^{2}}{2}}-1.8x)|_{0}^{5}=}
=
[
(
0.68
x
−
1.8
)
x
]
|
0
5
=
(
0.68
⋅
5
−
1.8
)
⋅
5
=
(
3.4
−
1.8
)
⋅
5
=
1.6
⋅
5
=
8.
{\displaystyle =[(0.68x-1.8)x]|_{0}^{5}=(0.68\cdot 5-1.8)\cdot 5=(3.4-1.8)\cdot 5=1.6\cdot 5=8.}
Nustatyti tiesės
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Pasinaudosime masės skaičiavimo formule
m
=
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
18
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {18x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
18
x
5
+
34
x
2
25
34
25
d
x
.
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {18x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx.}
Toliau integruodami gauname:
m
=
34
25
(
34
x
2
−
90
x
+
225
(
x
2
2
−
45
68
)
+
5625
arcsinh
34
x
−
45
75
68
34
)
|
0
5
=
{\displaystyle m={\frac {\sqrt {34}}{25}}\left({\sqrt {34x^{2}-90x+225}}\left({\frac {x^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34x-45}{75}}}{68{\sqrt {34}}}}\right)|_{0}^{5}=}
=
34
25
(
34
⋅
25
−
90
⋅
5
+
225
(
25
2
−
45
68
)
+
5625
arcsinh
34
⋅
5
−
45
75
68
34
)
−
34
25
(
34
⋅
0
−
90
⋅
0
+
225
(
0
2
2
−
45
68
)
+
5625
arcsinh
34
⋅
0
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 25-90\cdot 5+225}}\left({\frac {25}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 5-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {34\cdot 0-90\cdot 0+225}}\left({\frac {0^{2}}{2}}-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {34\cdot 0-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
850
−
450
+
225
25
⋅
68
−
90
2
⋅
68
+
5625
arcsinh
170
−
45
75
68
34
)
−
34
25
(
225
(
−
45
68
)
+
5625
arcsinh
−
45
75
68
34
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}\left({\sqrt {850-450+225}}{\frac {25\cdot 68-90}{2\cdot 68}}+{\frac {5625\operatorname {arcsinh} {\frac {170-45}{75}}}{68{\sqrt {34}}}}\right)-{\frac {\sqrt {34}}{25}}\left({\sqrt {225}}\left(-{\frac {45}{68}}\right)+{\frac {5625\operatorname {arcsinh} {\frac {-45}{75}}}{68{\sqrt {34}}}}\right)=}
=
34
25
(
295.95588235294
+
18.21252075
)
−
34
25
(
−
9.926470588
−
8.0696138597
)
=
{\displaystyle ={\frac {\sqrt {34}}{25}}(295.95588235294+18.21252075)-{\frac {\sqrt {34}}{25}}(-9.926470588-8.0696138597)=}
=
34
25
⋅
314.1684031
−
34
25
(
−
17.99608445
)
=
332.1644875
34
25
=
77.47340592.
{\displaystyle ={\frac {\sqrt {34}}{25}}\cdot 314.1684031-{\frac {\sqrt {34}}{25}}(-17.99608445)={\frac {332.1644875{\sqrt {34}}}{25}}=77.47340592.}
Wolframo integratoriui varžteliai atsisuko, nes integruojant tą patį taip ir taip nuo 0 iki 5 gauname skirtingus atsakymus atitinkamai:
16.43289751
{\displaystyle 16.43289751}
77.47340592
34
25
=
66.4328975.
{\displaystyle {\frac {77.47340592}{\sqrt {\frac {34}{25}}}}=66.4328975.}
Beje, integruojant taip nuo 0 iki 5 gauname:
[34^0.5]*[(45,955882352941176470588235294118+18,21252075324377514825874567249)-(-9,9264705882352941176470588235294-8,0696138597215246548555685967763)]/25=
=[34^0.5]*[64,168403106184951618846980966608-(-17,996084447956818772502627420306)]/25=
=[34^0.5]*82,164487554141770391349608386914/25=19,163886975712642545064105297031.
Wolframo integratorius duoda keistus sutapimus:
m
=
∫
0
5
γ
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
19.1638869757
34
=
3.286579502.
{\displaystyle m={\frac {\int _{0}^{5}\gamma dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.1638869757}{\sqrt {34}}}=3.286579502.}
Nors formulė (
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
{\displaystyle {\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}}
x
C
{\displaystyle x_{C}}
Tačiau, ko gero,
x
C
=
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
16.43289751
34
25
34
=
19.16388697
34
=
16.43289751
5
=
3.286579502.
{\displaystyle x_{C}=m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {16.43289751{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.16388697}{\sqrt {34}}}={\frac {16.43289751}{5}}=3.286579502.}
Kad išsiaiškinti teisybę pasinaudosime integralų lentele :
∫
a
x
2
+
b
x
+
c
d
x
=
b
+
2
a
x
4
a
a
x
2
+
b
x
+
c
+
4
a
c
−
b
2
8
a
3
/
2
ln
|
2
a
x
+
b
+
2
a
(
a
x
2
+
b
x
+
c
)
|
.
{\displaystyle \int {\sqrt {ax^{2}+bx+c}}\;dx={\frac {b+2ax}{4a}}{\sqrt {ax^{2}+bx+c}}+{\frac {4ac-b^{2}}{8a^{3/2}}}\ln |2ax+b+2{\sqrt {a(ax^{2}+bx+c)}}|.}
Tada integruojame:
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle \int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
18
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {18x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
18
x
5
+
34
x
2
25
34
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {18x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx=}
=
34
5
∫
0
5
34
25
x
2
−
18
5
x
+
9
d
x
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\int _{0}^{5}{\sqrt {{\frac {34}{25}}x^{2}-{\frac {18}{5}}x+9}}\;dx=}
=
34
5
∫
0
5
1.36
x
2
−
3.6
x
+
9
d
x
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\int _{0}^{5}{\sqrt {1.36x^{2}-3.6x+9}}\;dx=}
=
34
5
(
−
3.6
+
2
⋅
1.36
x
4
⋅
1.36
1.36
x
2
−
3.6
x
+
9
+
4
⋅
1.36
⋅
9
−
(
−
3.6
)
2
8
⋅
1.36
3
/
2
ln
|
2
⋅
1.36
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2\cdot 1.36x}{4\cdot 1.36}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {4\cdot 1.36\cdot 9-(-3.6)^{2}}{8\cdot 1.36^{3/2}}}\ln |2\cdot 1.36x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
x
5.44
1.36
x
2
−
3.6
x
+
9
+
48.96
−
12.96
8
⋅
2.515456
1
/
2
ln
|
2.72
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72x}{5.44}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {48.96-12.96}{8\cdot 2.515456^{1/2}}}\ln |2.72x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
x
5.44
1.36
x
2
−
3.6
x
+
9
+
36
12.68815132
ln
|
2.72
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72x}{5.44}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {36}{12.68815132}}\ln |2.72x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
⋅
5
5.44
1.36
⋅
25
−
3.6
⋅
5
+
9
+
2.8372927689
ln
|
2.72
⋅
5
−
3.6
+
2
1.36
(
1.36
⋅
25
−
3.6
⋅
5
+
9
)
|
)
−
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72\cdot 5}{5.44}}{\sqrt {1.36\cdot 25-3.6\cdot 5+9}}+2.8372927689\ln |2.72\cdot 5-3.6+2{\sqrt {1.36(1.36\cdot 25-3.6\cdot 5+9)}}|\right)-}
−
34
5
(
−
3.6
+
2.72
⋅
0
5.44
1.36
⋅
0
2
−
3.6
⋅
0
+
9
+
2.8372927689
ln
|
2.72
⋅
0
−
3.6
+
2
1.36
(
1.36
⋅
0
2
−
3.6
⋅
0
+
9
)
|
)
=
{\displaystyle -{\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72\cdot 0}{5.44}}{\sqrt {1.36\cdot 0^{2}-3.6\cdot 0+9}}+2.8372927689\ln |2.72\cdot 0-3.6+2{\sqrt {1.36(1.36\cdot 0^{2}-3.6\cdot 0+9)}}|\right)=}
=
34
5
(
10
5.44
34
−
18
+
9
+
2.8372927689
ln
|
13.6
−
3.6
+
2
1.36
(
34
−
18
+
9
)
|
)
−
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {10}{5.44}}{\sqrt {34-18+9}}+2.8372927689\ln |13.6-3.6+2{\sqrt {1.36(34-18+9)}}|\right)-}
−
34
5
(
−
3.6
5.44
9
+
2.8372927689
ln
|
−
3.6
+
2
1.36
⋅
9
|
)
=
{\displaystyle -{\frac {\sqrt {34}}{5}}\left({\frac {-3.6}{5.44}}{\sqrt {9}}+2.8372927689\ln |-3.6+2{\sqrt {1.36\cdot 9}}|\right)=}
=
34
5
(
10
5.44
25
+
2.8372927689
ln
|
10
+
2
1.36
⋅
25
|
)
−
34
5
(
−
3.6
5.44
⋅
3
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {10}{5.44}}{\sqrt {25}}+2.8372927689\ln |10+2{\sqrt {1.36\cdot 25}}|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-3.6}{5.44}}\cdot 3+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
50
5.44
+
2.8372927689
ln
|
10
+
2
34
|
)
−
34
5
(
−
10.8
5.44
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {50}{5.44}}+2.8372927689\ln |10+2{\sqrt {34}}|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-10.8}{5.44}}+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
9.191176471
+
2.8372927689
ln
|
21.66190379
|
)
−
34
5
(
−
10.8
5.44
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left(9.191176471+2.8372927689\ln |21.66190379|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-10.8}{5.44}}+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
9.191176471
+
8.726250336
)
−
34
5
(
−
1.985294118
+
3.469823414
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}(9.191176471+8.726250336)-{\frac {\sqrt {34}}{5}}(-1.985294118+3.469823414)=}
=
34
5
⋅
17.91742681
−
34
5
⋅
1.484529296
=
16.43289751
34
5
=
19.16388698.
{\displaystyle ={\frac {\sqrt {34}}{5}}\cdot 17.91742681-{\frac {\sqrt {34}}{5}}\cdot 1.484529296={\frac {16.43289751{\sqrt {34}}}{5}}=19.16388698.}
Vadinasi, randame tiesės masę kaip prašyta sąlygoje:
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
19.16388698
∫
0
5
1
+
(
−
3
5
)
2
d
x
=
19.16388698
∫
0
5
1
+
9
25
d
x
=
19.16388698
∫
0
5
34
25
d
x
=
{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {1+{\frac {9}{25}}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {\frac {34}{25}}}dx}}=}
=
19.16388698
34
5
x
|
0
5
=
19.16388698
34
5
⋅
5
−
34
5
⋅
0
=
19.16388698
34
=
3.286579502.
{\displaystyle ={\frac {19.16388698}{{\frac {\sqrt {34}}{5}}x|_{0}^{5}}}={\frac {19.16388698}{{\frac {\sqrt {34}}{5}}\cdot 5-{\frac {\sqrt {34}}{5}}\cdot 0}}={\frac {19.16388698}{\sqrt {34}}}=3.286579502.}
Nustatyti parabolės
y
=
x
2
{\displaystyle y=x^{2}}
x kinta nuo 0 iki 10. Tiesės tankis
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Iš pradžių apskaičiuosime be integralų apytiksliai:
m
=
1
10
∑
1
10
(
10
n
/
10
)
2
+
(
(
10
n
/
10
)
2
)
2
=
1
10
∑
1
10
n
2
+
n
4
=
{\displaystyle m={\frac {1}{10}}\sum _{1}^{10}{\sqrt {(10n/10)^{2}+((10n/10)^{2})^{2}}}={\frac {1}{10}}\sum _{1}^{10}{\sqrt {n^{2}+n^{4}}}=}
=
1
10
(
2
+
20
+
90
+
272
+
650
+
1332
+
2450
+
4160
+
6642
+
10100
)
=
{\displaystyle ={\frac {1}{10}}({\sqrt {2}}+{\sqrt {20}}+{\sqrt {90}}+{\sqrt {272}}+{\sqrt {650}}+{\sqrt {1332}}+{\sqrt {2450}}+{\sqrt {4160}}+{\sqrt {6642}}+{\sqrt {10100}})=}
=389,85003687399407318939051595541/10=38,985003687399407318939051595541.
