Trilypis integralas naudojamas tūriui apskaičiuoti ir mechanikoje – tose vietose, kur dvilypio integralo savybių neužtenka greitesniam apskaičiavimui.
keisti
∭
V
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y
,
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d
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∫
a
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∫
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1
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2
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{\displaystyle \iiint _{V}f(x,y,z)dxdydz=\int _{a}^{b}dx\int _{y_{1}(x)}^{y_{2}(x)}dy\int _{z_{1}(x,y)}^{z_{2}(x,y)}f(x,y,z)dz.}
Apskaičiuosime tūrį V tetraedro , apriboto plokštumų
x
=
0
,
{\displaystyle x=0,}
y
=
0
,
{\displaystyle y=0,}
z
=
0
,
{\displaystyle z=0,}
x
+
y
+
z
=
2.
{\displaystyle x+y+z=2.}
Integravimo sritis D projektuojama į plokštumą xOy . Tūrį V iš apačios riboja plokštuma
z
=
0
,
{\displaystyle z=0,}
z
=
2
−
x
−
y
.
{\displaystyle z=2-x-y.}
V
=
∭
V
d
x
d
y
d
z
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∫
0
2
d
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∫
0
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d
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0
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0
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x
−
y
d
y
=
{\displaystyle V=\iiint _{V}dxdydz=\int _{0}^{2}dx\int _{0}^{2-x}dy\int _{0}^{2-x-y}dz=\int _{0}^{2}dx\int _{0}^{2-x}z|_{0}^{2-x-y}dy=}
=
∫
0
2
d
x
∫
0
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−
x
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2
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x
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y
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d
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2
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|
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d
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∫
0
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4
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2
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x
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1
2
(
4
−
4
x
+
x
2
)
)
d
x
=
{\displaystyle =\int _{0}^{2}dx\int _{0}^{2-x}(2-x-y)dy=\int _{0}^{2}(2y-xy-{y^{2} \over 2})|_{0}^{2-x}dx=\int _{0}^{2}(4-2x-2x+x^{2}-{1 \over 2}(4-4x+x^{2}))dx=}
=
∫
0
2
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2
−
2
x
+
x
2
2
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d
x
=
(
2
x
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x
2
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x
3
6
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|
0
2
=
4
−
4
+
8
6
=
8
6
.
{\displaystyle =\int _{0}^{2}(2-2x+{x^{2} \over 2})dx=(2x-x^{2}+{x^{3} \over 6})|_{0}^{2}=4-4+{8 \over 6}={8 \over 6}.}
Šį atsakymą galima buvo gauti naudojantis mišriąja vektorių sandauga.
V
=
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a
×
b
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⋅
c
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8.
{\displaystyle V=(a\times b)\cdot c={\begin{vmatrix}a_{x}&a_{y}&a_{z}\\b_{z}&b_{y}&b_{z}\\c_{x}&c_{y}&c_{z}\end{vmatrix}}={\begin{vmatrix}2&0&0\\0&2&0\\0&0&2\end{vmatrix}}=8.}
Gretasienio tūris yra 8. Rasime piramidės (t. y. netaisyklingo tetraedro) su 4 viršūnėmis, kurios pagrindas yra trikampis, tūrį:
V
=
1
6
|
(
a
×
b
)
⋅
c
|
=
8
6
.
{\displaystyle V={\frac {1}{6}}|(a\times b)\cdot c|={\frac {8}{6}}.}
∭
V
d
x
d
y
d
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∫
−
1
1
d
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∫
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1
d
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d
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1
2
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1
1
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2
x
|
−
1
1
=
2
(
1
−
(
−
1
)
)
=
4.
{\displaystyle \iiint _{V}dxdydz=\int _{-1}^{1}dx\int _{0}^{1}dy\int _{0}^{2}dz=\int _{-1}^{1}dx\int _{0}^{1}2dy=\int _{-1}^{1}2dx=2x|_{-1}^{1}=2(1-(-1))=4.}
M (1-(-1); 1-0; 2-0)=M (2; 1; 2). Sudauginus kraštinių ilgius gauname stačiakampio gretasienio tūrį
V
=
2
⋅
1
⋅
2
=
4.
{\displaystyle V=2\cdot 1\cdot 2=4.}
V
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a
×
b
)
⋅
c
=
|
2
0
0
0
1
0
0
0
2
|
=
4.
{\displaystyle V=(a\times b)\cdot c={\begin{vmatrix}2&0&0\\0&1&0\\0&0&2\end{vmatrix}}=4.}
Apskaičiuosime tetraedro tūrį V , apriboto plokštumomis x+y+z=2, z=1, x=0, y=0. tetraedro trys kraštinės a=b=c=1 ir lygiagrečios atitinkamai x , y ir z ašims, o kitos trys kraštinės
d
=
e
=
f
=
1
2
+
1
2
=
2
.
{\displaystyle d=e=f={\sqrt {1^{2}+1^{2}}}={\sqrt {2}}.}
V
=
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V
d
x
d
y
d
z
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∬
D
d
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d
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1
2
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∫
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=
{\displaystyle V=\iiint _{V}dxdydz=\iint _{D}dxdy\int _{1}^{2-x-y}dz=\int _{0}^{1}dx\int _{0}^{1-x}dy\int _{1}^{2-x-y}dz=}
=
∫
0
1
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∫
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1
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1
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x
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=
{\displaystyle =\int _{0}^{1}dx\int _{0}^{1-x}(1-x-y)dy=\int _{0}^{1}(y-xy-{y^{2} \over 2})|_{0}^{1-x}dx=}
=
∫
0
1
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1
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x
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x
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x
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1
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2
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∫
0
1
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1
2
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x
+
x
2
2
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d
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=
{\displaystyle =\int _{0}^{1}(1-x-x+x^{2}-{1-2x+x^{2} \over 2})dx=\int _{0}^{1}({1 \over 2}-x+{x^{2} \over 2})dx=}
=
(
1
2
x
−
x
2
2
+
x
3
6
)
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0
1
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1
2
−
1
2
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1
6
=
1
6
.
{\displaystyle =({1 \over 2}x-{x^{2} \over 2}+{x^{3} \over 6})|_{0}^{1}={1 \over 2}-{1 \over 2}+{1 \over 6}={1 \over 6}.}
M (1-0; 1-0; 2-1)=M (1; 1; 1):
V
=
1
6
|
(
a
×
b
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⋅
c
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=
1
6
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1
0
0
0
1
0
0
0
1
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=
1
6
.
{\displaystyle V={1 \over 6}|(a\times b)\cdot c|={1 \over 6}{\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}}={1 \over 6}.}
Vaizdas:Trilypis1321.jpg 13.21. Pirmajame oktante esantį kūną riboja paviršiai
z
=
4
−
y
2
,
{\displaystyle z=4-y^{2},}
x
+
y
=
2
,
{\displaystyle x+y=2,}
2
x
+
y
=
2
,
{\displaystyle 2x+y=2,}
y
=
0
,
{\displaystyle y=0,}
z
=
0
{\displaystyle z=0}
V
=
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V
d
x
d
y
d
z
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∬
D
d
x
d
y
∫
0
4
−
y
2
.
{\displaystyle V=\iiint _{V}dxdydz=\iint _{D}dxdy\int _{0}^{4-y^{2}}.}
D yra kūno projekcija plokštumoje xOy . Parinkus vienokią integravimo tvarką, dvilypis integralas šioje srityje išreiškiamas vienu kartotiniu integralu, o pakeitus tą tvarką dviem kartotiniais integralais:
∬
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{\displaystyle \iint _{D}f(x,y)dxdy=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}f(x,y)dx}
∬
D
f
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∫
1
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0
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x
,
y
)
d
y
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{\displaystyle \iint _{D}f(x,y)dxdy=\int _{0}^{1}dx\int _{2-2x}^{2-x}f(x,y)dy+\int _{1}^{2}dx\int _{0}^{2-x}f(x,y)dy.}
V
=
∫
0
2
d
y
∫
2
−
y
2
2
−
y
d
x
∫
0
4
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d
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y
(
4
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)
d
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=
{\displaystyle V=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}dx\int _{0}^{4-y^{2}}dz=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}z|_{0}^{4-y^{2}}dx=\int _{0}^{2}dy\int _{2-y \over 2}^{2-y}(4-y^{2})dx=}
=
∫
0
2
(
4
−
y
2
)
x
|
2
−
y
2
2
−
y
d
y
=
∫
0
2
(
4
−
2
y
−
y
2
+
y
3
2
)
d
y
=
(
4
y
−
y
2
−
y
3
3
+
y
4
8
)
|
0
2
=
10
3
.
{\displaystyle =\int _{0}^{2}(4-y^{2})x|_{2-y \over 2}^{2-y}dy=\int _{0}^{2}(4-2y-y^{2}+{y^{3} \over 2})dy=(4y-y^{2}-{y^{3} \over 3}+{y^{4} \over 8})|_{0}^{2}={10 \over 3}.}
Apskaičiuosime tūrį kūno apriboto šiais paviršiais:
y
=
x
,
y
=
2
x
,
{\displaystyle y={\sqrt {x}},\;y=2{\sqrt {x}},}
z
=
0
{\displaystyle z=0}
x
+
z
=
6.
{\displaystyle x+z=6.}
x
+
z
=
6
,
{\displaystyle x+z=6,}
z lygi nuliui
x
=
6.
{\displaystyle x=6.}
x
=
6
{\displaystyle x=6}
y
=
6
{\displaystyle y={\sqrt {6}}}
y
=
2
6
.
{\displaystyle y=2{\sqrt {6}}.}
V
=
∫
0
6
d
x
∫
x
2
x
d
y
∫
0
6
−
x
d
z
=
∫
0
6
d
x
∫
x
2
x
(
6
−
x
)
d
y
=
{\displaystyle V=\int _{0}^{6}dx\int _{\sqrt {x}}^{2{\sqrt {x}}}dy\int _{0}^{6-x}dz=\int _{0}^{6}dx\int _{\sqrt {x}}^{2{\sqrt {x}}}(6-x)dy=}
=
∫
0
6
(
6
−
x
)
x
d
x
=
(
6
⋅
x
3
2
3
2
−
2
5
⋅
x
5
2
)
|
0
6
=
4
⋅
6
6
−
2
5
⋅
6
2
6
=
24
6
−
72
5
6
=
48
6
5
.
{\displaystyle =\int _{0}^{6}(6-x){\sqrt {x}}dx=(6\cdot {x^{3 \over 2} \over {3 \over 2}}-{2 \over 5}\cdot x^{5 \over 2})|_{0}^{6}=4\cdot 6{\sqrt {6}}-{2 \over 5}\cdot 6^{2}{\sqrt {6}}=24{\sqrt {6}}-{72 \over 5}{\sqrt {6}}={48{\sqrt {6}} \over 5}.}
Vaizdas:Integral379380.jpg 379. Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
y
=
x
2
{\displaystyle y=x^{2}}
xOy ),
y
=
1
,
{\displaystyle y=1,}
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
V
=
∫
0
1
d
x
∫
x
2
1
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
x
∫
x
2
1
d
y
z
|
0
x
2
+
y
2
=
∫
0
1
d
x
∫
x
2
1
(
(
x
2
+
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}{\mathsf {d}}y\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}{\mathsf {d}}y\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}((x^{2}+y^{2})-0){\mathsf {d}}y=}
=
∫
0
1
d
x
∫
x
2
1
(
x
2
+
y
2
)
d
y
=
∫
0
1
d
x
(
x
2
y
+
y
3
3
)
|
x
2
1
=
∫
0
1
d
x
[
(
x
2
⋅
1
+
1
3
3
)
−
(
x
2
⋅
x
2
+
(
x
2
)
3
3
)
]
=
{\displaystyle =\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}(x^{2}+y^{2}){\mathsf {d}}y=\int _{0}^{1}{\mathsf {d}}x\;(x^{2}y+{\frac {y^{3}}{3}})|_{x^{2}}^{1}=\int _{0}^{1}{\mathsf {d}}x\;[(x^{2}\cdot 1+{\frac {1^{3}}{3}})-(x^{2}\cdot x^{2}+{\frac {(x^{2})^{3}}{3}})]=}
=
∫
0
1
[
x
2
+
1
3
−
x
4
−
x
6
3
]
d
x
=
(
x
3
3
+
x
3
−
x
5
5
−
x
7
3
⋅
7
)
|
0
1
=
(
1
3
3
+
1
3
−
1
5
5
−
1
7
21
)
−
(
0
3
3
+
0
3
−
0
5
5
−
0
7
21
)
=
{\displaystyle =\int _{0}^{1}[x^{2}+{\frac {1}{3}}-x^{4}-{\frac {x^{6}}{3}}]{\mathsf {d}}x=\left({\frac {x^{3}}{3}}+{\frac {x}{3}}-{\frac {x^{5}}{5}}-{\frac {x^{7}}{3\cdot 7}}\right)|_{0}^{1}=\left({\frac {1^{3}}{3}}+{\frac {1}{3}}-{\frac {1^{5}}{5}}-{\frac {1^{7}}{21}}\right)-\left({\frac {0^{3}}{3}}+{\frac {0}{3}}-{\frac {0^{5}}{5}}-{\frac {0^{7}}{21}}\right)=}
=
(
1
3
+
1
3
−
1
5
−
1
21
)
=
(
2
3
−
1
5
−
1
21
)
=
(
2
⋅
7
−
1
21
−
1
5
)
=
(
13
21
−
1
5
)
=
13
⋅
5
−
21
21
⋅
5
=
65
−
21
105
=
44
105
=
0.419047619.
{\displaystyle =\left({\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}-{\frac {1}{21}}\right)=\left({\frac {2}{3}}-{\frac {1}{5}}-{\frac {1}{21}}\right)=\left({\frac {2\cdot 7-1}{21}}-{\frac {1}{5}}\right)=\left({\frac {13}{21}}-{\frac {1}{5}}\right)={\frac {13\cdot 5-21}{21\cdot 5}}={\frac {65-21}{105}}={\frac {44}{105}}=0.419047619.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Autoriaus manymu, tikrasis tūris gali buti apskaičiuotas (o kad geriau suprasti kaip apskaičiuoti, reikėtų įsigilinti į sukimo tūrio radimą) taip:
V
=
∫
0
1
y
⋅
y
2
d
y
=
y
1
2
+
2
d
y
=
∫
0
1
y
5
2
d
y
=
y
5
2
+
1
5
2
+
1
|
0
1
=
y
7
2
7
2
|
0
1
=
2
⋅
y
7
2
7
|
0
1
=
2
⋅
1
7
2
7
=
2
7
=
0.285714285
,
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}dy=y^{{\frac {1}{2}}+2}dy=\int _{0}^{1}y^{5 \over 2}dy={\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}|_{0}^{1}={\frac {y^{\frac {7}{2}}}{\frac {7}{2}}}|_{0}^{1}={\frac {2\cdot y^{\frac {7}{2}}}{7}}|_{0}^{1}={\frac {2\cdot 1^{\frac {7}{2}}}{7}}={\frac {2}{7}}=0.285714285,}
arba galbūt net taip:
V
=
∫
0
1
y
⋅
y
2
⋅
y
d
y
,
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}\cdot y\;dy,}
V
=
∫
0
1
y
⋅
y
2
⋅
y
d
y
.
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}\cdot {\sqrt {y}}\;dy.}
x
2
10000
+
y
2
=
z
{\displaystyle {\frac {x^{2}}{10000}}+y^{2}=z}
x turi būti 100, kai y =0, kad z būtų lygus 1), pakeitimu, šiame uždavinyje, tūris turėtų būti
V
=
∫
0
1
y
⋅
y
2
d
y
.
{\displaystyle V=\int _{0}^{1}{\sqrt {y}}\cdot y^{2}\;dy.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
y
=
x
2
{\displaystyle y=x^{2}}
y
=
1
,
{\displaystyle y=1,}
z
=
0
{\displaystyle z=0}
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
V
=
∫
0
1
d
y
∫
0
y
d
x
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
y
∫
0
y
d
x
z
|
0
x
2
+
y
2
=
∫
0
1
d
y
∫
0
y
(
(
x
2
+
y
2
)
−
0
)
d
x
=
{\displaystyle V=\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}((x^{2}+y^{2})-0){\mathsf {d}}x=}
=
∫
0
1
d
y
∫
0
y
(
x
2
+
y
2
)
d
x
=
∫
0
1
d
y
(
x
3
3
+
y
2
x
)
|
0
y
=
∫
0
1
d
y
[
(
(
y
)
3
3
+
y
2
y
)
−
(
0
3
3
+
y
2
⋅
0
)
]
=
{\displaystyle =\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}(x^{2}+y^{2}){\mathsf {d}}x=\int _{0}^{1}{\mathsf {d}}y\;({\frac {x^{3}}{3}}+y^{2}x)|_{0}^{\sqrt {y}}=\int _{0}^{1}{\mathsf {d}}y\;[({\frac {({\sqrt {y}})^{3}}{3}}+y^{2}{\sqrt {y}})-({\frac {0^{3}}{3}}+y^{2}\cdot 0)]=}
=
∫
0
1
[
y
3
2
3
+
y
5
2
]
d
y
=
(
y
3
2
+
1
3
(
3
2
+
1
)
+
y
5
2
+
1
5
2
+
1
)
|
0
1
=
(
y
5
2
3
⋅
5
2
+
y
7
2
5
+
2
2
)
|
0
1
=
(
2
y
5
2
15
+
2
y
7
2
7
)
|
0
1
=
{\displaystyle =\int _{0}^{1}[{\frac {y^{3 \over 2}}{3}}+y^{5 \over 2}]{\mathsf {d}}y=\left({\frac {y^{{\frac {3}{2}}+1}}{3({\frac {3}{2}}+1)}}+{\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}\right)|_{0}^{1}=\left({\frac {y^{\frac {5}{2}}}{3\cdot {\frac {5}{2}}}}+{\frac {y^{\frac {7}{2}}}{\frac {5+2}{2}}}\right)|_{0}^{1}=\left({\frac {2y^{\frac {5}{2}}}{15}}+{\frac {2y^{\frac {7}{2}}}{7}}\right)|_{0}^{1}=}
=
(
2
⋅
1
5
2
15
+
2
⋅
1
7
2
7
)
−
(
2
⋅
0
5
2
15
+
2
⋅
0
7
2
7
)
=
2
15
+
2
7
=
2
⋅
7
+
2
⋅
15
15
⋅
7
=
14
+
30
105
=
44
105
=
0.419047619.
{\displaystyle =\left({\frac {2\cdot 1^{\frac {5}{2}}}{15}}+{\frac {2\cdot 1^{\frac {7}{2}}}{7}}\right)-\left({\frac {2\cdot 0^{\frac {5}{2}}}{15}}+{\frac {2\cdot 0^{\frac {7}{2}}}{7}}\right)={\frac {2}{15}}+{\frac {2}{7}}={\frac {2\cdot 7+2\cdot 15}{15\cdot 7}}={\frac {14+30}{105}}={\frac {44}{105}}=0.419047619.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
y
=
x
2
{\displaystyle y=x^{2}}
xOy ),
y
=
9
,
{\displaystyle y=9,}
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
V
=
∫
0
3
d
x
∫
x
2
9
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
3
d
x
∫
x
2
9
d
y
z
|
0
x
2
+
y
2
=
∫
0
3
d
x
∫
x
2
9
(
(
x
2
+
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}{\mathsf {d}}y\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}{\mathsf {d}}y\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}((x^{2}+y^{2})-0){\mathsf {d}}y=}
=
∫
0
3
d
x
∫
x
2
9
(
x
2
+
y
2
)
d
y
=
∫
0
3
d
x
(
x
2
y
+
y
3
3
)
|
x
2
9
=
∫
0
1
d
x
[
(
x
2
⋅
9
+
9
3
3
)
−
(
x
2
⋅
x
2
+
(
x
2
)
3
3
)
]
=
{\displaystyle =\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}(x^{2}+y^{2}){\mathsf {d}}y=\int _{0}^{3}{\mathsf {d}}x\;(x^{2}y+{\frac {y^{3}}{3}})|_{x^{2}}^{9}=\int _{0}^{1}{\mathsf {d}}x\;[(x^{2}\cdot 9+{\frac {9^{3}}{3}})-(x^{2}\cdot x^{2}+{\frac {(x^{2})^{3}}{3}})]=}
=
∫
0
3
[
9
x
2
+
729
3
−
x
4
−
x
6
3
]
d
x
=
(
9
x
3
3
+
729
x
3
−
x
5
5
−
x
7
3
⋅
7
)
|
0
3
=
(
9
⋅
3
3
3
+
729
⋅
3
3
−
3
5
5
−
3
7
21
)
−
(
9
⋅
0
3
3
+
729
⋅
0
3
−
0
5
5
−
0
7
21
)
=
{\displaystyle =\int _{0}^{3}[9x^{2}+{\frac {729}{3}}-x^{4}-{\frac {x^{6}}{3}}]{\mathsf {d}}x=\left({\frac {9x^{3}}{3}}+{\frac {729x}{3}}-{\frac {x^{5}}{5}}-{\frac {x^{7}}{3\cdot 7}}\right)|_{0}^{3}=\left({\frac {9\cdot 3^{3}}{3}}+{\frac {729\cdot 3}{3}}-{\frac {3^{5}}{5}}-{\frac {3^{7}}{21}}\right)-\left({\frac {9\cdot 0^{3}}{3}}+{\frac {729\cdot 0}{3}}-{\frac {0^{5}}{5}}-{\frac {0^{7}}{21}}\right)=}
=
(
9
⋅
9
+
729
−
243
5
−
2187
21
)
=
81
+
729
−
243
5
−
729
7
=
810
−
243
5
−
729
7
=
810
⋅
5
⋅
7
−
243
⋅
7
−
729
⋅
5
5
⋅
7
=
{\displaystyle =\left(9\cdot 9+729-{\frac {243}{5}}-{\frac {2187}{21}}\right)=81+729-{\frac {243}{5}}-{\frac {729}{7}}=810-{\frac {243}{5}}-{\frac {729}{7}}={\frac {810\cdot 5\cdot 7-243\cdot 7-729\cdot 5}{5\cdot 7}}=}
=
28350
−
1701
−
3645
35
=
23004
35
=
657.2571429.
{\displaystyle ={\frac {28350-1701-3645}{35}}={\frac {23004}{35}}=657.2571429.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
y
=
x
2
{\displaystyle y=x^{2}}
y
=
9
,
{\displaystyle y=9,}
z
=
0
{\displaystyle z=0}
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
V
=
∫
0
9
d
y
∫
0
y
d
x
∫
0
x
2
+
y
2
d
z
=
∫
0
9
d
y
∫
0
y
d
x
z
|
0
x
2
+
y
2
=
∫
0
9
d
y
∫
0
y
(
(
x
2
+
y
2
)
−
0
)
d
x
=
{\displaystyle V=\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\int _{0}^{x^{2}+y^{2}}{\mathsf {d}}z=\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}{\mathsf {d}}x\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}((x^{2}+y^{2})-0){\mathsf {d}}x=}
=
∫
0
9
d
y
∫
0
y
(
x
2
+
y
2
)
d
x
=
∫
0
9
d
y
(
x
3
3
+
y
2
x
)
|
0
y
=
∫
0
9
d
y
[
(
(
y
)
3
3
+
y
2
y
)
−
(
0
3
3
+
y
2
⋅
0
)
]
=
{\displaystyle =\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}(x^{2}+y^{2}){\mathsf {d}}x=\int _{0}^{9}{\mathsf {d}}y\;({\frac {x^{3}}{3}}+y^{2}x)|_{0}^{\sqrt {y}}=\int _{0}^{9}{\mathsf {d}}y\;[({\frac {({\sqrt {y}})^{3}}{3}}+y^{2}{\sqrt {y}})-({\frac {0^{3}}{3}}+y^{2}\cdot 0)]=}
=
∫
0
9
[
y
3
2
3
+
y
5
2
]
d
y
=
(
y
3
2
+
1
3
(
3
2
+
1
)
+
y
5
2
+
1
5
2
+
1
)
|
0
9
=
(
y
5
2
3
⋅
5
2
+
y
7
2
5
+
2
2
)
|
0
9
=
(
2
y
5
2
15
+
2
y
7
2
7
)
|
0
9
=
{\displaystyle =\int _{0}^{9}[{\frac {y^{3 \over 2}}{3}}+y^{5 \over 2}]{\mathsf {d}}y=\left({\frac {y^{{\frac {3}{2}}+1}}{3({\frac {3}{2}}+1)}}+{\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}\right)|_{0}^{9}=\left({\frac {y^{\frac {5}{2}}}{3\cdot {\frac {5}{2}}}}+{\frac {y^{\frac {7}{2}}}{\frac {5+2}{2}}}\right)|_{0}^{9}=\left({\frac {2y^{\frac {5}{2}}}{15}}+{\frac {2y^{\frac {7}{2}}}{7}}\right)|_{0}^{9}=}
=
(
2
⋅
9
5
2
15
+
2
⋅
9
7
2
7
)
−
(
2
⋅
0
5
2
15
+
2
⋅
0
7
2
7
)
=
2
59049
15
+
2
4782969
7
=
2
⋅
243
15
+
2
⋅
2187
7
=
486
15
+
4374
7
=
{\displaystyle =\left({\frac {2\cdot 9^{\frac {5}{2}}}{15}}+{\frac {2\cdot 9^{\frac {7}{2}}}{7}}\right)-\left({\frac {2\cdot 0^{\frac {5}{2}}}{15}}+{\frac {2\cdot 0^{\frac {7}{2}}}{7}}\right)={\frac {2{\sqrt {59049}}}{15}}+{\frac {2{\sqrt {4782969}}}{7}}={\frac {2\cdot 243}{15}}+{\frac {2\cdot 2187}{7}}={\frac {486}{15}}+{\frac {4374}{7}}=}
=
486
⋅
7
+
4374
⋅
15
15
⋅
7
=
3402
+
65610
105
=
69012
105
=
23004
35
=
657.2571429.
{\displaystyle ={\frac {486\cdot 7+4374\cdot 15}{15\cdot 7}}={\frac {3402+65610}{105}}={\frac {69012}{105}}={\frac {23004}{35}}=657.2571429.}
Kad gauti tūrį dviejuose oktantuose, reikia padauginti iš 2.
Vaizdas:Integral379380.jpg 380. Pavyzdis . Rasti kūno tūrį V , išpjaunamą iš begalinės prizmės su kraštais
x
=
±
1
,
y
=
±
1
{\displaystyle x=\pm 1,\;y=\pm 1}
x
2
+
y
2
=
4
−
z
,
{\displaystyle x^{2}+y^{2}=4-z,}
x
2
+
y
2
=
4
(
z
+
2
)
{\displaystyle x^{2}+y^{2}=4(z+2)}
z
1
=
4
−
x
2
−
y
2
,
{\displaystyle z_{1}=4-x^{2}-y^{2},}
z
2
=
x
2
+
y
2
4
−
2.
{\displaystyle z_{2}={\frac {x^{2}+y^{2}}{4}}-2.}
x ir y yra 0, tai
z
1
=
4
{\displaystyle z_{1}=4}
z
2
=
−
2
,
{\displaystyle z_{2}=-2,}
V
=
∫
0
1
d
x
∫
0
1
d
y
∫
x
2
+
y
2
4
−
2
4
−
x
2
−
y
2
d
z
=
∫
0
1
d
x
∫
0
1
d
y
z
|
x
2
+
y
2
4
−
2
4
−
x
2
−
y
2
=
∫
0
1
d
x
∫
0
1
[
(
4
−
x
2
−
y
2
)
−
(
x
2
+
y
2
4
−
2
)
]
d
y
=
{\displaystyle V=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}{\mathsf {d}}y\int _{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}{\mathsf {d}}z=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}{\mathsf {d}}y\;z|_{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}[(4-x^{2}-y^{2})-({\frac {x^{2}+y^{2}}{4}}-2)]{\mathsf {d}}y=}
=
∫
0
1
d
x
∫
0
1
(
6
−
x
2
−
y
2
−
x
2
+
y
2
4
)
d
y
=
∫
0
1
d
x
(
6
y
−
x
2
y
−
y
3
3
−
x
2
y
4
−
y
3
4
⋅
3
)
|
0
1
=
∫
0
1
(
6
⋅
1
−
x
2
⋅
1
−
1
3
3
−
x
2
⋅
1
4
−
1
3
12
)
d
x
=
{\displaystyle =\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}(6-x^{2}-y^{2}-{\frac {x^{2}+y^{2}}{4}}){\mathsf {d}}y=\int _{0}^{1}{\mathsf {d}}x(6y-x^{2}y-{\frac {y^{3}}{3}}-{\frac {x^{2}y}{4}}-{\frac {y^{3}}{4\cdot 3}})|_{0}^{1}=\int _{0}^{1}(6\cdot 1-x^{2}\cdot 1-{\frac {1^{3}}{3}}-{\frac {x^{2}\cdot 1}{4}}-{\frac {1^{3}}{12}}){\mathsf {d}}x=}
=
∫
0
1
(
6
−
x
2
−
1
3
−
x
2
4
−
1
12
)
d
x
=
∫
0
1
(
6
⋅
12
−
1
⋅
4
−
1
12
−
x
2
−
x
2
4
)
d
x
=
∫
0
1
(
72
−
4
−
1
12
−
x
2
−
x
2
4
)
d
x
=
{\displaystyle =\int _{0}^{1}(6-x^{2}-{\frac {1}{3}}-{\frac {x^{2}}{4}}-{\frac {1}{12}}){\mathsf {d}}x=\int _{0}^{1}({\frac {6\cdot 12-1\cdot 4-1}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=\int _{0}^{1}({\frac {72-4-1}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=}
=
∫
0
1
(
67
12
−
x
2
−
x
2
4
)
d
x
=
(
67
x
12
−
x
3
3
−
x
3
4
⋅
3
)
|
0
1
=
67
⋅
1
12
−
1
3
3
−
1
3
12
=
67
−
4
−
1
12
=
67
−
5
12
=
62
12
=
31
6
=
5.166666667.