Integruojame (pasinaudodami integralų lentelės (31) formule ):
p
=
∫
0
10
(
x
)
2
+
(
x
2
)
2
d
x
=
∫
0
10
x
2
+
x
4
d
x
=
∫
0
10
x
1
+
x
2
d
x
=
{\displaystyle p=\int _{0}^{10}{\sqrt {(x)^{2}+(x^{2})^{2}}}dx=\int _{0}^{10}{\sqrt {x^{2}+x^{4}}}dx=\int _{0}^{10}x{\sqrt {1+x^{2}}}dx=}
=
1
3
(
x
2
+
1
)
3
/
2
|
0
10
=
1
3
(
10
2
+
1
)
3
/
2
−
1
3
(
0
2
+
1
)
3
/
2
=
{\displaystyle ={\frac {1}{3}}(x^{2}+1)^{3/2}|_{0}^{10}={\frac {1}{3}}(10^{2}+1)^{3/2}-{\frac {1}{3}}(0^{2}+1)^{3/2}=}
=
1
3
(
101
)
3
/
2
−
1
3
=
1
3
1030301
−
1
3
=
{\displaystyle ={\frac {1}{3}}(101)^{3/2}-{\frac {1}{3}}={\frac {1}{3}}{\sqrt {1030301}}-{\frac {1}{3}}=}
=338,01247924440330576404858539624.
Pastebime, kad masė galėtų būti
m
=
p
/
10
=
33.801247924440330576404858539624.
{\displaystyle m=p/10=33.801247924440330576404858539624.}
var
a:longint;
c:real;
begin
for a:=1 to 100 do
c:=c+sqrt(sqr(10*a/100)+sqr(sqr(10*a/100)))/100;
writeln(c);
readln;
end.
ir gauname atsakymą 34.3053251011728.
Panaudojus šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 10000 do
c:=c+sqrt(sqr(10*a/10000)+sqr(sqr(10*a/10000)))/10000;
writeln(c);
readln;
end.
gauname atsakymą 33.8062730205864.
Dar tikslesnį atsakymą
33.8012479746790
{\displaystyle 33.8012479746790}
var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(10.0*a/1000000000)+sqr(sqr(10.0*a/1000000000)))/1000000000;
writeln(c);
readln;
end.
po 38 sekundžių su 2,6 GHz procesoriumi.
Optimizuotas kodas: var
a:longint;
b,c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(0.00000001*a)+sqr(sqr(0.00000001*a)));
b:=0.000000001*c;
writeln(b);
readln;
end.
duoda atsakymą 33,8012479746835 po 18 sekundžių su 2,6 GHz procesoriumi.
Toliau, kad surasti kreivės masės skaičiavimo formulę (kai kiekviename kreivės taške tankis priklauso nuo tam tikros funkcijos). Todėl skaičiuojame:
m
=
∫
0
10
γ
1
+
[
y
′
]
2
d
x
∫
0
10
1
+
[
y
′
]
2
d
x
=
∫
0
10
x
2
+
y
2
1
+
[
(
x
2
)
′
]
2
d
x
∫
0
10
1
+
[
(
x
2
)
′
]
2
d
x
=
∫
0
10
x
2
+
(
x
2
)
2
1
+
[
2
x
]
2
d
x
∫
0
10
1
+
[
2
x
]
2
d
x
=
{\displaystyle m={\frac {\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[y']^{2}}}dx}}={\frac {\int _{0}^{10}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[(x^{2})']^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[(x^{2})']^{2}}}dx}}={\frac {\int _{0}^{10}{\sqrt {x^{2}+(x^{2})^{2}}}{\sqrt {1+[2x]^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[2x]^{2}}}dx}}=}
=
∫
0
10
x
2
+
x
4
1
+
4
x
2
d
x
∫
0
10
1
+
4
x
2
d
x
=
∫
0
10
x
1
+
x
2
1
+
4
x
2
d
x
∫
0
10
1
+
4
x
2
d
x
=
∫
0
10
x
4
x
4
+
5
x
2
+
1
d
x
∫
0
10
2
1
2
+
x
2
d
x
.
{\displaystyle ={\frac {\int _{0}^{10}{\sqrt {x^{2}+x^{4}}}{\sqrt {1+4x^{2}}}dx}{\int _{0}^{10}{\sqrt {1+4x^{2}}}dx}}={\frac {\int _{0}^{10}x{\sqrt {1+x^{2}}}{\sqrt {1+4x^{2}}}dx}{\int _{0}^{10}{\sqrt {1+4x^{2}}}dx}}={\frac {\int _{0}^{10}x{\sqrt {4x^{4}+5x^{2}+1}}dx}{\int _{0}^{10}2{\sqrt {{\frac {1}{2}}+x^{2}}}dx}}.}
Toliau integruodami taip arba taip ir įstačius x=1 gauname 1,0565457675431157081260089778614 ir 0,95907194527687339898921071078133 atitinkamai. Kita vertus, integruojant taip ir taip gauname tą patį rezultatą
1
12
x
2
(
8
x
4
+
15
x
2
+
6
)
.
{\displaystyle {\frac {1}{12}}x^{2}(8x^{4}+15x^{2}+6).}
∫
x
2
±
a
2
d
x
=
1
2
x
x
2
±
a
2
±
1
2
a
2
ln
|
x
+
x
2
±
a
2
|
{\displaystyle \int {\sqrt {x^{2}\pm a^{2}}}dx={\frac {1}{2}}x{\sqrt {x^{2}\pm a^{2}}}\pm {\frac {1}{2}}a^{2}\ln |x+{\sqrt {x^{2}\pm a^{2}}}|}
∫
0
10
x
4
x
4
+
5
x
2
+
1
d
x
2
∫
0
10
1
4
+
x
2
d
x
=
1
12
x
2
(
8
x
4
+
15
x
2
+
6
)
|
0
10
2
(
1
2
x
x
2
+
1
4
+
1
2
⋅
1
4
ln
|
x
+
x
2
+
1
4
|
)
|
0
10
=
{\displaystyle {\frac {\int _{0}^{10}x{\sqrt {4x^{4}+5x^{2}+1}}dx}{2\int _{0}^{10}{\sqrt {{\frac {1}{4}}+x^{2}}}dx}}={\frac {{\frac {1}{12}}x^{2}(8x^{4}+15x^{2}+6)|_{0}^{10}}{2({\frac {1}{2}}x{\sqrt {x^{2}+{\frac {1}{4}}}}+{\frac {1}{2}}\cdot {\frac {1}{4}}\ln |x+{\sqrt {x^{2}+{\frac {1}{4}}}}|)|_{0}^{10}}}=}
=
1
12
⋅
10
2
(
8
⋅
10
4
+
15
⋅
10
2
+
6
)
−
0
10
10
2
+
1
4
+
1
4
ln
|
10
+
10
2
+
1
4
|
−
(
0
⋅
0
2
+
1
4
+
1
4
ln
|
0
+
0
2
+
1
4
|
)
=
{\displaystyle ={\frac {{\frac {1}{12}}\cdot 10^{2}(8\cdot 10^{4}+15\cdot 10^{2}+6)-0}{10{\sqrt {10^{2}+{\frac {1}{4}}}}+{\frac {1}{4}}\ln |10+{\sqrt {10^{2}+{\frac {1}{4}}}}|-(0\cdot {\sqrt {0^{2}+{\frac {1}{4}}}}+{\frac {1}{4}}\ln |0+{\sqrt {0^{2}+{\frac {1}{4}}}}|)}}=}
=
1
12
⋅
100
(
80000
+
1500
+
6
)
10
100.25
+
1
4
ln
|
10
+
100.25
|
−
(
1
4
ln
|
1
2
|
)
=
{\displaystyle ={\frac {{\frac {1}{12}}\cdot 100(80000+1500+6)}{10{\sqrt {100.25}}+{\frac {1}{4}}\ln |10+{\sqrt {100.25}}|-({\frac {1}{4}}\ln |{\frac {1}{2}}|)}}=}
=
1
12
⋅
100
⋅
81506
100.12492197
+
0.7490891721
−
(
−
0.173286795
)
=
{\displaystyle ={\frac {{\frac {1}{12}}\cdot 100\cdot 81506}{100.12492197+0.7490891721-(-0.173286795)}}=}
=
679216.6667
101.0472979
=
6721.76971.
{\displaystyle ={\frac {679216.6667}{101.0472979}}=6721.76971.}
Nepriekaištingai tikslus atsakymas yra:
679216,66666666666666666666666667/(100,12492197250392863848606074161+0,74908917210724008285110536237573+0,17328679513998632735430803036454)=
=679216,66666666666666666666666667/101,04729793975115504869147413435=6721,7697109688724955439401205637.
Todėl belieka patikrinti ar Wolframo internetinio integratoriaus atsakymas teisingas paėmus išvėstinę ir paskui įstačius x=1 bei palyginti su neišintegruotu reiškiniu:
(
1
12
x
2
(
8
x
4
+
15
x
2
+
6
)
)
′
=
(
1
12
(
8
x
6
+
15
x
4
+
6
x
2
)
)
′
=
1
12
(
48
x
5
+
60
x
3
+
12
x
)
=
4
x
5
+
5
x
3
+
x
=
4
+
5
+
1
=
10
;
{\displaystyle ({\frac {1}{12}}x^{2}(8x^{4}+15x^{2}+6))'=({\frac {1}{12}}(8x^{6}+15x^{4}+6x^{2}))'={\frac {1}{12}}(48x^{5}+60x^{3}+12x)=4x^{5}+5x^{3}+x=4+5+1=10;}
x
4
x
4
+
5
x
2
+
1
=
4
x
6
+
5
x
4
+
1
=
4
+
5
+
1
=
10
.
{\displaystyle x{\sqrt {4x^{4}+5x^{2}+1}}={\sqrt {4x^{6}+5x^{4}+1}}={\sqrt {4+5+1}}={\sqrt {10}}.}
Pasirodo, kad nepridėta šaknis įvedimo formoje į integratorių, bet integruojant taip ir taip gauname tokį patį rezultatą, kuris yra labai sudetingas ir ilgas. Net didžiausioje integralų lentelėje nėra kaip išintegruoti
∫
x
1
+
x
2
1
+
4
x
2
d
x
.
{\displaystyle \int x{\sqrt {1+x^{2}}}{\sqrt {1+4x^{2}}}dx.}
∫
a
+
b
x
c
+
p
x
d
x
,
{\displaystyle \int {\sqrt {a+bx}}{\sqrt {c+px}}dx,}
Kai x kinta 0 iki 5, tai integruojant gauname
p
=
∫
0
5
(
x
)
2
+
(
x
2
)
2
d
x
=
∫
0
5
x
2
+
x
4
d
x
=
∫
0
10
x
1
+
x
2
d
x
=
{\displaystyle p=\int _{0}^{5}{\sqrt {(x)^{2}+(x^{2})^{2}}}dx=\int _{0}^{5}{\sqrt {x^{2}+x^{4}}}dx=\int _{0}^{10}x{\sqrt {1+x^{2}}}dx=}
=
1
3
(
x
2
+
1
)
3
/
2
|
0
5
=
1
3
(
5
2
+
1
)
3
/
2
−
1
3
(
0
2
+
1
)
3
/
2
=
{\displaystyle ={\frac {1}{3}}(x^{2}+1)^{3/2}|_{0}^{5}={\frac {1}{3}}(5^{2}+1)^{3/2}-{\frac {1}{3}}(0^{2}+1)^{3/2}=}
=
1
3
(
26
)
3
/
2
−
1
3
=
17576
−
1
3
=
{\displaystyle ={\frac {1}{3}}(26)^{3/2}-{\frac {1}{3}}={\frac {{\sqrt {17576}}-1}{3}}=}
=43,858169117804135193577942278197.