{\displaystyle =\int _{0}^{1}({\frac {67}{12}}-x^{2}-{\frac {x^{2}}{4}}){\mathsf {d}}x=({\frac {67x}{12}}-{\frac {x^{3}}{3}}-{\frac {x^{3}}{4\cdot 3}})|_{0}^{1}={\frac {67\cdot 1}{12}}-{\frac {1^{3}}{3}}-{\frac {1^{3}}{12}}={\frac {67-4-1}{12}}={\frac {67-5}{12}}={\frac {62}{12}}={\frac {31}{6}}=5.166666667.}
Kad gauti tūrį visuose 8-iuose oktantuose, reikia
31
6
{\displaystyle {\frac {31}{6}}}
Palyginimui, stačiakampio gretasienio tūris, kurio kraštinės a=1, b=1, c=6 yra lygus
V
b
i
g
=
1
⋅
1
⋅
6
=
6.
{\displaystyle V_{big}=1\cdot 1\cdot 6=6.}
Paraboloidas. Pavyzdis . Kūną riboja plokštuma xOy , cilindrinis paviršius
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1}
z
=
x
2
+
y
2
.
{\displaystyle z=x^{2}+y^{2}.}
z
=
1
{\displaystyle z=1}
Sprendimas . Kadangi kūnas yra simetriškas koordinačių plokštumų xOz ir yOz atžvilgiu, tai apskaičiuosime tik jo ketvirtadalio, esančio pirmajame oktante tūrį. Taigi
V
=
∫
0
1
d
x
∫
0
1
−
x
2
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
x
∫
0
1
−
x
2
(
x
2
+
y
2
)
d
y
=
∫
0
1
d
x
(
x
2
y
+
y
3
3
)
|
0
1
−
x
2
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{\sqrt {1-x^{2}}}dy\int _{0}^{x^{2}+y^{2}}dz=\int _{0}^{1}dx\int _{0}^{\sqrt {1-x^{2}}}(x^{2}+y^{2})dy=\int _{0}^{1}dx(x^{2}y+{\frac {y^{3}}{3}})|_{0}^{\sqrt {1-x^{2}}}=}
=
∫
0
1
(
x
2
1
−
x
2
+
(
1
−
x
2
)
3
3
)
d
x
=
∫
0
π
2
(
sin
2
(
t
)
1
−
sin
2
t
+
(
1
−
sin
2
t
)
3
3
)
cos
(
t
)
d
t
=
{\displaystyle =\int _{0}^{1}(x^{2}{\sqrt {1-x^{2}}}+{\frac {({\sqrt {1-x^{2}}})^{3}}{3}})dx=\int _{0}^{\pi \over 2}(\sin ^{2}(t){\sqrt {1-\sin ^{2}t}}+{\frac {({\sqrt {1-\sin ^{2}t}})^{3}}{3}})\cos(t)\;dt=}
=
∫
0
π
2
(
sin
2
(
t
)
cos
2
t
+
(
cos
2
t
)
3
3
)
cos
(
t
)
d
t
=
∫
0
π
2
(
sin
2
(
t
)
cos
t
+
1
3
cos
3
t
)
cos
(
t
)
d
t
=
∫
0
π
2
(
sin
2
(
t
)
cos
2
t
+
1
3
cos
4
t
)
d
t
=
{\displaystyle =\int _{0}^{\pi \over 2}(\sin ^{2}(t){\sqrt {\cos ^{2}t}}+{\frac {({\sqrt {\cos ^{2}t}})^{3}}{3}})\cos(t)\;dt=\int _{0}^{\pi \over 2}(\sin ^{2}(t)\cos t+{\frac {1}{3}}\cos ^{3}t)\cos(t)\;dt=\int _{0}^{\pi \over 2}(\sin ^{2}(t)\cos ^{2}t+{\frac {1}{3}}\cos ^{4}t)\;dt=}
=
∫
0
π
2
(
(
1
−
cos
2
t
)
cos
2
t
+
1
3
cos
4
t
)
d
t
=
∫
0
π
2
(
cos
2
t
−
cos
4
t
+
1
3
cos
4
t
)
d
t
=
∫
0
π
2
(
cos
2
t
−
2
3
cos
4
t
)
d
t
=
{\displaystyle =\int _{0}^{\pi \over 2}((1-\cos ^{2}t)\cos ^{2}t+{\frac {1}{3}}\cos ^{4}t)\;dt=\int _{0}^{\pi \over 2}(\cos ^{2}t-\cos ^{4}t+{\frac {1}{3}}\cos ^{4}t)\;dt=\int _{0}^{\pi \over 2}(\cos ^{2}t-{\frac {2}{3}}\cos ^{4}t)\;dt=}
=
∫
0
π
2
(
cos
(
2
t
)
+
1
2
−
2
3
⋅
cos
(
4
t
)
+
4
cos
(
2
t
)
+
3
8
)
d
t
=
(
1
2
sin
(
2
t
)
+
t
2
−
2
3
⋅
1
4
sin
(
4
t
)
+
2
sin
(
2
t
)
+
3
t
8
)
|
0
π
2
=
{\displaystyle =\int _{0}^{\pi \over 2}({\frac {\cos(2t)+1}{2}}-{\frac {2}{3}}\cdot {\cos(4t)+4\cos(2t)+3 \over 8})\;dt=({\frac {{\frac {1}{2}}\sin(2t)+t}{2}}-{\frac {2}{3}}\cdot {{\frac {1}{4}}\sin(4t)+2\sin(2t)+3t \over 8})|_{0}^{\pi \over 2}=}
=
(
1
2
sin
(
2
⋅
π
2
)
+
π
2
2
−
2
3
⋅
1
4
sin
(
4
⋅
π
2
)
+
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
8
)
−
(
1
2
sin
(
2
⋅
0
)
+
0
2
−
2
3
⋅
1
4
sin
(
4
⋅
0
)
+
2
sin
(
2
⋅
0
)
+
3
⋅
0
8
)
=
{\displaystyle =({\frac {{\frac {1}{2}}\sin(2\cdot {\frac {\pi }{2}})+{\frac {\pi }{2}}}{2}}-{\frac {2}{3}}\cdot {{\frac {1}{4}}\sin(4\cdot {\frac {\pi }{2}})+2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}} \over 8})-({\frac {{\frac {1}{2}}\sin(2\cdot 0)+0}{2}}-{\frac {2}{3}}\cdot {{\frac {1}{4}}\sin(4\cdot 0)+2\sin(2\cdot 0)+3\cdot 0 \over 8})=}
=
(
sin
π
+
π
4
−
2
⋅
1
4
sin
(
2
π
)
+
2
sin
(
π
)
+
3
π
2
24
)
−
(
1
2
sin
(
0
)
2
−
2
24
⋅
(
1
4
sin
(
0
)
+
2
sin
(
0
)
+
0
)
)
=
{\displaystyle =({\frac {\sin \pi +\pi }{4}}-2\cdot {{\frac {1}{4}}\sin(2\pi )+2\sin(\pi )+{\frac {3\pi }{2}} \over 24})-({\frac {{\frac {1}{2}}\sin(0)}{2}}-{\frac {2}{24}}\cdot ({\frac {1}{4}}\sin(0)+2\sin(0)+0))=}
=
(
0
+
π
4
−
2
⋅
1
4
⋅
0
+
2
⋅
0
+
3
π
2
24
)
−
0
=
π
4
−
3
π
2
12
=
π
4
−
3
π
24
=
π
4
−
π
8
=
π
8
=
0.392699081.
{\displaystyle =({\frac {0+\pi }{4}}-2\cdot {{\frac {1}{4}}\cdot 0+2\cdot 0+{\frac {3\pi }{2}} \over 24})-0={\frac {\pi }{4}}-{{\frac {3\pi }{2}} \over 12}={\frac {\pi }{4}}-{\frac {3\pi }{24}}={\frac {\pi }{4}}-{\frac {\pi }{8}}={\frac {\pi }{8}}=0.392699081.}
kur
x
=
sin
t
,
{\displaystyle x=\sin t,}
x
=
0
{\displaystyle x=0}
t
=
0
{\displaystyle t=0}
x
=
1
{\displaystyle x=1}
t
=
π
2
,
{\displaystyle t={\frac {\pi }{2}},}
d
x
=
cos
(
t
)
d
t
;
{\displaystyle dx=\cos(t)dt;}
cos
2
A
=
cos
(
2
A
)
+
1
2
,
cos
4
A
=
cos
(
4
A
)
+
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \cos ^{2}A={\frac {\cos(2A)+1}{2}},\quad \cos ^{4}A={\cos(4A)+4\cos(2A)+3 \over 8}.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
∫
0
π
2
(
cos
2
t
−
2
3
cos
4
t
)
d
t
=
(
2
−
1
)
!
!
2
!
!
⋅
π
2
−
2
3
⋅
(
4
−
1
)
!
!
4
!
!
⋅
π
2
=
1
!
!
2
!
!
⋅
π
2
−
1
3
⋅
3
!
!
4
!
!
⋅
π
=
1
2
⋅
π
2
−
1
3
⋅
3
4
⋅
2
⋅
π
=
π
4
−
1
8
⋅
π
=
π
8
.
{\displaystyle V=\int _{0}^{\pi \over 2}(\cos ^{2}t-{\frac {2}{3}}\cos ^{4}t)\;dt={\frac {(2-1)!!}{2!!}}\cdot {\frac {\pi }{2}}-{\frac {2}{3}}\cdot {\frac {(4-1)!!}{4!!}}\cdot {\frac {\pi }{2}}={\frac {1!!}{2!!}}\cdot {\frac {\pi }{2}}-{\frac {1}{3}}\cdot {\frac {3!!}{4!!}}\cdot \pi ={\frac {1}{2}}\cdot {\frac {\pi }{2}}-{\frac {1}{3}}\cdot {\frac {3}{4\cdot 2}}\cdot \pi ={\frac {\pi }{4}}-{\frac {1}{8}}\cdot \pi ={\frac {\pi }{8}}.}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš 4.
Pavyzdis . Rasti kūno tūrį V , išpjaunamą iš begalinės prizmės su kraštais
x
=
±
1
,
y
=
±
1
{\displaystyle x=\pm 1,\;y=\pm 1}
z
=
2
−
x
2
−
y
2
.
{\displaystyle z=2-x^{2}-y^{2}.}
V
=
∫
0
1
d
x
∫
0
1
d
y
∫
0
2
−
x
2
−
y
2
d
z
=
∫
0
1
d
x
∫
0
1
(
2
−
x
2
−
y
2
)
d
y
=
∫
0
1
d
x
(
2
y
−
x
2
y
−
y
3
3
)
|
0
1
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{1}dy\int _{0}^{2-x^{2}-y^{2}}dz=\int _{0}^{1}dx\int _{0}^{1}(2-x^{2}-y^{2})dy=\int _{0}^{1}dx\;(2y-x^{2}y-{\frac {y^{3}}{3}})|_{0}^{1}=}
=
∫
0
1
(
2
−
x
2
−
1
3
)
d
x
=
(
2
x
−
x
3
3
−
x
3
)
|
0
1
=
2
−
1
3
−
1
3
=
2
−
2
3
=
6
−
2
3
=
4
3
=
1.3
(
3
)
.
{\displaystyle =\int _{0}^{1}(2-x^{2}-{\frac {1}{3}})dx=(2x-{\frac {x^{3}}{3}}-{\frac {x}{3}})|_{0}^{1}=2-{\frac {1}{3}}-{\frac {1}{3}}=2-{\frac {2}{3}}={\frac {6-2}{3}}={\frac {4}{3}}=1.3(3).}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš keturių, tuomet tūris bus lygus
V
4
=
4
⋅
4
3
=
16
3
=
5.3
(
3
)
.
{\displaystyle V_{4}=4\cdot {\frac {4}{3}}={\frac {16}{3}}=5.3(3).}
Tūris esanti virš tūrio, kurį radome ir apribotas plokštuma
z
=
2
{\displaystyle z=2}
V
v
i
r
=
a
⋅
b
⋅
c
−
V
4
=
2
⋅
2
⋅
2
−
16
3
=
8
−
16
3
=
8
3
=
2.6
(
6
)
.
{\displaystyle V_{vir}=a\cdot b\cdot c-V_{4}=2\cdot 2\cdot 2-{\frac {16}{3}}=8-{\frac {16}{3}}={\frac {8}{3}}=2.6(6).}
Rasime kūno tūrį V , esantį po paraboloidu
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
x
=
1
,
y
=
1
{\displaystyle x=1,\;y=1}
V
=
∫
0
1
d
x
∫
0
1
d
y
∫
0
x
2
+
y
2
d
z
=
∫
0
1
d
x
∫
0
1
d
y
z
|
0
x
2
+
y
2
=
∫
0
1
d
x
∫
0
1
(
(
x
2
+
y
2
)
−
0
)
d
y
=
{\displaystyle V=\int _{0}^{1}dx\int _{0}^{1}dy\int _{0}^{x^{2}+y^{2}}dz=\int _{0}^{1}dx\int _{0}^{1}dy\;z|_{0}^{x^{2}+y^{2}}=\int _{0}^{1}dx\int _{0}^{1}((x^{2}+y^{2})-0)dy=}
=
∫
0
1
d
x
∫
0
1
(
x
2
+
y
2
)
d
y
=
∫
0
1
d
x
(
x
2
y
+
y
3
3
)
|
0
1
=
∫
0
1
(
x
2
⋅
1
+
1
3
3
)
d
x
=
(
x
3
3
+
x
3
)
|
0
1
=
1
−
1
3
3
+
1
3
=
2
3
=
0.6
(
6
)
.