Randame, m=p/5=43,858169117804135193577942278197/5=8,7716338235608270387155884556395. Tą patį atsakymą (8,77163383630657) gauname ir pasinaudodami Free Pascal kodu: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(5.0*a/1000000000)+sqr(sqr(5.0*a/1000000000)))/1000000000;
writeln(c);
readln;
end.
Nustatyti parabolės
y
=
x
2
{\displaystyle y=x^{2}}
x kinta nuo 0 iki 10. Tiesės tankis
γ
{\displaystyle \gamma }
O ) didėja proporcingai tik Ox kryptimi, t. y.
γ
=
x
.
{\displaystyle \gamma =x.}
Sprendimas . Greičiausias būdas apskaičiuoti, tai ko reikalauja sąlyga (uždavinys) yra toks:
m
=
∫
0
10
x
d
x
=
x
2
2
|
0
10
=
10
2
2
−
0
2
2
=
50.
{\displaystyle m=\int _{0}^{10}x\;dx={\frac {x^{2}}{2}}|_{0}^{10}={\frac {10^{2}}{2}}-{\frac {0^{2}}{2}}=50.}
Kitas būdas yra toks:
m
=
∫
0
10
γ
1
+
[
y
′
]
2
d
x
∫
0
10
1
+
[
y
′
]
2
d
x
=
∫
0
10
x
1
+
(
2
x
)
2
d
x
∫
0
10
1
+
(
2
x
)
2
d
x
=
∫
0
10
x
1
+
4
x
2
d
x
∫
0
10
1
+
4
x
2
d
x
=
{\displaystyle m={\frac {\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[y']^{2}}}dx}}={\frac {\int _{0}^{10}x{\sqrt {1+(2x)^{2}}}dx}{\int _{0}^{10}{\sqrt {1+(2x)^{2}}}dx}}={\frac {\int _{0}^{10}x{\sqrt {1+4x^{2}}}dx}{\int _{0}^{10}{\sqrt {1+4x^{2}}}dx}}=}
=
∫
0
10
2
x
1
4
+
x
2
d
x
2
∫
0
10
1
4
+
x
2
d
x
=
2
3
(
x
2
+
1
4
)
3
/
2
|
0
10
2
(
1
2
x
x
2
+
1
4
+
1
2
⋅
1
4
ln
|
x
+
x
2
+
1
4
|
)
|
0
10
=
{\displaystyle ={\frac {\int _{0}^{10}2x{\sqrt {{\frac {1}{4}}+x^{2}}}dx}{2\int _{0}^{10}{\sqrt {{\frac {1}{4}}+x^{2}}}dx}}={\frac {{\frac {2}{3}}(x^{2}+{\frac {1}{4}})^{3/2}|_{0}^{10}}{2\left({\frac {1}{2}}x{\sqrt {x^{2}+{\frac {1}{4}}}}+{\frac {1}{2}}\cdot {\frac {1}{4}}\ln |x+{\sqrt {x^{2}+{\frac {1}{4}}}}|\right)|_{0}^{10}}}=}
=
2
3
(
10
2
+
1
4
)
3
/
2
−
2
3
(
0
2
+
1
4
)
3
/
2
(
x
x
2
+
1
4
+
1
4
ln
|
x
+
x
2
+
1
4
|
)
|
0
10
=
{\displaystyle ={\frac {{\frac {2}{3}}(10^{2}+{\frac {1}{4}})^{3/2}-{\frac {2}{3}}(0^{2}+{\frac {1}{4}})^{3/2}}{\left(x{\sqrt {x^{2}+{\frac {1}{4}}}}+{\frac {1}{4}}\ln |x+{\sqrt {x^{2}+{\frac {1}{4}}}}|\right)|_{0}^{10}}}=}
=
2
3
⋅
100.25
3
/
2
−
2
3
(
1
2
)
3
10
100.25
+
1
4
ln
|
10
+
100.25
|
−
1
4
ln
|
1
2
|
=
{\displaystyle ={\frac {{\frac {2}{3}}\cdot 100.25^{3/2}-{\frac {2}{3}}\left({\frac {1}{2}}\right)^{3}}{10{\sqrt {100.25}}+{\frac {1}{4}}\ln |10+{\sqrt {100.25}}|-{\frac {1}{4}}\ln |{\frac {1}{2}}|}}=}
=(669,16822851623458973388183928978 - (2/3)*(1/8))/(100,12492197250392863848606074161 + 0,74908917210724008285110536237573 - (-0,17328679513998632735430803036454))=
=(669,16822851623458973388183928978 - 1/12)/101,04729793975115504869147413435=
=669,08489518290125640054850595645/101,04729793975115504869147413435=6,6215020967887643029775923955111;
čia pasinaudojome integralų lentele
∫
x
x
2
±
a
2
d
x
=
1
3
(
x
2
±
a
2
)
3
/
2
{\displaystyle \int x{\sqrt {x^{2}\pm a^{2}}}dx={\frac {1}{3}}(x^{2}\pm a^{2})^{3/2}}
∫
x
2
±
a
2
d
x
=
1
2
x
x
2
±
a
2
±
1
2
a
2
ln
|
x
+
x
2
±
a
2
|
.
{\displaystyle \int {\sqrt {x^{2}\pm a^{2}}}dx={\frac {1}{2}}x{\sqrt {x^{2}\pm a^{2}}}\pm {\frac {1}{2}}a^{2}\ln |x+{\sqrt {x^{2}\pm a^{2}}}|.}
Tuo atveju, jeigu x kinta nuo 0 iki 5 tada:
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
∫
0
5
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
1
+
4
x
2
d
x
∫
0
5
1
+
4
x
2
d
x
=
2
∫
0
5
x
1
4
+
x
2
d
x
2
∫
0
5
1
4
+
x
2
d
x
=
{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {\int _{0}^{5}x{\sqrt {1+4x^{2}}}dx}{\int _{0}^{5}{\sqrt {1+4x^{2}}}dx}}={\frac {2\int _{0}^{5}x{\sqrt {{\frac {1}{4}}+x^{2}}}dx}{2\int _{0}^{5}{\sqrt {{\frac {1}{4}}+x^{2}}}dx}}=}
=
2
3
(
5
2
+
1
4
)
3
/
2
−
2
3
(
0
2
+
1
4
)
3
/
2
(
x
x
2
+
1
4
+
1
4
ln
|
x
+
x
2
+
1
4
|
)
|
0
5
=
{\displaystyle ={\frac {{\frac {2}{3}}(5^{2}+{\frac {1}{4}})^{3/2}-{\frac {2}{3}}(0^{2}+{\frac {1}{4}})^{3/2}}{\left(x{\sqrt {x^{2}+{\frac {1}{4}}}}+{\frac {1}{4}}\ln |x+{\sqrt {x^{2}+{\frac {1}{4}}}}|\right)|_{0}^{5}}}=}
=
2
3
⋅
25.25
3
/
2
−
2
3
(
1
2
)
3
5
25.25
+
1
4
ln
|
5
+
25.25
|
−
1
4
ln
|
1
2
|
=
{\displaystyle ={\frac {{\frac {2}{3}}\cdot 25.25^{3/2}-{\frac {2}{3}}\left({\frac {1}{2}}\right)^{3}}{5{\sqrt {25.25}}+{\frac {1}{4}}\ln |5+{\sqrt {25.25}}|-{\frac {1}{4}}\ln |{\frac {1}{2}}|}}=}
=(84,586453144434159774345479682393-1/12)/(25,124689052802225675548162281899+0,57626894243450610735734085403457-(-0,17328679513998632735430803036454))=
=84,50311981110082644101214634906/25,874244790376718110259811166298=3,2659163772976925830147095531443.
Jėgos darbas padarytas judant kreive plokštumoje apskaičiuojamas pagal formulę
A
=
∫
B
C
P
d
x
+
Q
d
y
.
{\displaystyle A=\int _{BC}Pdx+Qdy.}
A
=
∫
B
C
P
d
x
+
Q
d
y
+
R
d
z
.
{\displaystyle A=\int _{BC}Pdx+Qdy+Rdz.}
Apskaičiuosime darbą jėgos
F
(
x
,
y
)
{\displaystyle F(x,y)}
Pagal sąlyga,
|
F
(
x
,
y
)
|
=
x
2
+
y
2
;
{\displaystyle |F(x,y)|={\sqrt {x^{2}+y^{2}}};}
P
=
−
x
,
{\displaystyle P=-x,}
Q
=
−
y
{\displaystyle Q=-y}
−
{\displaystyle -}
A
=
−
∮
L
x
d
x
+
y
d
y
,
{\displaystyle A=-\oint _{L}xdx+ydy,}
L - elipsė
x
=
a
cos
t
,
y
=
b
sin
t
,
{\displaystyle x=a\cos t,\;y=b\sin t,}
0
≤
t
≤
2
π
.
{\displaystyle 0\leq t\leq 2\pi .}
A
=
−
∫
0
2
π
a
cos
t
(
−
a
sin
t
)
d
t
+
b
sin
(
t
)
b
cos
(
t
)
d
t
=
−
∫
0
2
π
(
b
2
−
a
2
)
sin
(
t
)
cos
t
d
t
=
{\displaystyle A=-\int _{0}^{2\pi }a\cos t(-a\sin t)dt+b\sin(t)b\cos(t)dt=-\int _{0}^{2\pi }(b^{2}-a^{2})\sin(t)\cos t\;dt=}
=
a
2
−
b
2
2
∫
0
2
π
sin
(
2
t
)
d
t
=
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
2
π
=
0.
{\displaystyle ={a^{2}-b^{2} \over 2}\int _{0}^{2\pi }\sin(2t)dt={a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{2\pi }=0.}
Jei t keistusi nuo 0 iki
π
2
,
{\displaystyle {\pi \over 2},}
a
2
−
b
2
4
(
−
cos
(
2
t
)
)
|
0
π
2
=
a
2
−
b
2
4
(
−
cos
(
2
⋅
π
2
)
)
−
a
2
−
b
2
4
(
−
cos
(
2
⋅
0
)
)
=
{\displaystyle {a^{2}-b^{2} \over 4}(-\cos(2t))|_{0}^{\pi \over 2}={a^{2}-b^{2} \over 4}(-\cos(2\cdot {\pi \over 2}))-{a^{2}-b^{2} \over 4}(-\cos(2\cdot 0))=}
=
a
2
−
b
2
4
(
−
cos
π
+
cos
(
0
)
)
=
a
2
−
b
2
4
(
−
(
−
1
)
+
1
)
=
a
2
−
b
2
2
.
{\displaystyle ={a^{2}-b^{2} \over 4}(-\cos \pi +\cos(0))={a^{2}-b^{2} \over 4}(-(-1)+1)={a^{2}-b^{2} \over 2}.}
Tarkime, jei
a
=
5
,
b
=
3
,
{\displaystyle a=5,\;b=3,}
A
=
a
2
−
b
2
2
=
5
2
−
3
2
2
=
25
−
9
2
=
16
2
=
8.
{\displaystyle A={a^{2}-b^{2} \over 2}={5^{2}-3^{2} \over 2}={\frac {25-9}{2}}={\frac {16}{2}}=8.}
Kad patikrinti ar apskaičiuota teisingai reikia sudėti visas x reikšmes ant elipsės linijos pirmame ketvirtyje. Taip pat reikia sudėti visas y reikšmes ant elipsės linijos pirmame ketvirtyje. Galiausiai reikia sudėti sumas x ir y reikšmių, kad gauti darbą A pirmame ketvirtyje.
Arba tiesiog darbas yra lygus
A
=
1
100
∑
n
=
1
100
x
n
2
+
y
n
2
;
{\displaystyle A={\frac {1}{100}}\sum _{n=1}^{100}{\sqrt {x_{n}^{2}+y_{n}^{2}}};}
x
n
{\displaystyle x_{n}}
x reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki a ; analogiškai
y
n
{\displaystyle y_{n}}
y reikšmės ant elipsės linijos pirmame ketvirtyje nuo 0 iki b .
Kuris iš šių variantų yra darbas A , paaiškės pasumavus ir paskaičiavus. Sutapimas ar ne, bet
A
=
3
+
5
=
8
{\displaystyle A=3+5=8}
A
=
5
10
∑
n
=
1
10
(
n
(
3
/
10
)
)
2
+
(
(
10
−
n
)
(
5
/
10
)
)
2
=
{\displaystyle A={\frac {5}{10}}\sum _{n=1}^{10}{\sqrt {(n(3/10))^{2}+((10-n)(5/10))^{2}}}=}
A
=
(
0.3
2
+
5
2
+
0.6
2
+
4.5
2
+
0.9
2
+
4
2
+
1.2
2
+
3.5
2
+
1.5
2
+
3
2
+
{\displaystyle A=({\sqrt {0.3^{2}+5^{2}}}+{\sqrt {0.6^{2}+4.5^{2}}}+{\sqrt {0.9^{2}+4^{2}}}+{\sqrt {1.2^{2}+3.5^{2}}}+{\sqrt {1.5^{2}+3^{2}}}+}
+
1.8
2
+
2.5
2
+
2.1
2
+
2
2
+
2.4
2
+
1.5
2
+
2.7
2
+
1
2
+
3
2
+
0.5
2
)
/
10
=
{\displaystyle +{\sqrt {1.8^{2}+2.5^{2}}}+{\sqrt {2.1^{2}+2^{2}}}+{\sqrt {2.4^{2}+1.5^{2}}}+{\sqrt {2.7^{2}+1^{2}}}+{\sqrt {3^{2}+0.5^{2}}})/10=}
=
5
(
5.0089919
+
4.539823785
+
4.1
+
3.7
+
3.354101966
+
{\displaystyle =5(5.0089919+4.539823785+4.1+3.7+3.354101966+}
+
3.08058436
+
2.9
+
2.83019434
+
2.879236
+
3.041381265
)
/
10
=
{\displaystyle +3.08058436+2.9+2.83019434+2.879236+3.041381265)/10=}
=
5
⋅
35.43431361
/
10
=
17.717156805.
{\displaystyle =5\cdot 35.43431361/10=17.717156805.}
Iš tikro, ko gero, mes apskaičiuojame tokiu budu ne darbą atlikta apeinant elipsės liniją pirmame ketvirtyje, o darbą atlikta apeinant tiesę pirmame ketvirtyje. Štai kodas programos "Free Pascal" (FreePascal IDE for Win32 for i386; Target CPU: i386; Version 1.0.12 2011/04/23; <Compiler Version 2.4.4>; <Debugger GDB 7.2>; Copyright <C> 1998-2009): var
a:longint;
c:real;
begin
for a:=1 to 10
do
c:=c+0.5*sqrt(sqr(a*0.3)+sqr((11-a)*0.5));
writeln(c);
readln;
end.
gauname
A
=
17.7171568203085.
{\displaystyle A=17.7171568203085.}
Panaudojus šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+0.000000005*sqrt(sqr(a*0.000000003)+sqr((1000000001-a)*0.000000005));
writeln(c);
readln;
end.
gauname atsakymą
A
=
16.4328975236327
{\displaystyle A=16.4328975236327}
Panaudojus šį (teisingesnį) kodą: var
a:longint;
c:real;
begin
for a:=0 to 1000000000 do
c:=c+sqrt(sqr(a*0.000000003)+sqr((1000000000-a)*0.000000005))/1000000001;
writeln(5*c);
readln;
end.
gauname atsakymą
A
=
16.4328975142850
{\displaystyle A=16.4328975142850}
Integruojant gauname:
A
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
x
=
{\displaystyle A=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}dx=}
=
1
340
(
34
x
−
45
)
34
x
2
−
90
x
+
225
−
1125
arcsinh
(
3
5
−
34
x
75
)
68
34
|
0
5
=
{\displaystyle ={\frac {1}{340}}(34x-45){\sqrt {34x^{2}-90x+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34x}{75}})}{68{\sqrt {34}}}}|_{0}^{5}=}
=
34
⋅
5
−
45
340
34
⋅
5
2
−
90
⋅
5
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
5
75
)
68
34
−
(
34
⋅
0
−
45
340
34
⋅
0
2
−
90
⋅
0
+
225
−
1125
arcsinh
(
3
5
−
34
⋅
0
75
)
68
34
)
=
{\displaystyle ={\frac {34\cdot 5-45}{340}}{\sqrt {34\cdot 5^{2}-90\cdot 5+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 5}{75}})}{68{\sqrt {34}}}}-\left({\frac {34\cdot 0-45}{340}}{\sqrt {34\cdot 0^{2}-90\cdot 0+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {34\cdot 0}{75}})}{68{\sqrt {34}}}}\right)=}
=
170
−
45
340
850
−
450
+
225
−
1125
arcsinh
(
3
5
−
170
75
)
68
34
−
(
−
45
340
225
−
1125
arcsinh
(
3
5
)
68
34
)
=
{\displaystyle ={\frac {170-45}{340}}{\sqrt {850-450+225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}}-{\frac {170}{75}})}{68{\sqrt {34}}}}-\left({\frac {-45}{340}}{\sqrt {225}}-{\frac {1125\operatorname {arcsinh} ({\frac {3}{5}})}{68{\sqrt {34}}}}\right)=}
=
125
340
625
−
1125
arcsinh
(
3
⋅
75
−
170
⋅
5
5
⋅
75
)
68
34
−
(
−
9
68
⋅
15
−
1125
arcsinh
(
0.6
)
396.50472884948
)
=
{\displaystyle ={\frac {125}{340}}{\sqrt {625}}-{\frac {1125\operatorname {arcsinh} ({\frac {3\cdot 75-170\cdot 5}{5\cdot 75}})}{68{\sqrt {34}}}}-\left({\frac {-9}{68}}\cdot 15-{\frac {1125\operatorname {arcsinh} (0.6)}{396.50472884948}}\right)=}
=
25
68
⋅
25
−
1125
arcsinh
(
225
−
850
375
)
396.50472884948
−
(
−
135
68
−
1125
⋅
0.5688248987
396.50472884948
)
=
{\displaystyle ={\frac {25}{68}}\cdot 25-{\frac {1125\operatorname {arcsinh} ({\frac {225-850}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {1125\cdot 0.5688248987}{396.50472884948}}\right)=}
=
625
68
−
1125
arcsinh
(
−
625
375
)
396.50472884948
−
(
−
135
68
−
639.928011
396.50472884948
)
=
{\displaystyle ={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} ({\frac {-625}{375}})}{396.50472884948}}-\left({\frac {-135}{68}}-{\frac {639.928011}{396.50472884948}}\right)=}
=
9.19117647
−
1125
arcsinh
(
−
5
3
)
396.50472884948
−
(
−
1.98529412
−
1.613922772
)
=
{\displaystyle =9.19117647-{\frac {1125\operatorname {arcsinh} (-{\frac {5}{3}})}{396.50472884948}}-(-1.98529412-1.613922772)=}
=
9.19117647
−
1125
⋅
(
−
1.283795663
)
396.50472884948
−
(
−
3.599216892
)
=
{\displaystyle =9.19117647-{\frac {1125\cdot (-1.283795663)}{396.50472884948}}-(-3.599216892)=}
=
9.19117647
+
3.642504151
+
3.599216892
=
16.43289751.
{\displaystyle =9.19117647+3.642504151+3.599216892=16.43289751.}
Apskaičiuojame viską nepriekaištingai tiksliai su kompiuterio kalkuliatoriumi:
A
=
625
68
−
1125
arcsinh
(
−
5
/
3
)
68
34
−
(
−
135
68
−
1125
arcsinh
(
3
/
5
)
68
34
)
=
{\displaystyle A={\frac {625}{68}}-{\frac {1125\operatorname {arcsinh} (-5/3)}{68{\sqrt {34}}}}-\left({\frac {-135}{68}}-{\frac {1125\operatorname {arcsinh} (3/5)}{68{\sqrt {34}}}}\right)=}
=
12.833680621236990323769396193322
−
(
−
3.5992168895913637545005254840611
)
=
16.432897510828354078269921677383.
{\displaystyle =12.833680621236990323769396193322-(-3.5992168895913637545005254840611)=16.432897510828354078269921677383.}
Toliau bandome rasti atlikta darbą apeita tiese
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
A
=
−
∮
L
x
d
x
+
y
d
y
.
{\displaystyle A=-\oint _{L}xdx+ydy.}
d
y
=
−
3
5
d
x
.
{\displaystyle dy=-{\frac {3}{5}}\;dx.}
x integravimo ribos yra 0 iki 5, o y integravimo ribos yra nuo 0 iki 3. Gauname:
A
=
−
∮
L
x
d
x
+
y
d
y
=
−
(
∫
0
5
x
d
x
−
3
5
(
3
−
3
x
5
)
d
x
)
=
−
∫
0
5
(
x
−
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=-\oint _{L}xdx+ydy=-(\int _{0}^{5}x\;dx-{\frac {3}{5}}(3-{\frac {3x}{5}})dx)=-\int _{0}^{5}(x-{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
−
∫
0
5
(
x
−
9
5
+
9
x
25
)
d
x
=
−
∫
0
5
(
−
9
5
+
25
x
+
9
x
25
)
d
x
=
−
∫
0
5
(
34
x
25
−
9
5
)
d
x
=
{\displaystyle =-\int _{0}^{5}(x-{\frac {9}{5}}+{\frac {9x}{25}})dx=-\int _{0}^{5}(-{\frac {9}{5}}+{\frac {25x+9x}{25}})dx=-\int _{0}^{5}({\frac {34x}{25}}-{\frac {9}{5}})dx=}
=
−
(
34
x
2
25
⋅
2
−
9
5
x
)
|
0
5
=
−
[
(
17
⋅
5
2
25
−
9
5
⋅
5
)
−
(
17
⋅
0
2
25
−
9
5
⋅
0
)
]
=
−
(
17
−
9
)
=
−
8.
{\displaystyle =-({\frac {34x^{2}}{25\cdot 2}}-{\frac {9}{5}}x)|_{0}^{5}=-[({\frac {17\cdot 5^{2}}{25}}-{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}-{\frac {9}{5}}\cdot 0)]=-(17-9)=-8.}
Pastebime, kad jeigu elipsė būtų apskritimas tai pirmame ketvirtyje atliktas darbas A=0. Taip yra todėl, kad kai x didėja, tada y reikšmės mažėja. Todėl skaičiuoti darbą, ko gero, galėtų būti teisingiau
A
=
∮
L
x
d
x
−
y
d
y
{\displaystyle A=\oint _{L}xdx-ydy}
A
=
∮
L
−
x
d
x
+
y
d
y
.
{\displaystyle A=\oint _{L}-xdx+ydy.}
A
=
a
2
+
b
2
2
=
5
2
+
3
2
2
=
17.
{\displaystyle A={\frac {a^{2}+b^{2}}{2}}={\frac {5^{2}+3^{2}}{2}}=17.}
Tą patį darba gauname ir apeinant tiese:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
−
−
3
5
(
3
−
3
x
5
)
d
x
=
∫
0
5
(
x
+
3
5
(
3
−
3
x
5
)
)
d
x
=
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx-{\frac {-3}{5}}(3-{\frac {3x}{5}})dx=\int _{0}^{5}(x+{\frac {3}{5}}(3-{\frac {3x}{5}}))dx=}
=
∫
0
5
(
x
+
9
5
−
9
x
25
)
d
x
=
∫
0
5
(
9
5
+
25
x
−
9
x
25
)
d
x
=
∫
0
5
(
16
x
25
+
9
5
)
d
x
=
{\displaystyle =\int _{0}^{5}(x+{\frac {9}{5}}-{\frac {9x}{25}})dx=\int _{0}^{5}({\frac {9}{5}}+{\frac {25x-9x}{25}})dx=\int _{0}^{5}({\frac {16x}{25}}+{\frac {9}{5}})dx=}
=
(
16
x
2
25
⋅
2
+
9
5
x
)
|
0
5
=
(
8
⋅
5
2
25
+
9
5
⋅
5
)
−
(
17
⋅
0
2
25
+
9
5
⋅
0
)
=
8
+
9
=
17.