{\displaystyle =\int _{0}^{1}dx\int _{0}^{1}(x^{2}+y^{2})dy=\int _{0}^{1}dx(x^{2}y+{\frac {y^{3}}{3}})|_{0}^{1}=\int _{0}^{1}(x^{2}\cdot 1+{\frac {1^{3}}{3}})dx=({\frac {x^{3}}{3}}+{\frac {x}{3}})|_{0}^{1}=1-{\frac {1^{3}}{3}}+{\frac {1}{3}}={\frac {2}{3}}=0.6(6).}
Kad gauti tūrį keturiuose oktantuose, reikia padauginti iš keturių, tuomet tūris bus lygus
V
4
=
4
⋅
2
3
=
8
3
=
2.66666667.
{\displaystyle V_{4}=4\cdot {\frac {2}{3}}={\frac {8}{3}}=2.66666667.}
keisti
Su stačiakampėmis Dekarto koordinatėmis cilindrines koordinates sieja formulės
x
=
ρ
cos
ϕ
,
y
=
ρ
sin
ϕ
,
z
=
z
.
{\displaystyle x=\rho \cos \phi ,\;y=\rho \sin \phi ,\;z=z.}
∭
V
f
(
x
,
y
,
z
)
d
x
d
y
d
z
=
∭
V
f
(
ρ
cos
ϕ
,
ρ
sin
ϕ
,
z
)
ρ
d
ρ
d
ϕ
d
z
.
{\displaystyle \iiint _{V}f(x,y,z)dxdydz=\iiint _{V}f(\rho \cos \phi ,\rho \sin \phi ,z)\rho d\rho d\phi dz.}
V
=
∭
V
d
x
d
y
d
z
,
{\displaystyle V=\iiint _{V}dxdydz,}
V
=
∭
V
ρ
d
ρ
d
ϕ
d
z
.
{\displaystyle V=\iiint _{V}\rho d\rho d\phi dz.}
Kūną V riboja paviršiai
x
2
+
y
2
=
x
,
{\displaystyle x^{2}+y^{2}=x,}
x
2
+
y
2
=
2
x
,
{\displaystyle x^{2}+y^{2}=2x,}
z
=
4
−
x
2
+
y
2
,
{\displaystyle z=4-{\sqrt {x^{2}+y^{2}}},}
Kūnas V iš šonų apribotas dviejų cilindrų, kurių sudaromosios lygiagrečios ašiai Oz , o vedamosios - apskritimai
x
2
+
y
2
=
x
{\displaystyle x^{2}+y^{2}=x}
x
2
+
y
2
=
2
x
.
{\displaystyle x^{2}+y^{2}=2x.}
kūgis
z
=
4
−
x
2
+
y
2
,
{\displaystyle z=4-{\sqrt {x^{2}+y^{2}}},}
xOy atžvilgiu, tai apskaičiuosime
1
2
{\displaystyle {1 \over 2}}
D , t. y. kūno prjokecija plokštumoje xOy .
Cilindrinėje koordinačių sistemoje apskritimų lygtys yra
ρ
=
cos
ϕ
{\displaystyle \rho =\cos \phi }
ρ
=
2
cos
ϕ
,
{\displaystyle \rho =2\cos \phi ,}
z
=
4
−
ρ
.
{\displaystyle z=4-\rho .}
D gaunama, kai kampas
ϕ
{\displaystyle \phi }
π
2
,
{\displaystyle {\pi \over 2},}
ρ
{\displaystyle \rho }
cos
ϕ
{\displaystyle \cos \phi }
2
cos
ϕ
.
{\displaystyle 2\cos \phi .}
V
=
2
∫
0
π
2
d
ϕ
∫
cos
ϕ
2
cos
ϕ
ρ
d
ρ
∫
0
4
−
ρ
d
z
=
2
∫
0
π
2
d
ϕ
∫
cos
ϕ
2
cos
ϕ
ρ
z
|
0
4
−
ρ
d
ρ
=
2
∫
0
π
2
d
ϕ
∫
cos
ϕ
2
cos
ϕ
(
4
ρ
−
ρ
2
)
d
ρ
=
{\displaystyle V=2\int _{0}^{\pi \over 2}d\phi \int _{\cos \phi }^{2\cos \phi }\rho d\rho \int _{0}^{4-\rho }dz=2\int _{0}^{\pi \over 2}d\phi \int _{\cos \phi }^{2\cos \phi }\rho z|_{0}^{4-\rho }d\rho =2\int _{0}^{\pi \over 2}d\phi \int _{\cos \phi }^{2\cos \phi }(4\rho -\rho ^{2})d\rho =}
=
2
∫
0
π
2
(
2
ρ
2
−
ρ
3
3
)
|
cos
ϕ
2
cos
ϕ
d
ϕ
=
2
∫
0
π
2
(
6
cos
2
ϕ
−
7
3
cos
3
ϕ
)
d
ϕ
=
12
⋅
1
2
⋅
π
2
−
14
3
⋅
2
!
!
3
!
!
=
3
π
−
28
9
.
{\displaystyle =2\int _{0}^{\pi \over 2}(2\rho ^{2}-{\rho ^{3} \over 3})|_{\cos \phi }^{2\cos \phi }d\phi =2\int _{0}^{\pi \over 2}(6\cos ^{2}\phi -{7 \over 3}\cos ^{3}\phi )d\phi =12\cdot {1 \over 2}\cdot {\pi \over 2}-{14 \over 3}\cdot {2!! \over 3!!}=3\pi -{28 \over 9}.}
dvigubas faktorialas .
Kūną V riboja viršutinė sferos
z
=
6
−
x
2
−
y
2
{\displaystyle z={\sqrt {6-x^{2}-y^{2}}}}
z
=
x
2
+
y
2
.
{\displaystyle z=x^{2}+y^{2}.}
Kadangi kūnas yra simteriškas plokštumų xOz ir yOz atžvilgiu, tai apskaičiuosime
1
4
{\displaystyle {1 \over 4}}
D , turime suprojektuoti į plokštumą xOy sferos paraboloido susikirtimo kreivę, kurios lygtį gausime išsprendę jų lygčių sistemą. Į lygtį
z
=
6
−
x
2
−
y
2
{\displaystyle z={\sqrt {6-x^{2}-y^{2}}}}
z įrašome reiškinį
x
2
+
y
2
.
{\displaystyle x^{2}+y^{2}.}
x
2
+
y
2
=
6
−
x
2
−
y
2
;
{\displaystyle x^{2}+y^{2}={\sqrt {6-x^{2}-y^{2}}};}
r
=
6
−
r
;
{\displaystyle r={\sqrt {6-r}};}
r
2
+
r
−
6
=
0.
{\displaystyle r^{2}+r-6=0.}
D
=
b
2
−
4
a
c
=
1
−
4
(
−
6
)
=
25
;
{\displaystyle D=b^{2}-4ac=1-4(-6)=25;}
r
1
;
2
=
−
b
±
D
2
a
=
−
1
±
5
2
=
−
3
;
2.
{\displaystyle r_{1;2}={-b\pm {\sqrt {D}} \over 2a}={-1\pm 5 \over 2}=-3;2.}
Iš čia
r
=
ρ
2
=
x
2
+
y
2
=
2.
{\displaystyle r=\rho ^{2}=x^{2}+y^{2}=2.}
r yra susikirtimo parabaloido ir pusapskritimo koordinate z , o kadangi parabolės projekcija į plokštumą xOz yra nusakoma formule
z
=
x
2
,
{\displaystyle z=x^{2},}
z
=
r
=
2
=
x
2
{\displaystyle z=r=2=x^{2}}
z
=
r
=
2
=
y
2
{\displaystyle z=r=2=y^{2}}
x
=
r
=
2
,
{\displaystyle x={\sqrt {r}}={\sqrt {2}},}
Taigi viso kūno tūris
V
=
4
∫
0
π
2
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
6
−
ρ
d
z
=
4
∫
0
π
2
d
ϕ
∫
0
2
(
ρ
6
−
ρ
2
−
ρ
3
)
d
ρ
=
{\displaystyle V=4\int _{0}^{\pi \over 2}d\phi \int _{0}^{\sqrt {2}}\rho d\rho \int _{\rho ^{2}}^{\sqrt {6-\rho }}dz=4\int _{0}^{\pi \over 2}d\phi \int _{0}^{\sqrt {2}}(\rho {\sqrt {6-\rho ^{2}}}-\rho ^{3})d\rho =}
=
4
∫
0
π
2
(
−
(
6
−
ρ
2
)
3
3
−
ρ
4
4
)
|
0
2
d
ϕ
=
−
4
∫
0
π
2
[
(
(
6
−
2
)
3
3
+
4
4
)
−
(
(
6
−
0
)
3
3
+
0
4
)
]
d
ϕ
=
{\displaystyle =4\int _{0}^{\pi \over 2}(-{{\sqrt {(6-\rho ^{2})^{3}}} \over 3}-{\rho ^{4} \over 4})|_{0}^{\sqrt {2}}d\phi =-4\int _{0}^{\pi \over 2}[({{\sqrt {(6-2)^{3}}} \over 3}+{4 \over 4})-({{\sqrt {(6-0)^{3}}} \over 3}+{0 \over 4})]d\phi =}
=
−
4
∫
0
π
2
[
(
64
3
+
1
)
−
216
3
]
d
ϕ
=
−
4
∫
0
π
2
[
(
8
3
+
1
)
−
6
6
3
]
d
ϕ
=
4
3
(
6
6
−
11
)
∫
0
π
2
d
ϕ
=
4
π
6
(
6
6
−
11
)
.
{\displaystyle =-4\int _{0}^{\pi \over 2}[({{\sqrt {64}} \over 3}+1)-{{\sqrt {216}} \over 3}]d\phi =-4\int _{0}^{\pi \over 2}[({8 \over 3}+1)-{6{\sqrt {6}} \over 3}]d\phi ={4 \over 3}(6{\sqrt {6}}-11)\int _{0}^{\pi \over 2}d\phi ={4\pi \over 6}(6{\sqrt {6}}-11).}
Integralas integruojamas taip:
∫
ρ
6
−
ρ
2
d
ρ
=
∫
ρ
6
−
ρ
2
d
(
6
−
ρ
2
)
−
2
ρ
=
−
1
2
∫
6
−
ρ
2
d
(
6
−
ρ
2
)
=
−
1
2
⋅
(
6
−
ρ
2
)
1
2
+
1
1
2
+
1
+
C
=
{\displaystyle \int \rho {\sqrt {6-\rho ^{2}}}d\rho =\int \rho {\sqrt {6-\rho ^{2}}}{d(6-\rho ^{2}) \over -2\rho }=-{1 \over 2}\int {\sqrt {6-\rho ^{2}}}d(6-\rho ^{2})=-{1 \over 2}\cdot {(6-\rho ^{2})^{{1 \over 2}+1} \over {1 \over 2}+1}+C=}
=
−
1
2
⋅
(
6
−
ρ
2
)
3
2
3
2
+
C
=
−
(
6
−
ρ
2
)
3
3
+
C
,
{\displaystyle =-{1 \over 2}\cdot {(6-\rho ^{2})^{3 \over 2} \over {3 \over 2}}+C=-{{\sqrt {(6-\rho ^{2})^{3}}} \over 3}+C,}
d
(
6
−
ρ
2
)
=
−
2
ρ
d
ρ
,
{\displaystyle d(6-\rho ^{2})=-2\rho d\rho ,}
d
ρ
=
d
(
6
−
ρ
2
)
−
2
ρ
.
{\displaystyle d\rho ={d(6-\rho ^{2}) \over -2\rho }.}
Apskaičiuosime tūrį kūno V , apriboto paviršiais
x
2
+
y
2
=
z
,
{\displaystyle x^{2}+y^{2}=z,}
T erdvės sritį
ρ
ϕ
z
,
{\displaystyle \rho \phi z,}
ρ
2
=
z
,
{\displaystyle \rho ^{2}=z,}
z
=
1
,
{\displaystyle z=1,}
ϕ
=
0
,
{\displaystyle \phi =0,}
ϕ
=
2
π
.
{\displaystyle \phi =2\pi .}
V
=
∭
V
d
x
d
y
d
z
=
∭
T
ρ
d
ρ
d
ϕ
d
z
=
∫
0
2
π
d
ϕ
∫
0
1
ρ
d
ρ
∫
ρ
2
1
d
z
=
{\displaystyle V=\iiint _{V}dxdydz=\iiint _{T}\rho d\rho d\phi dz=\int _{0}^{2\pi }d\phi \int _{0}^{1}\rho d\rho \int _{\rho ^{2}}^{1}dz=}
=
∫
0
2
π
d
ϕ
∫
0
1
ρ
(
1
−
ρ
2
)
d
ρ
=
∫
0
2
π
(
ρ
2
2
−
ρ
4
4
)
|
0
1
d
ϕ
=
∫
0
2
π
1
4
d
ϕ
=
π
2
.
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{1}\rho (1-\rho ^{2})d\rho =\int _{0}^{2\pi }({\rho ^{2} \over 2}-{\rho ^{4} \over 4})|_{0}^{1}d\phi =\int _{0}^{2\pi }{1 \over 4}d\phi ={\pi \over 2}.}
Apskaičiuosime tūrį kūno V , apriboto paviršiais
x
2
+
y
2
=
z
,
{\displaystyle x^{2}+y^{2}=z,}
T erdvės sritį
ρ
ϕ
z
,
{\displaystyle \rho \phi z,}
ρ
2
=
z
,
{\displaystyle \rho ^{2}=z,}
z
=
100
,
{\displaystyle z=100,}
ϕ
=
0
,
{\displaystyle \phi =0,}
ϕ
=
2
π
.
{\displaystyle \phi =2\pi .}
ρ
=
10
{\displaystyle \rho =10}
V
=
∭
V
d
x
d
y
d
z
=
∭
T
ρ
d
ρ
d
ϕ
d
z
=
∫
0
2
π
d
ϕ
∫
0
10
ρ
d
ρ
∫
ρ
2
100
d
z
=
{\displaystyle V=\iiint _{V}dxdydz=\iiint _{T}\rho d\rho d\phi dz=\int _{0}^{2\pi }d\phi \int _{0}^{10}\rho d\rho \int _{\rho ^{2}}^{100}dz=}
=
∫
0
2
π
d
ϕ
∫
0
10
ρ
(
100
−
ρ
2
)
d
ρ
=
∫
0
2
π
(
100
ρ
2
2
−
ρ
4
4
)
|
0
10
d
ϕ
=
∫
0
2
π
(
50
⋅
10
2
−
10
4
4
)
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{10}\rho (100-\rho ^{2})d\rho =\int _{0}^{2\pi }({100\rho ^{2} \over 2}-{\rho ^{4} \over 4})|_{0}^{10}d\phi =\int _{0}^{2\pi }(50\cdot 10^{2}-{10^{4} \over 4})d\phi =}
=
∫
0
2
π
(
5000
−
2500
)
d
ϕ
=
2500
∫
0
2
π
d
ϕ
=
5000
π
.