{\displaystyle =({\frac {16x^{2}}{25\cdot 2}}+{\frac {9}{5}}x)|_{0}^{5}=({\frac {8\cdot 5^{2}}{25}}+{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}+{\frac {9}{5}}\cdot 0)=8+9=17.}
Lygiai tą patį darbą gausime ir integruojant taip:
A
=
∮
L
x
d
x
−
y
d
y
=
∫
0
5
x
d
x
+
∫
0
3
y
d
y
=
x
2
2
|
0
5
+
y
2
2
|
0
3
=
25
2
+
9
2
=
17.
{\displaystyle A=\oint _{L}xdx-ydy=\int _{0}^{5}x\;dx+\int _{0}^{3}y\;dy={\frac {x^{2}}{2}}|_{0}^{5}+{\frac {y^{2}}{2}}|_{0}^{3}={\frac {25}{2}}+{\frac {9}{2}}=17.}
Tą patį gausime ir taip integruojant nuo 0 iki 5:
A
=
∫
F
d
F
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
d
(
x
2
+
(
3
−
3
x
5
)
2
)
=
{\displaystyle A=\int F\;\mathbf {d} F=\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\;\mathbf {d} \left({\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}\right)=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
0.5
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
2
x
+
2
(
3
−
3
x
/
5
)
(
−
0.6
)
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}0.5(x^{2}+(3-3x/5)^{2})^{-0.5}(2x+2(3-3x/5)(-0.6))\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
2
+
(
3
−
3
x
/
5
)
2
)
−
0.5
(
x
−
0.6
(
3
−
3
x
/
5
)
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x^{2}+(3-3x/5)^{2})^{-0.5}(x-0.6(3-3x/5)\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
(
x
−
1.8
+
0.36
x
)
x
2
+
(
3
−
3
x
/
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\frac {{\sqrt {x^{2}+(3-{\frac {3x}{5}})^{2}}}(x-1.8+0.36x)}{\sqrt {x^{2}+(3-3x/5)^{2}}}}\;dx=}
=
∫
0
5
(
1.36
x
−
1.8
)
d
x
=
(
1.36
x
2
2
−
1.8
x
)
|
0
5
=
{\displaystyle =\int _{0}^{5}(1.36x-1.8)\;dx=(1.36{\frac {x^{2}}{2}}-1.8x)|_{0}^{5}=}
=
[
(
0.68
x
−
1.8
)
x
]
|
0
5
=
(
0.68
⋅
5
−
1.8
)
⋅
5
=
(
3.4
−
1.8
)
⋅
5
=
1.6
⋅
5
=
8.
{\displaystyle =[(0.68x-1.8)x]|_{0}^{5}=(0.68\cdot 5-1.8)\cdot 5=(3.4-1.8)\cdot 5=1.6\cdot 5=8.}
Nustatyti tiesės
y
=
3
−
3
x
5
{\displaystyle y=3-{\frac {3x}{5}}}
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas . Pasinaudosime masės skaičiavimo formule
m
=
∫
L
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}
=
∫
0
5
x
2
+
(
3
−
3
x
5
)
2
1
+
(
−
3
5
)
2
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+\left(3-{\frac {3x}{5}}\right)^{2}}}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}\;dx=}
=
∫
0
5
x
2
+
9
−
18
x
5
+
9
x
2
25
1
+
9
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {x^{2}+9-{\frac {18x}{5}}+{\frac {9x^{2}}{25}}}}{\sqrt {1+{\frac {9}{25}}}}\;dx=}
=
∫
0
5
9
−
18
x
5
+
34
x
2
25
34
25
d
x
=
{\displaystyle =\int _{0}^{5}{\sqrt {9-{\frac {18x}{5}}+{\frac {34x^{2}}{25}}}}{\sqrt {\frac {34}{25}}}\;dx=}
=
34
5
∫
0
5
34
25
x
2
−
18
5
x
+
9
d
x
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\int _{0}^{5}{\sqrt {{\frac {34}{25}}x^{2}-{\frac {18}{5}}x+9}}\;dx=}
=
34
5
∫
0
5
1.36
x
2
−
3.6
x
+
9
d
x
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\int _{0}^{5}{\sqrt {1.36x^{2}-3.6x+9}}\;dx=}
=
34
5
(
−
3.6
+
2
⋅
1.36
x
4
⋅
1.36
1.36
x
2
−
3.6
x
+
9
+
4
⋅
1.36
⋅
9
−
(
−
3.6
)
2
8
⋅
1.36
3
/
2
ln
|
2
⋅
1.36
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2\cdot 1.36x}{4\cdot 1.36}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {4\cdot 1.36\cdot 9-(-3.6)^{2}}{8\cdot 1.36^{3/2}}}\ln |2\cdot 1.36x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
x
5.44
1.36
x
2
−
3.6
x
+
9
+
48.96
−
12.96
8
⋅
2.515456
1
/
2
ln
|
2.72
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72x}{5.44}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {48.96-12.96}{8\cdot 2.515456^{1/2}}}\ln |2.72x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
x
5.44
1.36
x
2
−
3.6
x
+
9
+
36
12.68815132
ln
|
2.72
x
−
3.6
+
2
1.36
(
1.36
x
2
−
3.6
x
+
9
)
|
)
|
0
5
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72x}{5.44}}{\sqrt {1.36x^{2}-3.6x+9}}+{\frac {36}{12.68815132}}\ln |2.72x-3.6+2{\sqrt {1.36(1.36x^{2}-3.6x+9)}}|\right)|_{0}^{5}=}
=
34
5
(
−
3.6
+
2.72
⋅
5
5.44
1.36
⋅
25
−
3.6
⋅
5
+
9
+
2.8372927689
ln
|
2.72
⋅
5
−
3.6
+
2
1.36
(
1.36
⋅
25
−
3.6
⋅
5
+
9
)
|
)
−
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72\cdot 5}{5.44}}{\sqrt {1.36\cdot 25-3.6\cdot 5+9}}+2.8372927689\ln |2.72\cdot 5-3.6+2{\sqrt {1.36(1.36\cdot 25-3.6\cdot 5+9)}}|\right)-}
−
34
5
(
−
3.6
+
2.72
⋅
0
5.44
1.36
⋅
0
2
−
3.6
⋅
0
+
9
+
2.8372927689
ln
|
2.72
⋅
0
−
3.6
+
2
1.36
(
1.36
⋅
0
2
−
3.6
⋅
0
+
9
)
|
)
=
{\displaystyle -{\frac {\sqrt {34}}{5}}\left({\frac {-3.6+2.72\cdot 0}{5.44}}{\sqrt {1.36\cdot 0^{2}-3.6\cdot 0+9}}+2.8372927689\ln |2.72\cdot 0-3.6+2{\sqrt {1.36(1.36\cdot 0^{2}-3.6\cdot 0+9)}}|\right)=}
=
34
5
(
10
5.44
34
−
18
+
9
+
2.8372927689
ln
|
13.6
−
3.6
+
2
1.36
(
34
−
18
+
9
)
|
)
−
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {10}{5.44}}{\sqrt {34-18+9}}+2.8372927689\ln |13.6-3.6+2{\sqrt {1.36(34-18+9)}}|\right)-}
−
34
5
(
−
3.6
5.44
9
+
2.8372927689
ln
|
−
3.6
+
2
1.36
⋅
9
|
)
=
{\displaystyle -{\frac {\sqrt {34}}{5}}\left({\frac {-3.6}{5.44}}{\sqrt {9}}+2.8372927689\ln |-3.6+2{\sqrt {1.36\cdot 9}}|\right)=}
=
34
5
(
10
5.44
25
+
2.8372927689
ln
|
10
+
2
1.36
⋅
25
|
)
−
34
5
(
−
3.6
5.44
⋅
3
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {10}{5.44}}{\sqrt {25}}+2.8372927689\ln |10+2{\sqrt {1.36\cdot 25}}|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-3.6}{5.44}}\cdot 3+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
50
5.44
+
2.8372927689
ln
|
10
+
2
34
|
)
−
34
5
(
−
10.8
5.44
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left({\frac {50}{5.44}}+2.8372927689\ln |10+2{\sqrt {34}}|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-10.8}{5.44}}+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
9.191176471
+
2.8372927689
ln
|
21.66190379
|
)
−
34
5
(
−
10.8
5.44
+
2.8372927689
ln
|
−
3.6
+
2
12.24
|
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}\left(9.191176471+2.8372927689\ln |21.66190379|\right)-{\frac {\sqrt {34}}{5}}\left({\frac {-10.8}{5.44}}+2.8372927689\ln |-3.6+2{\sqrt {12.24}}|\right)=}
=
34
5
(
9.191176471
+
8.726250336
)
−
34
5
(
−
1.985294118
+
3.469823414
)
=
{\displaystyle ={\frac {\sqrt {34}}{5}}(9.191176471+8.726250336)-{\frac {\sqrt {34}}{5}}(-1.985294118+3.469823414)=}
=
34
5
⋅
17.91742681
−
34
5
⋅
1.484529296
=
16.43289751
34
5
=
19.16388698.
{\displaystyle ={\frac {\sqrt {34}}{5}}\cdot 17.91742681-{\frac {\sqrt {34}}{5}}\cdot 1.484529296={\frac {16.43289751{\sqrt {34}}}{5}}=19.16388698.}
Kad tą patį apskaičiuoti su programa "Free Pascal" reikia surasti tiesės ilgį, kai x kinta nuo 0 iki 5 (tai yra tiesės ilgis tik pirmame ketviryje):
l
=
5
2
+
3
2
=
25
+
9
=
34
=
5.830951895.
{\displaystyle l={\sqrt {5^{2}+3^{2}}}={\sqrt {25+9}}={\sqrt {34}}=5.830951895.}
Todėl "Free Pascal" kodas yra toks: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(a*0.000000003)+sqr((1000000001-a)*0.000000005));
writeln(sqrt(sqr(3)+sqr(5))*c/1000000000);
readln;
end.
duodantis rezultatą 19,163886990613093 po 18 sekundžių su 2,6 GHz procesoriumi.
Nustatyti parabolės
y
=
x
2
{\displaystyle y=x^{2}}
x kinta nuo 0 iki 10. Tiesės tankis
γ
{\displaystyle \gamma }
O ) didėja proporcingai, t. y.
γ
=
x
2
+
y
2
.
{\displaystyle \gamma ={\sqrt {x^{2}+y^{2}}}.}
Sprendimas .Integruojame (pasinaudodami integralų lentelės (31) formule ):
p
=
∫
0
10
(
x
)
2
+
(
x
2
)
2
d
x
=
∫
0
10
x
2
+
x
4
d
x
=
∫
0
10
x
1
+
x
2
d
x
=
{\displaystyle p=\int _{0}^{10}{\sqrt {(x)^{2}+(x^{2})^{2}}}dx=\int _{0}^{10}{\sqrt {x^{2}+x^{4}}}dx=\int _{0}^{10}x{\sqrt {1+x^{2}}}dx=}
=
1
3
(
x
2
+
1
)
3
/
2
|
0
10
=
1
3
(
10
2
+
1
)
3
/
2
−
1
3
(
0
2
+
1
)
3
/
2
=
{\displaystyle ={\frac {1}{3}}(x^{2}+1)^{3/2}|_{0}^{10}={\frac {1}{3}}(10^{2}+1)^{3/2}-{\frac {1}{3}}(0^{2}+1)^{3/2}=}
=
1
3
(
101
)
3
/
2
−
1
3
=
1
3
1030301
−
1
3
=
{\displaystyle ={\frac {1}{3}}(101)^{3/2}-{\frac {1}{3}}={\frac {1}{3}}{\sqrt {1030301}}-{\frac {1}{3}}=}
=338,01247924440330576404858539624.