{\displaystyle =\int _{0}^{2\pi }(5000-2500)d\phi =2500\int _{0}^{2\pi }d\phi =5000\pi .}
Pavyzdis . Apskaičiuoti integralą
I
=
∫
V
d
V
,
{\displaystyle I=\int _{V}\;{\mathsf {d}}V,}
xOy ir xOz , cilindru
x
2
+
y
2
=
a
x
{\displaystyle x^{2}+y^{2}=ax}
x
2
+
y
2
+
z
2
=
a
2
.
{\displaystyle x^{2}+y^{2}+z^{2}=a^{2}.}
f
=
1
{\displaystyle f=1}
Sprendimas. Pereidami į cilindrinę koordinačių sistemą, gauname
z
1
=
0
{\displaystyle z_{1}=0}
z
2
=
a
2
−
x
2
−
y
2
=
a
2
−
ρ
2
;
{\displaystyle z_{2}={\sqrt {a^{2}-x^{2}-y^{2}}}={\sqrt {a^{2}-\rho ^{2}}};}
ρ
1
=
0
,
{\displaystyle \rho _{1}=0,}
ρ
2
=
a
cos
ϕ
,
{\displaystyle \rho _{2}=a\cos \phi ,}
x
2
+
y
2
=
ρ
2
,
{\displaystyle x^{2}+y^{2}=\rho ^{2},}
a
x
=
a
ρ
cos
ϕ
;
{\displaystyle ax=a\rho \cos \phi ;}
ϕ
1
=
0
,
{\displaystyle \phi _{1}=0,}
ϕ
2
=
π
2
.
{\displaystyle \phi _{2}={\frac {\pi }{2}}.}
V
=
I
=
∫
0
π
2
∫
0
a
cos
ϕ
∫
0
a
2
−
ρ
2
ρ
d
z
d
ρ
d
ϕ
=
{\displaystyle V=I=\int _{0}^{\pi \over 2}\int _{0}^{a\cos \phi }\int _{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho {\mathsf {d}}z{\mathsf {d}}\rho {\mathsf {d}}\phi =}
=
∫
0
π
2
[
∫
0
a
cos
ϕ
(
∫
0
a
2
−
ρ
2
ρ
d
z
)
d
ρ
]
d
ϕ
=
∫
0
π
2
[
∫
0
a
cos
ϕ
(
z
|
0
a
2
−
ρ
2
ρ
)
d
ρ
]
d
ϕ
=
∫
0
π
2
[
∫
0
a
cos
ϕ
(
(
a
2
−
ρ
2
−
0
)
ρ
)
d
ρ
]
d
ϕ
=
{\displaystyle =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(\int _{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho {\mathsf {d}}z\right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(z|_{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho \right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(({\sqrt {a^{2}-\rho ^{2}}}-0)\rho \right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =}
=
∫
0
π
2
[
∫
0
a
cos
ϕ
ρ
a
2
−
ρ
2
d
ρ
]
d
ϕ
=
∫
0
π
2
[
∫
0
a
cos
ϕ
ρ
a
2
−
ρ
2
d
(
a
2
−
ρ
2
)
−
2
ρ
]
d
ϕ
=
−
1
2
∫
0
π
2
[
∫
0
a
cos
ϕ
a
2
−
ρ
2
d
(
a
2
−
ρ
2
)
]
d
ϕ
=
{\displaystyle =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\rho {\sqrt {a^{2}-\rho ^{2}}}\,{\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\rho {\sqrt {a^{2}-\rho ^{2}}}\,{\frac {{\mathsf {d}}(a^{2}-\rho ^{2})}{-2\rho }}\right]{\mathsf {d}}\phi =-{\frac {1}{2}}\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }{\sqrt {a^{2}-\rho ^{2}}}\,{\mathsf {d}}(a^{2}-\rho ^{2})\right]{\mathsf {d}}\phi =}
=
−
1
2
∫
0
π
2
[
(
a
2
−
ρ
2
)
3
2
3
2
|
0
a
cos
ϕ
]
d
ϕ
=
−
1
2
⋅
2
3
⋅
∫
0
π
2
[
(
a
2
−
(
a
cos
ϕ
)
2
)
3
−
(
a
2
−
0
2
)
3
]
d
ϕ
=
−
1
3
∫
0
π
2
[
(
a
2
(
1
−
cos
2
ϕ
)
)
3
−
a
6
]
d
ϕ
=
{\displaystyle =-{\frac {1}{2}}\int _{0}^{\pi \over 2}\left[{\frac {(a^{2}-\rho ^{2})^{3 \over 2}}{3 \over 2}}|_{0}^{a\cos \phi }\right]{\mathsf {d}}\phi =-{\frac {1}{2}}\cdot {\frac {2}{3}}\cdot \int _{0}^{\pi \over 2}\left[{\sqrt {(a^{2}-(a\cos \phi )^{2})^{3}}}-{\sqrt {(a^{2}-0^{2})^{3}}}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[{\sqrt {(a^{2}(1-\cos ^{2}\phi ))^{3}}}-{\sqrt {a^{6}}}\right]{\mathsf {d}}\phi =}
=
−
1
3
∫
0
π
2
[
a
6
(
1
−
cos
2
ϕ
)
3
−
a
3
]
d
ϕ
=
−
1
3
∫
0
π
2
[
a
3
(
sin
2
ϕ
)
3
−
a
3
]
d
ϕ
=
−
1
3
∫
0
π
2
[
a
3
sin
3
ϕ
−
a
3
]
d
ϕ
=
−
a
3
3
∫
0
π
2
[
sin
3
ϕ
−
1
]
d
ϕ
=
{\displaystyle =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[{\sqrt {a^{6}(1-\cos ^{2}\phi )^{3}}}-a^{3}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[a^{3}{\sqrt {(\sin ^{2}\phi )^{3}}}-a^{3}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[a^{3}\sin ^{3}\phi -a^{3}\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =}
=
−
a
3
3
∫
0
π
2
[
sin
3
ϕ
−
1
]
d
ϕ
=
−
a
3
3
∫
0
π
2
[
1
4
(
3
sin
ϕ
−
sin
(
3
ϕ
)
)
−
1
]
d
ϕ
=
−
a
3
12
∫
0
π
2
[
3
sin
ϕ
−
sin
(
3
ϕ
)
−
4
]
d
ϕ
=
{\displaystyle =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[{\frac {1}{4}}(3\sin \phi -\sin(3\phi ))-1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{12}}\int _{0}^{\pi \over 2}\left[3\sin \phi -\sin(3\phi )-4\right]{\mathsf {d}}\phi =}
=
−
a
3
12
[
−
3
cos
ϕ
+
cos
(
3
ϕ
)
3
−
4
ϕ
]
|
0
π
2
=
−
a
3
12
[
−
3
cos
π
2
+
cos
(
3
⋅
π
2
)
3
−
4
⋅
π
2
−
(
−
3
cos
0
+
cos
(
3
⋅
0
)
3
−
4
⋅
0
)
]
=
{\displaystyle =-{\frac {a^{3}}{12}}\left[-3\cos \phi +{\frac {\cos(3\phi )}{3}}-4\phi \right]|_{0}^{\pi \over 2}=-{\frac {a^{3}}{12}}\left[-3\cos {\frac {\pi }{2}}+{\frac {\cos(3\cdot {\frac {\pi }{2}})}{3}}-4\cdot {\frac {\pi }{2}}-(-3\cos 0+{\frac {\cos(3\cdot 0)}{3}}-4\cdot 0)\right]=}
=
−
a
3
12
[
−
3
⋅
0
+
0
3
−
2
π
−
(
−
3
⋅
1
+
1
3
)
]
=
−
a
3
12
[
−
2
π
+
3
−
1
3
]
=
−
a
3
12
⋅
−
6
π
+
9
−
1
3
=
−
a
3
12
⋅
−
6
π
+
8
3
=
{\displaystyle =-{\frac {a^{3}}{12}}\left[-3\cdot 0+{\frac {0}{3}}-2\pi -(-3\cdot 1+{\frac {1}{3}})\right]=-{\frac {a^{3}}{12}}\left[-2\pi +3-{\frac {1}{3}}\right]=-{\frac {a^{3}}{12}}\cdot {\frac {-6\pi +9-1}{3}}=-{\frac {a^{3}}{12}}\cdot {\frac {-6\pi +8}{3}}=}
=
a
3
12
⋅
6
π
−
8
3
=
a
3
(
6
π
−
8
)
36
=
a
3
(
3
π
−
4
)
18
=
0.301376553
⋅
a
3
.
{\displaystyle ={\frac {a^{3}}{12}}\cdot {\frac {6\pi -8}{3}}={\frac {a^{3}(6\pi -8)}{36}}={\frac {a^{3}(3\pi -4)}{18}}=0.301376553\cdot a^{3}.}
čia
d
(
a
2
−
ρ
2
)
=
−
2
ρ
d
ρ
,
d
ρ
=
d
(
a
2
−
ρ
2
)
−
2
ρ
;
sin
(
3
A
)
=
3
sin
A
−
4
sin
3
A
,
sin
3
A
=
1
4
(
3
sin
A
−
sin
(
3
A
)
)
.
{\displaystyle {\mathsf {d}}(a^{2}-\rho ^{2})=-2\rho \,{\mathsf {d}}\rho ,\;\;{\mathsf {d}}\rho ={\frac {{\mathsf {d}}(a^{2}-\rho ^{2})}{-2\rho }};\quad \sin(3A)=3\sin A-4\sin ^{3}A,\;\;\sin ^{3}A={\frac {1}{4}}(3\sin A-\sin(3A)).}
Kai
a
=
3
{\displaystyle a=3}
V
=
0.301376553
⋅
a
3
=
0.301376553
⋅
3
3
=
8.137166941.
{\displaystyle V=0.301376553\cdot a^{3}=0.301376553\cdot 3^{3}=8.137166941.}
Kad įsivaizduoti kaip atrodo kūnas, galima pasakyti, kad sferos centras yra (0; 0; 0), o sferos sindulys
R
=
a
{\displaystyle R=a}
Ox . Cilindro [pagrindo] spindulys
r
=
a
2
{\displaystyle r={\frac {a}{2}}}
d
=
a
{\displaystyle d=a}
Ox ir vienas jo pagrindo kraštas liečiasi su koordinačiu pradžios tašku O , o kitas liečiasi su tašku a ant Ox ašies. Sfera, kurios lygtis, priminimui, yra
x
2
+
y
2
+
z
2
=
a
2
{\displaystyle x^{2}+y^{2}+z^{2}=a^{2}}
Žinodami cilindro tūrio formulę
V
c
i
l
=
π
r
2
h
,
{\displaystyle V_{cil}=\pi r^{2}h\,,}
V
≈
V
c
i
l
2
=
π
r
2
h
2
=
π
(
a
2
)
2
h
2
=
π
(
3
2
)
2
⋅
3
2
=
1
2
⋅
9
4
⋅
3
π
=
27
π
8
=
3.375
π
=
10.60287521.
{\displaystyle V\approx {\frac {V_{cil}}{2}}={\frac {\pi r^{2}h}{2}}={\frac {\pi ({\frac {a}{2}})^{2}h}{2}}={\frac {\pi ({\frac {3}{2}})^{2}\cdot 3}{2}}={\frac {1}{2}}\cdot {\frac {9}{4}}\cdot 3\pi ={\frac {27\pi }{8}}=3.375\pi =10.60287521.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
−
a
3
3
∫
0
π
2
[
sin
3
ϕ
−
1
]
d
ϕ
=
−
a
3
3
(
∫
0
π
2
sin
3
ϕ
d
ϕ
−
∫
0
π
2
d
ϕ
)
=
−
a
3
3
(
(
3
−
1
)
!
!
3
!
!
−
(
π
2
−
0
)
)
=
{\displaystyle V=-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}(\int _{0}^{\pi \over 2}\sin ^{3}\phi \;{\mathsf {d}}\phi -\int _{0}^{\pi \over 2}{\mathsf {d}}\phi )=-{\frac {a^{3}}{3}}({\frac {(3-1)!!}{3!!}}-({\pi \over 2}-0))=}
=
−
a
3
3
(
2
!
!
3
!
!
−
π
2
)
=
−
a
3
3
(
2
3
−
π
2
)
=
a
3
3
(
π
2
−
2
3
)
=
3
3
3
(
π
2
−
2
3
)
=
9
⋅
0.90412966
=
8.13716694.
{\displaystyle =-{\frac {a^{3}}{3}}({\frac {2!!}{3!!}}-{\pi \over 2})=-{\frac {a^{3}}{3}}({\frac {2}{3}}-{\pi \over 2})={\frac {a^{3}}{3}}({\pi \over 2}-{\frac {2}{3}})={\frac {3^{3}}{3}}({\pi \over 2}-{\frac {2}{3}})=9\cdot 0.90412966=8.13716694.}
Pasinaudodami analitiniu mąstymu, pabandysime parodyti, kad tūris rastas teisingai. Apskritimo spindulys R=3, todėl ketvirtadalis skritulio ploto yra
S
s
k
r
=
π
R
2
4
=
π
⋅
3
2
4
=
9
π
4
=
2.25
π
=
7.068583471.
{\displaystyle S_{skr}={\frac {\pi R^{2}}{4}}={\frac {\pi \cdot 3^{2}}{4}}={\frac {9\pi }{4}}=2.25\pi =7.068583471.}
S
k
v
=
a
2
=
3
2
=
9.
{\displaystyle S_{kv}=a^{2}=3^{2}=9.}
Dabar randame kvadrato ir 1/4 skritulio santykį:
S
k
v
S
s
k
r
=
9
9
4
⋅
π
=
4
π
=
1.273239545.
{\displaystyle {\frac {S_{kv}}{S_{skr}}}={\frac {9}{{\frac {9}{4}}\cdot \pi }}={\frac {4}{\pi }}=1.273239545.}
Akivaizdu, kad padalinus visą cilindro tūrį iš tūrio, kurį riboja cilindras ir sfera, turėtume gautį santykį didesnį nei kvadrato ir ketvirtadalio skritulio, o santykis yra:
V
c
i
l
2
V
=
10.60287521
8.137166941
=
1.303018027.