Toliau, kad surasti kreivės masę (kai kiekviename kreivės taške tankis priklauso nuo tam tikros funkcijos), skaičiuojame:
m
=
∫
0
10
γ
1
+
[
y
′
]
2
d
x
=
∫
0
10
x
2
+
y
2
1
+
[
(
x
2
)
′
]
2
d
x
=
∫
0
10
x
2
+
(
x
2
)
2
1
+
[
2
x
]
2
d
x
=
{\displaystyle m=\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx=\int _{0}^{10}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[(x^{2})']^{2}}}dx=\int _{0}^{10}{\sqrt {x^{2}+(x^{2})^{2}}}{\sqrt {1+[2x]^{2}}}dx=}
=
∫
0
10
x
2
+
x
4
1
+
4
x
2
d
x
=
∫
0
10
x
1
+
x
2
1
+
4
x
2
d
x
=
∫
0
10
x
4
x
4
+
5
x
2
+
1
d
x
.
{\displaystyle =\int _{0}^{10}{\sqrt {x^{2}+x^{4}}}{\sqrt {1+4x^{2}}}dx=\int _{0}^{10}x{\sqrt {1+x^{2}}}{\sqrt {1+4x^{2}}}dx=\int _{0}^{10}x{\sqrt {4x^{4}+5x^{2}+1}}dx.}
Toliau integruodami taip arba taip ir įstačius x=1 gauname 1,0565457675431157081260089778614 ir 0,95907194527687339898921071078133 atitinkamai. Kita vertus, integruojant taip ir taip gauname tą patį rezultatą
1
12
x
2
(
8
x
4
+
15
x
2
+
6
)
.
{\displaystyle {\frac {1}{12}}x^{2}(8x^{4}+15x^{2}+6).}
∫
0
10
x
4
x
4
+
5
x
2
+
1
d
x
=
1
12
x
2
(
8
x
4
+
15
x
2
+
6
)
|
0
10
=
{\displaystyle \int _{0}^{10}x{\sqrt {4x^{4}+5x^{2}+1}}dx={\frac {1}{12}}x^{2}(8x^{4}+15x^{2}+6)|_{0}^{10}=}
=
1
12
⋅
10
2
(
8
⋅
10
4
+
15
⋅
10
2
+
6
)
−
0
=
{\displaystyle ={\frac {1}{12}}\cdot 10^{2}(8\cdot 10^{4}+15\cdot 10^{2}+6)-0=}
=
1
12
⋅
100
(
80000
+
1500
+
6
)
=
{\displaystyle ={\frac {1}{12}}\cdot 100(80000+1500+6)=}
=
1
12
⋅
100
⋅
81506
=
679216.6667.
{\displaystyle ={\frac {1}{12}}\cdot 100\cdot 81506=679216.6667.}
Nepriekaištingai tikslus atsakymas yra:
679216,66666666666666666666666667.
Todėl belieka patikrinti ar Wolframo internetinio integratoriaus atsakymas teisingas paėmus išvėstinę ir paskui įstačius x=1 bei palyginti su neišintegruotu reiškiniu:
(
1
12
x
2
(
8
x
4
+
15
x
2
+
6
)
)
′
=
(
1
12
(
8
x
6
+
15
x
4
+
6
x
2
)
)
′
=
1
12
(
48
x
5
+
60
x
3
+
12
x
)
=
4
x
5
+
5
x
3
+
x
=
4
+
5
+
1
=
10
;
{\displaystyle ({\frac {1}{12}}x^{2}(8x^{4}+15x^{2}+6))'=({\frac {1}{12}}(8x^{6}+15x^{4}+6x^{2}))'={\frac {1}{12}}(48x^{5}+60x^{3}+12x)=4x^{5}+5x^{3}+x=4+5+1=10;}
x
4
x
4
+
5
x
2
+
1
=
4
x
6
+
5
x
4
+
1
=
4
+
5
+
1
=
10
.
{\displaystyle x{\sqrt {4x^{4}+5x^{2}+1}}={\sqrt {4x^{6}+5x^{4}+1}}={\sqrt {4+5+1}}={\sqrt {10}}.}
Pasirodo, kad nepridėta šaknis įvedimo formoje į integratorių, bet integruojant taip ir taip gauname tokį patį rezultatą, kuris yra labai sudetingas ir ilgas. Net didžiausioje integralų lentelėje nėra kaip išintegruoti
∫
x
1
+
x
2
1
+
4
x
2
d
x
.
{\displaystyle \int x{\sqrt {1+x^{2}}}{\sqrt {1+4x^{2}}}dx.}
∫
a
+
b
x
c
+
p
x
d
x
,
{\displaystyle \int {\sqrt {a+bx}}{\sqrt {c+px}}dx,}
Free Pascal kodu: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+(sqrt(sqr(a*0.00000001)+sqr(sqr(a*0.00000001)))-sqrt(sqr((a-1.0)*0.00000001)+sqr(sqr((a-1.0)*0.00000001))))*sqrt(sqr(a*0.00000001)+sqr(sqr(a*0.00000001)));
writeln(c);
readln;
end.
kuris duoda atsakymą m=5050,00000667382 po 53 sekundžių su 2,6 GHz procesoriumi.
Kai x kinta 0 iki 5, tai integruojant gauname
p
=
∫
0
5
(
x
)
2
+
(
x
2
)
2
d
x
=
∫
0
5
x
2
+
x
4
d
x
=
∫
0
10
x
1
+
x
2
d
x
=
{\displaystyle p=\int _{0}^{5}{\sqrt {(x)^{2}+(x^{2})^{2}}}dx=\int _{0}^{5}{\sqrt {x^{2}+x^{4}}}dx=\int _{0}^{10}x{\sqrt {1+x^{2}}}dx=}
=
1
3
(
x
2
+
1
)
3
/
2
|
0
5
=
1
3
(
5
2
+
1
)
3
/
2
−
1
3
(
0
2
+
1
)
3
/
2
=
{\displaystyle ={\frac {1}{3}}(x^{2}+1)^{3/2}|_{0}^{5}={\frac {1}{3}}(5^{2}+1)^{3/2}-{\frac {1}{3}}(0^{2}+1)^{3/2}=}
=
1
3
(
26
)
3
/
2
−
1
3
=
17576
−
1
3
=
{\displaystyle ={\frac {1}{3}}(26)^{3/2}-{\frac {1}{3}}={\frac {{\sqrt {17576}}-1}{3}}=}
=43,858169117804135193577942278197.
Tą patį atsakymą (43,8581691815329) gauname ir pasinaudodami Free Pascal kodu: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(5.0*a/1000000000)+sqr(sqr(5.0*a/1000000000)))/1000000000;
writeln(c*5);
readln;
end.
Kad rasti masę, kai x kinta 0 iki 5, pasinaudojame Free Pascal kodu: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+(sqrt(sqr(5.0*a/1000000000)+sqr(sqr(5.0*a/1000000000)))-sqrt(sqr(5.0*(a-1.0)*0.000000001)+sqr(sqr(5.0*(a-1.0)*0.000000001))))*sqrt(sqr(5.0*a/1000000000)+sqr(sqr(5.0*a/1000000000)));
writeln(c);
readln;
end.
kuris duoda atsakymą
m
=
325.000000425112
{\displaystyle m=325.000000425112}
Alternatyvus Free Pascal kodas, kuris skaičiuoja pagal formulę
m
=
∫
0
5
x
2
+
y
2
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
2
+
(
x
2
)
2
1
+
(
2
x
)
2
d
x
,
{\displaystyle m=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}dx=\int _{0}^{5}{\sqrt {x^{2}+(x^{2})^{2}}}{\sqrt {1+(2x)^{2}}}dx,}
var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(1+sqr(2*5.0*a/1000000000))*sqrt(sqr(5.0*a/1000000000)+sqr(sqr(5.0*a/1000000000)));
writeln(5*c/1000000000);
readln;
end
ir duoda atsakymą
m
=
327.860390075605
{\displaystyle m=327.860390075605}
var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(1+sqr(0.00000001*a))*sqrt(sqr(0.000000005*a)+sqr(sqr(0.000000005*a)));
writeln(0.000000005*c);
readln;
end.
duoda atsakymą
m
=
327.86039007560539
{\displaystyle m=327.86039007560539}
Kitoks kreivės masės apskaičiavimo Free Pascal kodas yra: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(0.000000005*a-0.000000005*(a-1))+sqr(sqr(0.000000005*a)-sqr(0.000000005*(a-1))))*sqrt(sqr(0.000000005*a)+sqr(sqr(0.000000005*a)));
writeln(c);
readln;
end.
kuris duoda atsakymą
m
=
327.860389859764
{\displaystyle m=327.860389859764}
var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(0.000000005)+sqr(sqr(0.000000005*a)-sqr(0.000000005*(a-1))))*sqrt(sqr(0.000000005*a)+sqr(sqr(0.000000005*a)));
writeln(c);
readln;
end
kuris duoda atsakymą
m
=
327.860389859763
{\displaystyle m=327.860389859763}
var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(0.000000000000000025+sqr(sqr(0.000000005*a)-sqr(0.000000005*(a-1))))*sqrt(sqr(0.000000005*a)+sqr(sqr(0.000000005*a)));
writeln(c);
readln;
end.
kuris duoda atsakymą
m
=
327.860389859763
{\displaystyle m=327.860389859763}
Free Pascal automatiškai optimizuoja kodą pakeldamas konstantą 0,000000005 kvadratu ir visoms iteracijoms naudodamas gautą 0,000000000000000025 reikšmę).
Nustatyti parabolės
y
=
x
2
{\displaystyle y=x^{2}}
x kinta nuo 0 iki 10. Tiesės tankis
γ
{\displaystyle \gamma }
O ) didėja proporcingai tik Ox kryptimi, t. y.
γ
=
x
.
{\displaystyle \gamma =x.}
Sprendimas . Greičiausias būdas apskaičiuoti, tai ko reikalauja sąlyga (uždavinys) yra toks:
m
=
∫
0
10
x
d
x
=
x
2
2
|
0
10
=
10
2
2
−
0
2
2
=
50.
{\displaystyle m=\int _{0}^{10}x\;dx={\frac {x^{2}}{2}}|_{0}^{10}={\frac {10^{2}}{2}}-{\frac {0^{2}}{2}}=50.}
Kitas būdas yra toks:
m
=
∫
0
10
γ
1
+
[
y
′
]
2
d
x
=
∫
0
10
x
1
+
(
2
x
)
2
d
x
=
∫
0
10
x
1
+
4
x
2
d
x
=
{\displaystyle m=\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx=\int _{0}^{10}x{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{10}x{\sqrt {1+4x^{2}}}dx=}
=
∫
0
10
2
x
1
4
+
x
2
d
x
=
2
3
(
x
2
+
1
4
)
3
/
2
|
0
10
=
{\displaystyle =\int _{0}^{10}2x{\sqrt {{\frac {1}{4}}+x^{2}}}dx={\frac {2}{3}}(x^{2}+{\frac {1}{4}})^{3/2}|_{0}^{10}=}
=
2
3
(
10
2
+
1
4
)
3
/
2
−
2
3
(
0
2
+
1
4
)
3
/
2
=
{\displaystyle ={\frac {2}{3}}(10^{2}+{\frac {1}{4}})^{3/2}-{\frac {2}{3}}(0^{2}+{\frac {1}{4}})^{3/2}=}
=
2
3
⋅
100.25
3
/
2
−
2
3
(
1
2
)
3
=
{\displaystyle ={\frac {2}{3}}\cdot 100.25^{3/2}-{\frac {2}{3}}\left({\frac {1}{2}}\right)^{3}=}
=669,16822851623458973388183928978 - (2/3)*(1/8)=
=669,16822851623458973388183928978 - 1/12=
=669,08489518290125640054850595645;
čia pasinaudojome integralų lentele
∫
x
x
2
±
a
2
d
x
=
1
3
(
x
2
±
a
2
)
3
/
2
.
{\displaystyle \int x{\sqrt {x^{2}\pm a^{2}}}dx={\frac {1}{3}}(x^{2}\pm a^{2})^{3/2}.}
Tuo atveju, jeigu x kinta nuo 0 iki 5 tada:
m
=
∫
0
5
γ
1
+
[
y
′
]
2
d
x
=
∫
0
5
x
1
+
4
x
2
d
x
=
2
∫
0
5
x
1
4
+
x
2
d
x
=
{\displaystyle m=\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx=\int _{0}^{5}x{\sqrt {1+4x^{2}}}dx=2\int _{0}^{5}x{\sqrt {{\frac {1}{4}}+x^{2}}}dx=}
=
2
3
(
5
2
+
1
4
)
3
/
2
−
2
3
(
0
2
+
1
4
)
3
/
2
=
{\displaystyle ={\frac {2}{3}}(5^{2}+{\frac {1}{4}})^{3/2}-{\frac {2}{3}}(0^{2}+{\frac {1}{4}})^{3/2}=}
=
2
3
⋅
25.25
3
/
2
−
2
3
(
1
2
)
3
=
{\displaystyle ={\frac {2}{3}}\cdot 25.25^{3/2}-{\frac {2}{3}}\left({\frac {1}{2}}\right)^{3}=}
=84,586453144434159774345479682393-1/12=
=84,50311981110082644101214634906.
Free Pascal kodas duodą tokį patį rezultatą (kai x kinta nuo 0 iki 5):var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(0.000000005)+sqr(sqr(0.000000005*a)-sqr(0.000000005*(a-1))))*0.000000005*a;
writeln(c);
readln;
end.
m=84,5031198757743 po 25 sekundžių su 2,6 GHz procesoriumi. Optimizuotas jo variantas: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(0.000000005)+sqr(sqr(0.000000005*a)-sqr(0.000000005*(a-1))))*a;
writeln(c*0.000000005);
readln;
end
duoda rezultatą m=84,503119875777473 po 23 sekundžių su 2,6 GHz procesoriumi (daugybos operaciją padaro 2-3 sekundėmis greičiau, tačiau darant vien tik daugybą: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+a*1.0*a;
writeln(c);
readln;
end.
gauname rezultatą
3.33333333833246
⋅
10
26
{\displaystyle 3.33333333833246\cdot 10^{26}}
a kvadratu, kaip šiame kode: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqr(a);
writeln(c);
readln;
end.
gauname rezultatą
−
1.49157006236160
⋅
10
13
{\displaystyle -1.49157006236160\cdot 10^{13}}
var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqr(1.0*a);
writeln(c);
readln;
end.
duodantį rezultatą
3.33333333833246
⋅
10
26
{\displaystyle 3.33333333833246\cdot 10^{26}}
2.50000000499896
⋅
10
35
{\displaystyle 2.50000000499896\cdot 10^{35}}
2.00000000499915
⋅
10
44
{\displaystyle 2.00000000499915\cdot 10^{44}}
a 17 laipsniu per kiekvieną iteraciją): var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+a*sqr(sqr(sqr(a*1.0*a)));
writeln(c);
readln;
end.
gauname rezultatą
5.55555560555633
⋅
10
160
{\displaystyle 5.55555560555633\cdot 10^{160}}
∫
1
10
9
x
17
d
x
=
x
18
18
|
1
10
9
=
(
10
9
)
18
18
−
1
18
18
=
10
162
18
−
1
18
=
{\displaystyle \int _{1}^{10^{9}}x^{17}\;dx={\frac {x^{18}}{18}}|_{1}^{10^{9}}={\frac {(10^{9})^{18}}{18}}-{\frac {1^{18}}{18}}={\frac {10^{162}}{18}}-{\frac {1}{18}}=}
=
5.5555555555555555555555555555556
⋅
10
160
−
0.05555555555555555555555555555556
=
5.5555555555555555555555555555556
⋅
10
160
{\displaystyle =5.5555555555555555555555555555556\cdot 10^{160}-0.05555555555555555555555555555556=5.5555555555555555555555555555556\cdot 10^{160}}
Su Gryno formule apskaičiuosime kreivinį integralą
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
,
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy,}
L - apskritimas
x
2
+
y
2
=
R
2
.
{\displaystyle x^{2}+y^{2}=R^{2}.}
Funkcijos
P
(
x
,
y
)
=
x
−
y
,
Q
(
x
,
y
)
=
x
+
y
{\displaystyle P(x,y)=x-y,\;Q(x,y)=x+y}
∂
P
∂
y
=
−
1
,
∂
Q
∂
x
=
1
{\displaystyle {\partial P \over \partial y}=-1,\;{\partial Q \over \partial x}=1}
x
2
+
y
2
=
R
2
.
{\displaystyle x^{2}+y^{2}=R^{2}.}
ρ
2
=
R
2
,
{\displaystyle \rho ^{2}=R^{2},}
ρ
=
R
{\displaystyle \rho =R}
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
=
∬
D
[
1
−
(
−
1
)
]
d
x
d
y
=
2
∬
D
d
x
d
y
=
2
s
=
2
∫
0
2
π
d
ϕ
∫
0
R
ρ
d
ρ
=
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy=\iint _{D}[1-(-1)]dxdy=2\iint _{D}dxdy=2s=2\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho d\rho =}
=
∫
0
2
π
ρ
2
|
0
R
d
ϕ
=
R
2
∫
0
2
π
d
ϕ
=
R
2
ϕ
|
0
2
π
=
2
π
R
2
.
{\displaystyle =\int _{0}^{2\pi }\rho ^{2}|_{0}^{R}d\phi =R^{2}\int _{0}^{2\pi }d\phi =R^{2}\phi |_{0}^{2\pi }=2\pi R^{2}.}
Patikrinimas . Iš apskritimo lygties
x
2
+
y
2
=
R
2
{\displaystyle x^{2}+y^{2}=R^{2}}
y
=
R
2
−
x
2
,
{\displaystyle y={\sqrt {R^{2}-x^{2}}},}
x
=
R
2
−
y
2
.
{\displaystyle x={\sqrt {R^{2}-y^{2}}}.}
Todėl
P
(
x
,
y
)
=
x
−
y
=
x
−
R
2
−
x
2
,
{\displaystyle P(x,y)=x-y=x-{\sqrt {R^{2}-x^{2}}},}
Q
(
x
,
y
)
=
x
+
y
=
R
2
−
y
2
+
y
.
{\displaystyle Q(x,y)=x+y={\sqrt {R^{2}-y^{2}}}+y.}
Pasinaudodami internetiniu integratoriumi, gauname , kad
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
=
∫
0
R
(
x
−
R
2
−
x
2
)
d
x
+
∫
0
R
(
R
2
−
y
2
+
y
)
d
y
=
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy=\int _{0}^{R}(x-{\sqrt {R^{2}-x^{2}}})dx+\int _{0}^{R}({\sqrt {R^{2}-y^{2}}}+y)dy=}
(
−
1
2
x
R
2
−
x
2
+
1
2
R
2
arctan
x
R
2
−
x
2
x
2
−
R
2
+
x
2
2
)
|
0
R
+
1
2
(
y
(
R
2
−
y
2
+
y
)
+
R
2
arctan
y
R
2
−
y
2
)
|
0
R
=
{\displaystyle \left(-{\frac {1}{2}}x{\sqrt {R^{2}-x^{2}}}+{\frac {1}{2}}R^{2}\arctan {\frac {x{\sqrt {R^{2}-x^{2}}}}{x^{2}-R^{2}}}+{\frac {x^{2}}{2}}\right)|_{0}^{R}+{\frac {1}{2}}\left(y({\sqrt {R^{2}-y^{2}}}+y)+R^{2}\arctan {\frac {y}{R^{2}-y^{2}}}\right)|_{0}^{R}=}
(
−
1
2
R
R
2
−
R
2
+
1
2
R
2
arctan
R
R
2
−
R
2
R
2
−
R
2
+
R
2
2
)
+
1
2
(
R
(
R
2
−
R
2
+
R
)
+
R
2
arctan
R
R
2
−
R
2
)
=
{\displaystyle \left(-{\frac {1}{2}}R{\sqrt {R^{2}-R^{2}}}+{\frac {1}{2}}R^{2}\arctan {\frac {R{\sqrt {R^{2}-R^{2}}}}{R^{2}-R^{2}}}+{\frac {R^{2}}{2}}\right)+{\frac {1}{2}}\left(R({\sqrt {R^{2}-R^{2}}}+R)+R^{2}\arctan {\frac {R}{R^{2}-R^{2}}}\right)=}
(
1
2
R
2
arctan
(
∞
)
+
R
2
2
)
+
1
2
(
R
(
0
+
R
)
+
R
2
arctan
(
∞
)
)
=
{\displaystyle \left({\frac {1}{2}}R^{2}\arctan(\infty )+{\frac {R^{2}}{2}}\right)+{\frac {1}{2}}\left(R(0+R)+R^{2}\arctan(\infty )\right)=}
(
1
2
R
2
π
+
R
2
2
)
+
1
2
(
R
2
+
R
2
π
)
=
{\displaystyle \left({\frac {1}{2}}R^{2}\pi +{\frac {R^{2}}{2}}\right)+{\frac {1}{2}}\left(R^{2}+R^{2}\pi \right)=}
R
2
π
+
R
2
=
R
2
(
π
+
1
)
.
{\displaystyle R^{2}\pi +R^{2}=R^{2}(\pi +1).}
Riba
lim
x
→
R
arctan
x
R
2
−
x
2
x
2
−
R
2
=
lim
x
→
R
arctan
x
x
2
−
R
2
=
lim
z
→
0
arctan
R
z
=
arctan
(
∞
)
=
π
.
{\displaystyle \lim _{x\to R}\arctan {\frac {x{\sqrt {R^{2}-x^{2}}}}{x^{2}-R^{2}}}=\lim _{x\to R}\arctan {\frac {x}{\sqrt {x^{2}-R^{2}}}}=\lim _{z\to 0}\arctan {\frac {R}{z}}=\arctan(\infty )=\pi .}
arctan
(
0
)
=
0.
{\displaystyle \arctan(0)=0.}
Atsakymą
R
2
(
π
+
1
)
{\displaystyle R^{2}(\pi +1)}
2
R
2
(
π
+
1
)
,
{\displaystyle 2R^{2}(\pi +1),}
2
π
R
2
,
{\displaystyle 2\pi R^{2},}
R yra didelis.
Patikrinimas nesuveikė.
Su Gryno formule apskaičiuosime kreivinį integralą
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
,
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy,}
L - apskritimas
x
2
+
y
2
=
R
2
.
{\displaystyle x^{2}+y^{2}=R^{2}.}
Funkcijos
P
(
x
,
y
)
=
x
−
y
,
Q
(
x
,
y
)
=
x
+
y
{\displaystyle P(x,y)=x-y,\;Q(x,y)=x+y}
∂
P
∂
y
=
−
1
,
∂
Q
∂
x
=
1
{\displaystyle {\partial P \over \partial y}=-1,\;{\partial Q \over \partial x}=1}
x
2
+
y
2
=
R
2
.
{\displaystyle x^{2}+y^{2}=R^{2}.}
ρ
2
=
R
2
,
{\displaystyle \rho ^{2}=R^{2},}
ρ
=
R
{\displaystyle \rho =R}
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
=
∬
D
[
1
−
(
−
1
)
]
d
x
d
y
=
2
∬
D
d
x
d
y
=
2
s
=
2
∫
0
2
π
d
ϕ
∫
0
R
ρ
d
ρ
=
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy=\iint _{D}[1-(-1)]dxdy=2\iint _{D}dxdy=2s=2\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho d\rho =}
=
∫
0
2
π
ρ
2
|
0
R
d
ϕ
=
R
2
∫
0
2
π
d
ϕ
=
R
2
ϕ
|
0
2
π
=
2
π
R
2
.