{\displaystyle {\frac {\frac {V_{cil}}{2}}{V}}={\frac {10.60287521}{8.137166941}}=1.303018027.}
Taip ir yra, tolstant nuo Ox ašies, z reikšmės mažėja, kas ir užtikrina didesnį santykį. Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
y
{\displaystyle x^{2}+y^{2}=y}
xOy , kurio centro koordinatės (0; 0.5), o spindulys r=1/2),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=\rho \sin \phi ,}
ρ
=
sin
ϕ
.
{\displaystyle \rho =\sin \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
sin
ϕ
d
ϕ
=
∫
0
π
2
(
sin
4
ϕ
4
−
0
4
4
)
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {\sin ^{4}\phi }{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =}
=
1
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
1
4
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
1
32
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {1}{32}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
1
32
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
1
32
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
1
32
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
1
32
⋅
3
π
2
=
3
π
64
=
0.147262155.
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {1}{32}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{64}}=0.147262155.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
1
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
1
4
⋅
(
4
−
1
)
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
⋅
1
4
⋅
2
⋅
π
2
=
1
4
⋅
3
8
⋅
π
2
=
3
π
64
=
0.147262155.
{\displaystyle V={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\cdot {(4-1)!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3\cdot 1 \over 4\cdot 2}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3 \over 8}\cdot {\pi \over 2}={\frac {3\pi }{64}}=0.147262155.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
3
π
64
{\displaystyle {\frac {3\pi }{64}}}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
x
{\displaystyle x^{2}+y^{2}=x}
xOy , kurio centro koordinatės (0.5; 0), o spindulys r=1/2),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
ρ
cos
ϕ
,
{\displaystyle \rho ^{2}=\rho \cos \phi ,}
ρ
=
cos
ϕ
.
{\displaystyle \rho =\cos \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
cos
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
cos
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
cos
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
cos
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
cos
ϕ
d
ϕ
=
∫
0
π
2
(
cos
4
ϕ
4
−
0
4
4
)
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{\cos \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{\cos \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {\cos ^{4}\phi }{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =}
=
1
4
∫
0
π
2
cos
4
ϕ
d
ϕ
=
1
4
∫
0
π
2
cos
(
4
ϕ
)
+
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
1
32
(
1
4
⋅
sin
(
4
ϕ
)
+
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\cos ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )+4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {1}{32}}({\frac {1}{4}}\cdot \sin(4\phi )+2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
1
32
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
+
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
+
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})+2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)+2\sin(2\cdot 0)+3\cdot 0)]=}
=
1
32
[
(
1
4
⋅
sin
(
2
π
)
+
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
+
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )+2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)+2\sin(0))]=}
=
1
32
[
(
1
4
⋅
0
+
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
+
2
⋅
0
)
]
=
1
32
⋅
3
π
2
=
3
π
64
=
0.147262155.
{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot 0+2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0+2\cdot 0)]={\frac {1}{32}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{64}}=0.147262155.}
kur
cos
4
A
=
cos
(
4
A
)
+
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \cos ^{4}A={\cos(4A)+4\cos(2A)+3 \over 8}.}
Pasinaudojant dvigubu faktorialu gauname tą patį atsakymą:
V
=
1
4
∫
0
π
2
cos
4
ϕ
d
ϕ
=
1
4
⋅
(
4
−
1
)
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
!
!
4
!
!
⋅
π
2
=
1
4
⋅
3
⋅
1
4
⋅
2
⋅
π
2
=
1
4
⋅
3
8
⋅
π
2
=
3
π
64
=
0.147262155.
{\displaystyle V={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\cos ^{4}\phi {\mathsf {d}}\phi ={\frac {1}{4}}\cdot {(4-1)!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3!! \over 4!!}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3\cdot 1 \over 4\cdot 2}\cdot {\pi \over 2}={\frac {1}{4}}\cdot {3 \over 8}\cdot {\pi \over 2}={\frac {3\pi }{64}}=0.147262155.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
3
π
64
{\displaystyle {\frac {3\pi }{64}}}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
2
y
{\displaystyle x^{2}+y^{2}=2y}
xOy , kurio centro koordinatės (0; 1), o spindulys r=1),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
2
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=2\rho \sin \phi ,}
ρ
=
2
sin
ϕ
.
{\displaystyle \rho =2\sin \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
2
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
2
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
2
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
2
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
2
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
2
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
16
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{2\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{2\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(2\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {16\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
4
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
1
2
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =4\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =4\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {1}{2}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
1
2
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
1
2
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
1
2
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
1
2
⋅
3
π
2
=
3
π
4
=
2.35619449.
{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {1}{2}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{4}}=2.35619449.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
3
π
4
{\displaystyle {\frac {3\pi }{4}}}
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=1, aukštis
h
=
2
2
=
4
{\displaystyle h=2^{2}=4}
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
1
2
⋅
4
2
=
2
π
=
6.283185307.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot 1^{2}\cdot 4}{2}}=2\pi =6.283185307.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
8
y
{\displaystyle x^{2}+y^{2}=8y}
xOy , kurio centro koordinatės (0; 4), o spindulys r=4),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
8
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=8\rho \sin \phi ,}
ρ
=
8
sin
ϕ
.
{\displaystyle \rho =8\sin \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
8
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
8
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
8
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
8
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
8
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
8
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
4096
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{8\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{8\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(8\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {4096\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
1024
∫
0
π
2
sin
4
ϕ
d
ϕ
=
1024
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
128
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =1024\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =1024\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi =128({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
128
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle =128[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
128
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle =128[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
128
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
128
⋅
3
π
2
=
192
π
=
603.1857895.
{\displaystyle =128[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]=128\cdot {\frac {3\pi }{2}}=192\pi =603.1857895.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
192
π
{\displaystyle 192\pi }
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=4, aukštis
h
=
8
2
=
64
{\displaystyle h=8^{2}=64}
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
4
2
⋅
64
2
=
π
⋅
16
⋅
64
2
=
π
⋅
16
⋅
32
=
512
π
=
1608.495439.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot 4^{2}\cdot 64}{2}}={\frac {\pi \cdot 16\cdot 64}{2}}=\pi \cdot 16\cdot 32=512\pi =1608.495439.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
9
y
{\displaystyle x^{2}+y^{2}=9y}
xOy , kurio centro koordinatės (0; 4.5), o spindulys r=9/2),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
9
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=9\rho \sin \phi ,}
ρ
=
9
sin
ϕ
.
{\displaystyle \rho =9\sin \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
9
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
9
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
9
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
9
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
9
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
9
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
6561
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{9\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{9\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(9\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {6561\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
6561
4
∫
0
π
2
sin
4
ϕ
d
ϕ
=
6561
4
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
6561
32
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle ={\frac {6561}{4}}\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi ={\frac {6561}{4}}\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {6561}{32}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
6561
32
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
6561
32
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
6561
32
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
6561
32
⋅
3
π
2
=
19683
π
64
=
307.546875
π
=
966.1870031.
{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {6561}{32}}\cdot {\frac {3\pi }{2}}={\frac {19683\pi }{64}}=307.546875\pi =966.1870031.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
19683
π
64
{\displaystyle {\frac {19683\pi }{64}}}
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=9/2, aukštis
h
=
9
2
=
81
{\displaystyle h=9^{2}=81}
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
(
9
2
)
2
⋅
81
2
=
π
⋅
81
4
⋅
81
2
=
π
⋅
81
⋅
81
4
⋅
2
=
6561
π
8
=
820.125
π
=
2576.498675.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot \left({\frac {9}{2}}\right)^{2}\cdot 81}{2}}={\frac {\pi \cdot {\frac {81}{4}}\cdot 81}{2}}={\frac {\pi \cdot 81\cdot 81}{4\cdot 2}}={\frac {6561\pi }{8}}=820.125\pi =2576.498675.}
Pavyzdis . Rasti kūno tūrį V , apriboto paviršiais
x
2
+
y
2
=
10
y
{\displaystyle x^{2}+y^{2}=10y}
xOy , kurio centro koordinatės (0; 5), o spindulys r=5),
z
=
0
{\displaystyle z=0}
xOy ),
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Sprendimas . Pereidami į polinę koordinačių sistemą, turime apskritimo lytį
ρ
2
=
10
ρ
sin
ϕ
,
{\displaystyle \rho ^{2}=10\rho \sin \phi ,}
ρ
=
10
sin
ϕ
.
{\displaystyle \rho =10\sin \phi .}
z
=
ρ
2
.
{\displaystyle z=\rho ^{2}.}
ϕ
{\displaystyle \phi }
π
2
.
{\displaystyle {\frac {\pi }{2}}.}
V
=
∭
V
d
x
d
y
d
z
=
∭
V
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
10
sin
ϕ
∫
0
ρ
2
ρ
d
ϕ
d
ρ
d
z
=
∫
0
π
2
∫
0
10
sin
ϕ
z
|
0
ρ
2
ρ
d
ϕ
d
ρ
=
{\displaystyle V=\iiint _{V}{\mathsf {d}}x{\mathsf {d}}y{\mathsf {d}}z=\iiint _{V}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }\int _{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho {\mathsf {d}}z=\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }z|_{0}^{\rho ^{2}}\rho {\mathsf {d}}\phi {\mathsf {d}}\rho =}
=
∫
0
π
2
∫
0
10
sin
ϕ
ρ
(
ρ
2
−
0
)
d
ϕ
d
ρ
=
∫
0
π
2
∫
0
10
sin
ϕ
ρ
3
d
ϕ
d
ρ
=
∫
0
π
2
ρ
4
4
|
0
10
sin
ϕ
d
ϕ
=
∫
0
π
2
(
(
10
sin
ϕ
)
4
4
−
0
4
4
)
d
ϕ
=
∫
0
π
2
10000
sin
4
ϕ
4
d
ϕ
=
{\displaystyle =\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }\rho (\rho ^{2}-0){\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}\int _{0}^{10\sin \phi }\rho ^{3}{\mathsf {d}}\phi {\mathsf {d}}\rho =\int _{0}^{\frac {\pi }{2}}{\frac {\rho ^{4}}{4}}|_{0}^{10\sin \phi }{\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}({\frac {(10\sin \phi )^{4}}{4}}-{\frac {0^{4}}{4}}){\mathsf {d}}\phi =\int _{0}^{\frac {\pi }{2}}{\frac {10000\sin ^{4}\phi }{4}}{\mathsf {d}}\phi =}
=
2500
∫
0
π
2
sin
4
ϕ
d
ϕ
=
2500
∫
0
π
2
cos
(
4
ϕ
)
−
4
cos
(
2
ϕ
)
+
3
8
d
ϕ
=
625
2
(
1
4
⋅
sin
(
4
ϕ
)
−
2
sin
(
2
ϕ
)
+
3
ϕ
)
|
0
π
2
=
{\displaystyle =2500\int _{0}^{\frac {\pi }{2}}\sin ^{4}\phi {\mathsf {d}}\phi =2500\int _{0}^{\frac {\pi }{2}}{\cos(4\phi )-4\cos(2\phi )+3 \over 8}{\mathsf {d}}\phi ={\frac {625}{2}}({\frac {1}{4}}\cdot \sin(4\phi )-2\sin(2\phi )+3\phi )|_{0}^{\frac {\pi }{2}}=}
=
625
2
[
(
1
4
⋅
sin
(
4
⋅
π
2
)
−
2
sin
(
2
⋅
π
2
)
+
3
⋅
π
2
)
−
(
1
4
⋅
sin
(
4
⋅
0
)
−
2
sin
(
2
⋅
0
)
+
3
⋅
0
)
]
=
{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}
=
625
2
[
(
1
4
⋅
sin
(
2
π
)
−
2
sin
(
π
)
+
3
π
2
)
−
(
1
4
⋅
sin
(
0
)
−
2
sin
(
0
)
)
]
=
{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}
=
625
2
[
(
1
4
⋅
0
−
2
⋅
0
+
3
π
2
)
−
(
1
4
⋅
0
−
2
⋅
0
)
]
=
625
2
⋅
3
π
2
=
1875
π
4
=
468.75
π
=
1472.621556.
{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {625}{2}}\cdot {\frac {3\pi }{2}}={\frac {1875\pi }{4}}=468.75\pi =1472.621556.}
kur
sin
4
A
=
cos
(
4
A
)
−
4
cos
(
2
A
)
+
3
8
.
{\displaystyle \sin ^{4}A={\cos(4A)-4\cos(2A)+3 \over 8}.}
Kad gauti tūrį dviejuose oktantuose, reikia gautą turį
1875
π
4
{\displaystyle {\frac {1875\pi }{4}}}
Palyginimui, cilindro tūris viename oktante, kurio spindulys r=5, aukštis
h
=
10
2
=
100
{\displaystyle h=10^{2}=100}
V
c
i
l
1
=
π
⋅
r
2
⋅
h
2
=
π
⋅
5
2
⋅
100
2
=
π
⋅
25
⋅
100
2
=
π
⋅
25
⋅
50
=
1250
π
=
3926.990817.
{\displaystyle V_{cil1}={\frac {\pi \cdot r^{2}\cdot h}{2}}={\frac {\pi \cdot 5^{2}\cdot 100}{2}}={\frac {\pi \cdot 25\cdot 100}{2}}=\pi \cdot 25\cdot 50=1250\pi =3926.990817.}
Galime pabandyti suprasti ar integravimo budu gautas atsakymas yra teisingas. Kai
x
=
5
{\displaystyle x=5}
y
=
5
{\displaystyle y=5}
z reikšmė lygi
z
=
x
2
+
y
2
=
5
2
+
5
2
=
25
+
25
=
50.
{\displaystyle z=x^{2}+y^{2}=5^{2}+5^{2}=25+25=50.}
x
=
0
{\displaystyle x=0}
y
=
10
{\displaystyle y=10}
z reikšmė yra
z
=
x
2
+
y
2
=
0
2
+
10
2
=
0
+
100
=
100.
{\displaystyle z=x^{2}+y^{2}=0^{2}+10^{2}=0+100=100.}
y reikšmei, z reikšmė apskritimo srityje didėja kvadratu. O kai apskritimo srityje y reikšmė mažesnė už 10, tada ir z reikšmė visoje apskritimo (
x
2
+
y
2
=
10
y
{\displaystyle x^{2}+y^{2}=10y}
x ir y niekada neduos didesnės z , reikšmės už tą atvejį, kai R=y=10. Cilindru iš praboloido iškerpamas tūris yra tik 2,66667 karto mažesnis už viso cilindro tūrį. Kitaip tariant, jei viso cilindro tūris yra 1, tai tūris, kurį gauname integravimo budu dviejuose oktantuose yra 0.375 visais atvejais. Dar palyginimui, plotas po parabolės
y
=
x
2
{\displaystyle y=x^{2}}
x
⋅
y
=
x
⋅
y
2
=
x
⋅
x
2
.
{\displaystyle x\cdot y=x\cdot y^{2}=x\cdot x^{2}.}
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
Dar pastebėjimas, kad z reikšmė yra didesnė, kai
x
=
5
{\displaystyle x=5}
y
=
5
{\displaystyle y=5}
z
=
x
2
+
y
2
=
5
2
+
5
2
=
50
{\displaystyle z=x^{2}+y^{2}=5^{2}+5^{2}=50}
y
=
5
{\displaystyle y=5}
x
=
0
,
{\displaystyle x=0,}
z
=
x
2
+
y
2
=
0
2
+
5
2
=
25
{\displaystyle z=x^{2}+y^{2}=0^{2}+5^{2}=25}
Oy ašis, dominuoja didesnės z reikšmės, negu centre, tačiau didžiausia z reikšmė vis tiek, kai y=10, x=0.
Pereidami į polinę koordinačių sistemą rasime tūrį po paraboloidu
z
=
1
+
x
2
+
y
2
,
{\displaystyle z=1+x^{2}+y^{2},}
x
2
+
y
2
=
1.
{\displaystyle x^{2}+y^{2}=1.}
V
=
∫
0
π
2
d
ϕ
∫
0
1
ρ
d
ρ
∫
0
1
+
ρ
2
d
z
=
∫
0
π
2
d
ϕ
∫
0
1
ρ
(
1
+
ρ
2
)
d
ρ
=
∫
0
π
2
d
ϕ
∫
0
1
(
ρ
+
ρ
3
)
d
ρ
=
∫
0
π
2
d
ϕ
(
ρ
2
2
+
ρ
4
4
)
|
0
1
=
{\displaystyle V=\int _{0}^{\pi \over 2}d\phi \int _{0}^{1}\rho d\rho \int _{0}^{1+\rho ^{2}}dz=\int _{0}^{\pi \over 2}d\phi \int _{0}^{1}\rho (1+\rho ^{2})d\rho =\int _{0}^{\pi \over 2}d\phi \int _{0}^{1}(\rho +\rho ^{3})d\rho =\int _{0}^{\pi \over 2}d\phi ({\frac {\rho ^{2}}{2}}+{\frac {\rho ^{4}}{4}})|_{0}^{1}=}
=
∫
0
π
2
(
1
2
2
+
1
4
4
)
d
ϕ
=
(
1
2
+
1
4
)
ϕ
|
0
π
2
=
2
+
1
4
ϕ
|
0
π
2
=
3
4
⋅
π
2
=
3
π
8
=
1.178097245.
{\displaystyle =\int _{0}^{\pi \over 2}({\frac {1^{2}}{2}}+{\frac {1^{4}}{4}})d\phi =({\frac {1}{2}}+{\frac {1}{4}})\phi |_{0}^{\pi \over 2}={\frac {2+1}{4}}\phi |_{0}^{\pi \over 2}={\frac {3}{4}}\cdot {\pi \over 2}={\frac {3\pi }{8}}=1.178097245.}
Šis tūris keturiuose oktantuose yra lygus
3
π
2
=
4.71238898.
{\displaystyle {\frac {3\pi }{2}}=4.71238898.}
Trilypio integralo taikymas mechanikoje
keisti
keisti
Kai tam tikros masės tankis lygus
γ
(
x
,
y
,
z
)
,
{\displaystyle \gamma (x,y,z),}
x
c
=
∭
V
x
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
∭
V
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
,
y
c
=
∭
V
y
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
∭
V
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
,
z
c
=
∭
V
z
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
∭
V
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
.
{\displaystyle x_{c}={\iiint _{V}x\gamma (x,y,z)dxdydz \over \iiint _{V}\gamma (x,y,z)dxdydz},\;y_{c}={\iiint _{V}y\gamma (x,y,z)dxdydz \over \iiint _{V}\gamma (x,y,z)dxdydz},\;z_{c}={\iiint _{V}z\gamma (x,y,z)dxdydz \over \iiint _{V}\gamma (x,y,z)dxdydz}.}
Pavyzdžiai
Kūną riboja paviršiai
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
z
=
4.
{\displaystyle z=4.}
γ
=
c
o
n
s
t
.
{\displaystyle \gamma =const.}
Kadangi kūnas simteriškas plokštumų xOy ir yOz atžvilgiu, tai
x
c
=
y
c
=
0.
{\displaystyle x_{c}=y_{c}=0.}
z
c
{\displaystyle z_{c}}
γ
=
c
o
n
s
t
,
{\displaystyle \gamma =const,}
z
c
=
∭
V
z
d
z
d
y
d
z
∭
V
d
x
d
y
d
z
.
{\displaystyle z_{c}={\iiint _{V}zdzdydz \over \iiint _{V}dxdydz}.}
z
c
=
∫
0
2
π
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
4
z
d
z
∫
0
2
π
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
4
d
z
=
1
2
∫
0
2
π
d
ϕ
∫
0
2
(
16
−
ρ
4
)
ρ
d
ρ
∫
0
2
π
d
ϕ
∫
0
2
(
4
−
ρ
2
)
ρ
d
ρ
=
{\displaystyle z_{c}={\int _{0}^{2\pi }d\phi \int _{0}^{2}\rho d\rho \int _{\rho ^{2}}^{4}z\;dz \over \int _{0}^{2\pi }d\phi \int _{0}^{2}\rho d\rho \int _{\rho ^{2}}^{4}dz}={{1 \over 2}\int _{0}^{2\pi }d\phi \int _{0}^{2}(16-\rho ^{4})\rho d\rho \over \int _{0}^{2\pi }d\phi \int _{0}^{2}(4-\rho ^{2})\rho d\rho }=}
=
1
2
∫
0
2
π
(
16
ρ
2
2
−
ρ
6
6
)
|
0
2
d
ϕ
∫
0
2
π
(
4
ρ
2
2
−
ρ
4
4
)
|
0
2
d
ϕ
=
1
2
∫
0
2
π
(
32
−
32
3
)
d
ϕ
∫
0
2
π
(
8
−
4
)
d
ϕ
=
1
2
⋅
64
3
⋅
2
π
4
⋅
2
π
=
8
3
.
{\displaystyle ={{1 \over 2}\int _{0}^{2\pi }(16{\rho ^{2} \over 2}-{\rho ^{6} \over 6})|_{0}^{2}d\phi \over \int _{0}^{2\pi }(4{\rho ^{2} \over 2}-{\rho ^{4} \over 4})|_{0}^{2}d\phi }={{1 \over 2}\int _{0}^{2\pi }(32-{32 \over 3})d\phi \over \int _{0}^{2\pi }(8-4)d\phi }={{1 \over 2}\cdot {64 \over 3}\cdot 2\pi \over 4\cdot 2\pi }={8 \over 3}.}
keisti
Taško M (x; y; z), kurio masė m , inercijos momentai koordinačių plokštumų xOy , xOz ir yOz atžvilgiu išreiškiami formulėmis
I
x
O
y
=
z
2
m
,
{\displaystyle I_{xOy}=z^{2}m,}
I
x
O
z
=
y
2
m
,
{\displaystyle I_{xOz}=y^{2}m,}
I
y
O
z
=
x
2
m
,
{\displaystyle I_{yOz}=x^{2}m,}
ašių Ox , Oy , Oz atžvilgiu - formulėmis
I
x
x
=
(
y
2
+
z
2
)
m
,
{\displaystyle I_{xx}=(y^{2}+z^{2})m,}
I
y
y
=
(
x
2
+
z
2
)
m
,
{\displaystyle I_{yy}=(x^{2}+z^{2})m,}
I
z
z
=
(
x
2
+
y
2
)
m
,
{\displaystyle I_{zz}=(x^{2}+y^{2})m,}
koordinačių pradžios atžvilgiu - formule
I
0
=
(
x
2
+
y
2
+
z
2
)
m
.
{\displaystyle I_{0}=(x^{2}+y^{2}+z^{2})m.}
Kūno inercijos momentai išreiškiami atitinkamais trilypiais integralais. Pavyzdžiui, tam tikros masės kūno, kurio tankis
γ
(
x
,
y
,
z
)
,
{\displaystyle \gamma (x,y,z),}
xOy atžvilgiu apskaičiuojamas pagal formulę
I
x
O
y
=
∭
V
z
2
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
,
{\displaystyle I_{xOy}=\iiint _{V}z^{2}\gamma (x,y,z)dxdydz,}
ašies Oz atžvilgiu - pagal formulę
I
z
z
=
∭
V
(
x
2
+
y
2
)
γ
(
x
,
y
,
z
)
d
x
d
y
d
z
{\displaystyle I_{zz}=\iiint _{V}(x^{2}+y^{2})\gamma (x,y,z)dxdydz}
Pavyzdžiai
Apskaičiuokime kūno, kurį riboja paraboloidas
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
z
=
4
{\displaystyle z=4}
γ
=
1
{\displaystyle \gamma =1}
Koordinačių ašis parinkime taip, kad jų pradžios taškas sutaptų su paraboloido masės centru, o ašis Ox būtų statmena paraboloido sukimosi ašiai. Tuomet turėsime rasti
I
x
x
{\displaystyle I_{xx}}
I
y
y
{\displaystyle I_{yy}}
Paraboloido lygtis tokioje koordinačių sistemoje yra
z
+
8
3
=
x
2
+
y
2
,
{\displaystyle z+{8 \over 3}=x^{2}+y^{2},}
xOy - sritis, apribota apskritimo
x
2
+
y
2
=
4.
{\displaystyle x^{2}+y^{2}=4.}
I
x
x
=
∭
V
(
y
2
+
z
2
)
d
x
d
y
d
z
.
{\displaystyle I_{xx}=\iiint _{V}(y^{2}+z^{2})dxdydz.}
J
x
x
=
∫
0
2
π
d
ϕ
∫
0
2
ρ
d
ρ
∫
ρ
2
−
8
/
3
4
/
3
(
ρ
2
sin
2
ϕ
+
z
2
)
d
z
=
∫
0
2
π
d
ϕ
∫
0
2
ρ
(
z
ρ
2
sin
2
ϕ
+
z
3
3
)
|
ρ
2
−
8
3
4
3
d
ρ
=
{\displaystyle J_{xx}=\int _{0}^{2\pi }d\phi \int _{0}^{2}\rho d\rho \int _{\rho ^{2}-8/3}^{4/3}(\rho ^{2}\sin ^{2}\phi +z^{2})dz=\int _{0}^{2\pi }d\phi \int _{0}^{2}\rho (z\rho ^{2}\sin ^{2}\phi +{z^{3} \over 3})|_{\rho ^{2}-{8 \over 3}}^{4 \over 3}d\rho =}
=
∫
0
2
π
d
ϕ
∫
0
2
[
(
4
ρ
3
−
ρ
5
)
sin
2
ϕ
+
64
9
ρ
−
ρ
7
3
+
8
ρ
5
3
−
64
9
ρ
3
]
d
ρ
=
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{2}[(4\rho ^{3}-\rho ^{5})\sin ^{2}\phi +{64 \over 9}\rho -{\rho ^{7} \over 3}+{8\rho ^{5} \over 3}-{64 \over 9}\rho ^{3}]d\rho =}
=
∫
0
2
π
[
(
ρ
4
−
ρ
6
6
)
sin
2
ϕ
+
32
9
ρ
2
−
ρ
8
24
+
4
ρ
6
9
−
16
9
ρ
4
]
|
0
2
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }[(\rho ^{4}-{\rho ^{6} \over 6})\sin ^{2}\phi +{32 \over 9}\rho ^{2}-{\rho ^{8} \over 24}+{4\rho ^{6} \over 9}-{16 \over 9}\rho ^{4}]|_{0}^{2}d\phi =}
=
∫
0
2
π
(
16
3
sin
2
ϕ
+
128
9
−
64
6
+
256
9
−
256
9
)
d
ϕ
=
∫
0
2
π
(
16
3
sin
2
ϕ
+
32
9
)
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }({16 \over 3}\sin ^{2}\phi +{128 \over 9}-{64 \over 6}+{256 \over 9}-{256 \over 9})d\phi =\int _{0}^{2\pi }({16 \over 3}\sin ^{2}\phi +{32 \over 9})d\phi =}
=
16
3
∫
0
2
π
1
−
cos
(
2
ϕ
)
2
d
ϕ
+
64
π
9
=
16
π
3
−
8
3
∫
0
2
π
cos
(
2
ϕ
)
d
(
2
ϕ
)
2
+
64
π
9
=
112
π
9
−
4
3
sin
(
2
ϕ
)
|
0
2
π
=
112
π
9
.
{\displaystyle ={16 \over 3}\int _{0}^{2\pi }{1-\cos(2\phi ) \over 2}d\phi +{64\pi \over 9}={16\pi \over 3}-{8 \over 3}\int _{0}^{2\pi }\cos(2\phi ){d(2\phi ) \over 2}+{64\pi \over 9}={112\pi \over 9}-{4 \over 3}\sin(2\phi )|_{0}^{2\pi }={112\pi \over 9}.}
Apskaičiuosime kūno sritį V , kuri apribota paviršiais
z
=
x
2
+
y
2
{\displaystyle z=x^{2}+y^{2}}
z
=
1
{\displaystyle z=1}
Oz ašies atžvilgiu
∭
V
(
x
2
+
y
2
)
d
z
d
y
d
z
.
{\displaystyle \iiint _{V}(x^{2}+y^{2})dzdydz.}
V į plokštumą xOy projektuojasi į skritulį
x
2
+
y
2
≤
1
,
{\displaystyle x^{2}+y^{2}\leq 1,}
ϕ
{\displaystyle \phi }
2
π
{\displaystyle 2\pi }
ρ
{\displaystyle \rho }
ρ
=
0
{\displaystyle \rho =0}
ρ
=
1
{\displaystyle \rho =1}
ρ
{\displaystyle \rho }
(
0
≤
ρ
≤
1
)
{\displaystyle (0\leq \rho \leq 1)}
Oxyz atitinka cilindras
x
2
+
y
2
=
ρ
2
.
{\displaystyle x^{2}+y^{2}=\rho ^{2}.}
V , gauname kitimą koordinčių z nuo reikšmės taškams gulinčių ant paraboloido
z
=
x
2
+
y
2
,
{\displaystyle z=x^{2}+y^{2},}
z
=
1
{\displaystyle z=1}
z
=
ρ
2
{\displaystyle z=\rho ^{2}}
z
=
1.