{\displaystyle =\int _{0}^{2\pi }\rho ^{2}|_{0}^{R}d\phi =R^{2}\int _{0}^{2\pi }d\phi =R^{2}\phi |_{0}^{2\pi }=2\pi R^{2}.}
Patikrinimas . Iš apskritimo lygties
x
2
+
y
2
=
R
2
{\displaystyle x^{2}+y^{2}=R^{2}}
y
=
R
2
−
x
2
,
{\displaystyle y={\sqrt {R^{2}-x^{2}}},}
x
=
R
2
−
y
2
.
{\displaystyle x={\sqrt {R^{2}-y^{2}}}.}
Todėl
P
(
x
,
y
)
=
x
−
y
=
x
−
R
2
−
x
2
,
{\displaystyle P(x,y)=x-y=x-{\sqrt {R^{2}-x^{2}}},}
Q
(
x
,
y
)
=
x
+
y
=
R
2
−
y
2
+
y
.
{\displaystyle Q(x,y)=x+y={\sqrt {R^{2}-y^{2}}}+y.}
Pasinaudodami internetiniu integratoriumi, gauname , kad
∮
L
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
=
∫
−
R
R
(
x
−
R
2
−
x
2
)
d
x
+
∫
−
R
R
(
R
2
−
y
2
+
y
)
d
y
=
{\displaystyle \oint _{L}(x-y)dx+(x+y)dy=\int _{-R}^{R}(x-{\sqrt {R^{2}-x^{2}}})dx+\int _{-R}^{R}({\sqrt {R^{2}-y^{2}}}+y)dy=}
(
−
1
2
x
R
2
−
x
2
+
1
2
R
2
arctan
x
R
2
−
x
2
x
2
−
R
2
+
x
2
2
)
|
−
R
R
+
1
2
(
y
(
R
2
−
y
2
+
y
)
+
R
2
arctan
y
R
2
−
y
2
)
|
−
R
R
=
{\displaystyle \left(-{\frac {1}{2}}x{\sqrt {R^{2}-x^{2}}}+{\frac {1}{2}}R^{2}\arctan {\frac {x{\sqrt {R^{2}-x^{2}}}}{x^{2}-R^{2}}}+{\frac {x^{2}}{2}}\right)|_{-R}^{R}+{\frac {1}{2}}\left(y({\sqrt {R^{2}-y^{2}}}+y)+R^{2}\arctan {\frac {y}{R^{2}-y^{2}}}\right)|_{-R}^{R}=}
(
−
1
2
R
R
2
−
R
2
+
1
2
R
2
arctan
R
R
2
−
R
2
R
2
−
R
2
+
R
2
2
)
−
(
−
1
2
(
−
R
)
R
2
−
(
−
R
)
2
+
1
2
R
2
arctan
−
R
R
2
−
(
−
R
)
2
(
−
R
)
2
−
R
2
+
(
−
R
)
2
2
)
+
{\displaystyle \left(-{\frac {1}{2}}R{\sqrt {R^{2}-R^{2}}}+{\frac {1}{2}}R^{2}\arctan {\frac {R{\sqrt {R^{2}-R^{2}}}}{R^{2}-R^{2}}}+{\frac {R^{2}}{2}}\right)-\left(-{\frac {1}{2}}(-R){\sqrt {R^{2}-(-R)^{2}}}+{\frac {1}{2}}R^{2}\arctan {\frac {-R{\sqrt {R^{2}-(-R)^{2}}}}{(-R)^{2}-R^{2}}}+{\frac {(-R)^{2}}{2}}\right)+}
+
1
2
(
R
(
R
2
−
R
2
+
R
)
+
R
2
arctan
R
R
2
−
R
2
)
−
1
2
(
−
R
(
R
2
−
(
−
R
)
2
−
R
)
+
R
2
arctan
−
R
R
2
−
(
−
R
)
2
)
=
{\displaystyle +{\frac {1}{2}}\left(R({\sqrt {R^{2}-R^{2}}}+R)+R^{2}\arctan {\frac {R}{R^{2}-R^{2}}}\right)-{\frac {1}{2}}\left(-R({\sqrt {R^{2}-(-R)^{2}}}-R)+R^{2}\arctan {\frac {-R}{R^{2}-(-R)^{2}}}\right)=}
(
1
2
R
2
arctan
(
∞
)
+
R
2
2
)
−
(
1
2
R
2
arctan
(
−
∞
)
+
R
2
2
)
+
{\displaystyle \left({\frac {1}{2}}R^{2}\arctan(\infty )+{\frac {R^{2}}{2}}\right)-\left({\frac {1}{2}}R^{2}\arctan(-\infty )+{\frac {R^{2}}{2}}\right)+}
+
1
2
(
R
(
0
+
R
)
+
R
2
arctan
(
∞
)
)
−
1
2
(
−
R
(
0
−
R
)
+
R
2
arctan
(
−
∞
)
)
=
{\displaystyle +{\frac {1}{2}}\left(R(0+R)+R^{2}\arctan(\infty )\right)-{\frac {1}{2}}\left(-R(0-R)+R^{2}\arctan(-\infty )\right)=}
=
(
1
2
R
2
π
+
R
2
2
)
−
(
1
2
R
2
(
−
π
)
+
R
2
2
)
+
{\displaystyle =\left({\frac {1}{2}}R^{2}\pi +{\frac {R^{2}}{2}}\right)-\left({\frac {1}{2}}R^{2}(-\pi )+{\frac {R^{2}}{2}}\right)+}
+
1
2
(
R
2
+
R
2
π
)
−
1
2
(
R
2
+
R
2
(
−
π
)
)
=
{\displaystyle +{\frac {1}{2}}\left(R^{2}+R^{2}\pi \right)-{\frac {1}{2}}\left(R^{2}+R^{2}(-\pi )\right)=}
=
R
2
π
+
R
2
π
=
2
π
R
2
.
{\displaystyle =R^{2}\pi +R^{2}\pi =2\pi R^{2}.}
Riba
lim
x
→
R
arctan
x
R
2
−
x
2
x
2
−
R
2
=
lim
x
→
R
arctan
x
x
2
−
R
2
=
lim
z
→
0
arctan
R
z
=
arctan
(
∞
)
=
π
.
{\displaystyle \lim _{x\to R}\arctan {\frac {x{\sqrt {R^{2}-x^{2}}}}{x^{2}-R^{2}}}=\lim _{x\to R}\arctan {\frac {x}{\sqrt {x^{2}-R^{2}}}}=\lim _{z\to 0}\arctan {\frac {R}{z}}=\arctan(\infty )=\pi .}
arctan
(
0
)
=
0
,
arctan
(
−
∞
)
=
−
π
.
{\displaystyle \arctan(0)=0,\;\arctan(-\infty )=-\pi .}
![{\displaystyle =-({\frac {34x^{2}}{25\cdot 2}}-{\frac {9}{5}}x)|_{0}^{5}=-[({\frac {17\cdot 5^{2}}{25}}-{\frac {9}{5}}\cdot 5)-({\frac {17\cdot 0^{2}}{25}}-{\frac {9}{5}}\cdot 0)]=-(17-9)=-8.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e6e118d84ff8de5fd58ec7cedaba1fa788dd7d3)
![{\displaystyle =[(0.68x-1.8)x]|_{0}^{5}=(0.68\cdot 5-1.8)\cdot 5=(3.4-1.8)\cdot 5=1.6\cdot 5=8.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b32eb4056b29824208af261acd667dd509a52c3b)
![{\displaystyle m=\int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22e88773ce0d7fc6b0c1dca842c7a19b69d5dd6c)
![{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {20.23112447{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {23.59334271}{\sqrt {34}}}={\frac {20.23112447}{5}}=4.046224894.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/beb94512dace09eb2d94b0fe38cb98a7096200e0)
![{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.1638869757{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.1638869757}{5}}=3.832777395.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b91b79576bd4e04baacb9a910342780177a78739)
![{\displaystyle m={\frac {\int _{0}^{5}\gamma dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.1638869757}{\sqrt {34}}}=3.286579502.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c33a1f52cc77b42e1f8eef3dc7a1c4a0f866ebb)
![{\displaystyle {\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2e796a3bda7d0d8e0dbbe95a9dc48d27fb2b061)
![{\displaystyle x_{C}=m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {16.43289751{\sqrt {\frac {34}{25}}}}{\sqrt {34}}}={\frac {19.16388697}{\sqrt {34}}}={\frac {16.43289751}{5}}=3.286579502.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8af10676d3b7b448f267abe93287933b71f119d8)
![{\displaystyle \int _{L}\gamma {\sqrt {1+[y']^{2}}}\;dx=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}\;dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a41c71c25ef25086cc4954765fda3aa7b1cc6c3b)
![{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {1+\left(-{\frac {3}{5}}\right)^{2}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {1+{\frac {9}{25}}}}dx}}={\frac {19.16388698}{\int _{0}^{5}{\sqrt {\frac {34}{25}}}dx}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7937a5ec3e84eee07bb9f7be9c539634cd4cb463)
![{\displaystyle m={\frac {\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[y']^{2}}}dx}}={\frac {\int _{0}^{10}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[(x^{2})']^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[(x^{2})']^{2}}}dx}}={\frac {\int _{0}^{10}{\sqrt {x^{2}+(x^{2})^{2}}}{\sqrt {1+[2x]^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[2x]^{2}}}dx}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a5dbe7e12f87ce9a20a60ff6e5ccf403a599091)
![{\displaystyle m={\frac {\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{10}{\sqrt {1+[y']^{2}}}dx}}={\frac {\int _{0}^{10}x{\sqrt {1+(2x)^{2}}}dx}{\int _{0}^{10}{\sqrt {1+(2x)^{2}}}dx}}={\frac {\int _{0}^{10}x{\sqrt {1+4x^{2}}}dx}{\int _{0}^{10}{\sqrt {1+4x^{2}}}dx}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92e1c559ac8541e1eb60bef5bc69e2a842285297)
![{\displaystyle m={\frac {\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx}{\int _{0}^{5}{\sqrt {1+[y']^{2}}}dx}}={\frac {\int _{0}^{5}x{\sqrt {1+4x^{2}}}dx}{\int _{0}^{5}{\sqrt {1+4x^{2}}}dx}}={\frac {2\int _{0}^{5}x{\sqrt {{\frac {1}{4}}+x^{2}}}dx}{2\int _{0}^{5}{\sqrt {{\frac {1}{4}}+x^{2}}}dx}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ca76a1b5afc1ba4e848105d9d39acf9485bb57f)
![{\displaystyle m=\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx=\int _{0}^{10}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[(x^{2})']^{2}}}dx=\int _{0}^{10}{\sqrt {x^{2}+(x^{2})^{2}}}{\sqrt {1+[2x]^{2}}}dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4fc2cb836c3705b062c2bfac552a141c0cf9e8a)
![{\displaystyle m=\int _{0}^{5}{\sqrt {x^{2}+y^{2}}}{\sqrt {1+[y']^{2}}}dx=\int _{0}^{5}{\sqrt {x^{2}+(x^{2})^{2}}}{\sqrt {1+(2x)^{2}}}dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbe481bd7f13080ddd7035c7ff58392c14db3cb7)
![{\displaystyle m=\int _{0}^{10}\gamma {\sqrt {1+[y']^{2}}}dx=\int _{0}^{10}x{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{10}x{\sqrt {1+4x^{2}}}dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b56e13098551a0566a29f08dc923fad4df5e7f06)
![{\displaystyle m=\int _{0}^{5}\gamma {\sqrt {1+[y']^{2}}}dx=\int _{0}^{5}x{\sqrt {1+4x^{2}}}dx=2\int _{0}^{5}x{\sqrt {{\frac {1}{4}}+x^{2}}}dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c235d2f0757c3838de8c29cbceed5f82b81fffd7)
![{\displaystyle \oint _{L}(x-y)dx+(x+y)dy=\iint _{D}[1-(-1)]dxdy=2\iint _{D}dxdy=2s=2\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho d\rho =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db7fb9ab38401a6634b1f0a6a88616c99461a25a)