{\displaystyle z=1.}
I
z
z
=
∭
V
(
x
2
+
y
2
)
d
x
d
y
d
z
=
∫
0
2
π
d
ϕ
∫
0
1
d
ρ
∫
ρ
2
1
ρ
2
⋅
ρ
d
z
=
{\displaystyle I_{zz}=\iiint _{V}(x^{2}+y^{2})dxdydz=\int _{0}^{2\pi }d\phi \int _{0}^{1}d\rho \int _{\rho ^{2}}^{1}\rho ^{2}\cdot \rho dz=}
=
∫
0
2
π
d
ϕ
∫
0
1
ρ
3
z
|
ρ
2
1
d
ρ
=
∫
0
2
π
(
ρ
4
4
−
ρ
6
6
)
|
0
1
d
ϕ
=
1
12
∫
0
2
π
d
ϕ
=
π
6
.
{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{1}\rho ^{3}z|_{\rho ^{2}}^{1}d\rho =\int _{0}^{2\pi }({\rho ^{4} \over 4}-{\rho ^{6} \over 6})|_{0}^{1}d\phi ={1 \over 12}\int _{0}^{2\pi }d\phi ={\pi \over 6}.}
keisti
x
=
ρ
sin
θ
cos
ϕ
,
y
=
ρ
sin
θ
sin
ϕ
,
z
=
ρ
cos
θ
(
0
≤
ρ
<
∞
,
0
≤
ϕ
≤
2
π
,
0
≤
θ
≤
π
)
.
{\displaystyle x=\rho \sin \theta \cos \phi ,\;y=\rho \sin \theta \sin \phi ,\;z=\rho \cos \theta \;(0\leq \rho <\infty ,\;0\leq \phi \leq 2\pi ,\;0\leq \theta \leq \pi ).}
∭
V
f
(
x
,
y
,
z
)
d
x
d
y
d
z
=
∭
T
f
[
ρ
sin
θ
cos
ϕ
,
ρ
sin
θ
sin
ϕ
,
ρ
cos
θ
]
ρ
2
sin
θ
d
ρ
d
ϕ
d
θ
.
{\displaystyle \iiint _{V}f(x,y,z)dxdydz=\iiint _{T}f[\rho \sin \theta \cos \phi ,\;\rho \sin \theta \sin \phi ,\;\rho \cos \theta ]\rho ^{2}\sin \theta d\rho d\phi d\theta .}
x
2
+
y
2
+
z
2
=
ρ
2
sin
2
θ
cos
2
ϕ
+
ρ
2
sin
2
θ
sin
2
ϕ
+
ρ
2
cos
2
θ
=
ρ
2
sin
2
θ
+
ρ
2
cos
2
θ
=
ρ
2
.
{\displaystyle x^{2}+y^{2}+z^{2}=\rho ^{2}\sin ^{2}\theta \cos ^{2}\phi +\rho ^{2}\sin ^{2}\theta \sin ^{2}\phi +\rho ^{2}\cos ^{2}\theta =\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta =\rho ^{2}.}
Pavyzdžiai
Apskaičiuosime rutulio
x
2
+
y
2
+
z
2
≤
R
2
{\displaystyle x^{2}+y^{2}+z^{2}\leq R^{2}}
tūrį V :
V
=
∭
V
d
x
d
y
d
z
=
∭
T
ρ
2
sin
θ
d
ρ
d
θ
d
ϕ
=
∫
0
R
d
ρ
∫
0
π
d
θ
∫
0
2
π
ρ
2
sin
θ
d
ϕ
=
{\displaystyle V=\iiint _{V}dxdydz=\iiint _{T}\rho ^{2}\sin \theta d\rho d\theta d\phi =\int _{0}^{R}d\rho \int _{0}^{\pi }d\theta \int _{0}^{2\pi }\rho ^{2}\sin \theta d\phi =}
=
2
π
∫
0
R
ρ
2
d
ρ
∫
0
π
sin
θ
d
θ
=
−
2
π
∫
0
R
ρ
2
cos
θ
|
0
π
d
ρ
=
4
π
∫
0
R
ρ
2
d
ρ
=
4
π
ρ
3
3
|
0
R
=
4
π
R
3
3
.
{\displaystyle =2\pi \int _{0}^{R}\rho ^{2}d\rho \int _{0}^{\pi }\sin \theta d\theta =-2\pi \int _{0}^{R}\rho ^{2}\cos \theta |_{0}^{\pi }d\rho =4\pi \int _{0}^{R}\rho ^{2}d\rho =4\pi {\rho ^{3} \over 3}|_{0}^{R}={4\pi R^{3} \over 3}.}
Apskaičiuosime rutulio
x
2
+
y
2
+
z
2
≤
R
2
{\displaystyle x^{2}+y^{2}+z^{2}\leq R^{2}}
x
2
+
y
2
+
z
2
=
ρ
2
,
{\displaystyle x^{2}+y^{2}+z^{2}=\rho ^{2},}
I
0
=
∭
V
(
x
2
+
y
2
+
z
2
)
d
x
d
y
d
z
=
∭
T
ρ
2
ρ
2
sin
θ
d
ρ
d
θ
d
ϕ
=
∫
0
R
ρ
4
d
ρ
∫
0
π
sin
θ
d
θ
∫
0
2
π
d
ϕ
=
{\displaystyle I_{0}=\iiint _{V}(x^{2}+y^{2}+z^{2})dxdydz=\iiint _{T}\rho ^{2}\rho ^{2}\sin \theta d\rho d\theta d\phi =\int _{0}^{R}\rho ^{4}d\rho \int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =}
=
2
π
∫
0
R
ρ
4
d
ρ
∫
0
π
sin
θ
d
θ
=
2
π
∫
0
R
ρ
4
(
−
cos
θ
)
|
0
π
d
ρ
=
2
π
∫
0
R
ρ
4
(
1
+
1
)
d
ρ
=
4
π
ρ
5
5
|
0
R
=
4
π
R
5
5
.
{\displaystyle =2\pi \int _{0}^{R}\rho ^{4}d\rho \int _{0}^{\pi }\sin \theta d\theta =2\pi \int _{0}^{R}\rho ^{4}(-\cos \theta )|_{0}^{\pi }d\rho =2\pi \int _{0}^{R}\rho ^{4}(1+1)d\rho =4\pi {\rho ^{5} \over 5}|_{0}^{R}={4\pi R^{5} \over 5}.}
Nustatysime masės centro koordinates viršutinės pusės vienalyčio rutulio V spindulio R esančio centre koordinačių pradžios. Duotas pusrutulis apribotas paviršiais
z
=
R
2
−
x
2
−
y
2
{\displaystyle z={\sqrt {R^{2}-x^{2}-y^{2}}}}
z
=
0.
{\displaystyle z=0.}
x
c
=
y
c
=
0.
{\displaystyle x_{c}=y_{c}=0.}
z
c
,
{\displaystyle z_{c},}
z
c
=
∭
V
z
d
x
d
y
d
z
∭
V
d
x
d
y
d
z
=
∭
V
z
d
x
d
y
d
z
2
3
π
R
3
.
{\displaystyle z_{c}={\iiint _{V}zdxdydz \over \iiint _{V}dxdydz}={\iiint _{V}zdxdydz \over {2 \over 3}\pi R^{3}}.}
Pereidami į sferines koordinates, gauname
z
c
=
∫
0
2
π
d
ϕ
∫
0
π
/
2
sin
θ
cos
θ
d
θ
∫
0
R
ρ
3
d
ρ
2
3
π
R
3
=
R
4
4
∫
0
2
π
d
ϕ
∫
0
π
/
2
sin
θ
d
(
sin
θ
)
2
3
π
R
3
=
R
4
4
∫
0
2
π
sin
2
θ
2
|
0
π
/
2
d
ϕ
2
3
π
R
3
=
{\displaystyle z_{c}={\int _{0}^{2\pi }d\phi \int _{0}^{\pi /2}\sin \theta \cos \theta d\theta \int _{0}^{R}\rho ^{3}d\rho \over {2 \over 3}\pi R^{3}}={{R^{4} \over 4}\int _{0}^{2\pi }d\phi \int _{0}^{\pi /2}\sin \theta d(\sin \theta ) \over {2 \over 3}\pi R^{3}}={{R^{4} \over 4}\int _{0}^{2\pi }{\sin ^{2}\theta \over 2}|_{0}^{\pi /2}d\phi \over {2 \over 3}\pi R^{3}}=}
=
R
4
4
∫
0
2
π
1
2
d
ϕ
2
3
π
R
3
=
R
4
4
⋅
1
2
⋅
2
π
2
3
π
R
3
=
3
8
R
.
{\displaystyle ={{R^{4} \over 4}\int _{0}^{2\pi }{1 \over 2}d\phi \over {2 \over 3}\pi R^{3}}={{R^{4} \over 4}\cdot {1 \over 2}\cdot 2\pi \over {2 \over 3}\pi R^{3}}={3 \over 8}R.}
Apskaičiuosime masę pusrutulio V spindulio R , jeigu masės pasiskirstimas tankis kiekviename jo taške proporcingas atstumui taško nuo tam tikro fiksuoto taško O ant krašto pusrutulio pagrindo. Išrinksime koordinačių pradžią taške O , o plokštumą xOy pusrutulio taip, kad pusrutulio centras gulėtų ant ašies Oy .
Tada lygtys paviršiaus, apribojančio kūną V iš viršaus, užsirašis pavidale:
x
2
+
(
y
−
R
)
2
+
z
2
=
R
2
,
{\displaystyle x^{2}+(y-R)^{2}+z^{2}=R^{2},}
x
2
+
y
2
+
z
2
=
2
R
y
,
{\displaystyle x^{2}+y^{2}+z^{2}=2Ry,}
ρ
=
2
R
sin
θ
sin
ϕ
,
{\displaystyle \rho =2R\sin \theta \sin \phi ,}
masės pasiskirstimo tankis nustatomas formule
γ
=
k
x
2
+
y
2
+
z
2
,
{\displaystyle \gamma =k{\sqrt {x^{2}+y^{2}+z^{2}}},}
masės nustatymas reiškia apskaičiavimą integralo
m
=
k
∭
V
x
2
+
y
2
+
z
2
d
x
d
y
d
z
=
k
∭
T
ρ
⋅
ρ
2
sin
θ
d
ρ
d
ϕ
d
θ
=
{\displaystyle m=k\iiint _{V}{\sqrt {x^{2}+y^{2}+z^{2}}}dxdydz=k\iiint _{T}\rho \cdot \rho ^{2}\sin \theta d\rho d\phi d\theta =}
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{\displaystyle =k\int _{0}^{\pi }d\phi \int _{0}^{\pi \over 2}\sin \theta d\theta \int _{0}^{2R\sin \theta \sin \phi }\rho ^{3}d\rho =4kR^{4}\int _{0}^{\pi }\sin ^{4}\phi d\phi \int _{0}^{\pi \over 2}\sin ^{5}\theta d\theta =4kR^{4}\int _{0}^{\pi }\sin ^{4}\phi {4!! \over 5!!}d\phi =}
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{\displaystyle =4kR^{4}\cdot {4\cdot 2 \over 5\cdot 3}\int _{0}^{\pi }\sin ^{4}\phi d\phi ={32kR^{4} \over 15}\int _{0}^{\pi \over 2}2\sin ^{4}\phi d\phi ={32kR^{4} \over 15}\cdot 2\cdot {3 \over 4\cdot 2}\cdot {\pi \over 2}={32kR^{4} \over 15}\cdot {3\pi \over 8}={4k\pi R^{4} \over 5}.}
dvigubu faktorialu trigonometrijoje:
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{\displaystyle \int _{0}^{\pi \over 2}\sin ^{n}x\;dx=\int _{0}^{\pi \over 2}\cos ^{n}x\;dx={(n-1)!! \over n!!}\cdot {\pi \over 2},}
n lyginis;
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{\displaystyle \int _{0}^{\pi \over 2}\sin ^{n}x\;dx=\int _{0}^{\pi \over 2}\cos ^{n}x\;dx={(n-1)!! \over n!!},}
n nelyginis.
![{\displaystyle =\int _{0}^{1}{\mathsf {d}}x\int _{x^{2}}^{1}(x^{2}+y^{2}){\mathsf {d}}y=\int _{0}^{1}{\mathsf {d}}x\;(x^{2}y+{\frac {y^{3}}{3}})|_{x^{2}}^{1}=\int _{0}^{1}{\mathsf {d}}x\;[(x^{2}\cdot 1+{\frac {1^{3}}{3}})-(x^{2}\cdot x^{2}+{\frac {(x^{2})^{3}}{3}})]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22a70a331aa2fc8d883f773525cead370ec15dfa)
![{\displaystyle =\int _{0}^{1}[x^{2}+{\frac {1}{3}}-x^{4}-{\frac {x^{6}}{3}}]{\mathsf {d}}x=\left({\frac {x^{3}}{3}}+{\frac {x}{3}}-{\frac {x^{5}}{5}}-{\frac {x^{7}}{3\cdot 7}}\right)|_{0}^{1}=\left({\frac {1^{3}}{3}}+{\frac {1}{3}}-{\frac {1^{5}}{5}}-{\frac {1^{7}}{21}}\right)-\left({\frac {0^{3}}{3}}+{\frac {0}{3}}-{\frac {0^{5}}{5}}-{\frac {0^{7}}{21}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab7953ed363c1fd3869fce6c4def9427952bbf27)
![{\displaystyle =\int _{0}^{1}{\mathsf {d}}y\int _{0}^{\sqrt {y}}(x^{2}+y^{2}){\mathsf {d}}x=\int _{0}^{1}{\mathsf {d}}y\;({\frac {x^{3}}{3}}+y^{2}x)|_{0}^{\sqrt {y}}=\int _{0}^{1}{\mathsf {d}}y\;[({\frac {({\sqrt {y}})^{3}}{3}}+y^{2}{\sqrt {y}})-({\frac {0^{3}}{3}}+y^{2}\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9175bc7f8ca0cd8a29ab718051a73f575af0fdec)
![{\displaystyle =\int _{0}^{1}[{\frac {y^{3 \over 2}}{3}}+y^{5 \over 2}]{\mathsf {d}}y=\left({\frac {y^{{\frac {3}{2}}+1}}{3({\frac {3}{2}}+1)}}+{\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}\right)|_{0}^{1}=\left({\frac {y^{\frac {5}{2}}}{3\cdot {\frac {5}{2}}}}+{\frac {y^{\frac {7}{2}}}{\frac {5+2}{2}}}\right)|_{0}^{1}=\left({\frac {2y^{\frac {5}{2}}}{15}}+{\frac {2y^{\frac {7}{2}}}{7}}\right)|_{0}^{1}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e44bb85c9b6aa68a299152098e97478eba747ca)
![{\displaystyle =\int _{0}^{3}{\mathsf {d}}x\int _{x^{2}}^{9}(x^{2}+y^{2}){\mathsf {d}}y=\int _{0}^{3}{\mathsf {d}}x\;(x^{2}y+{\frac {y^{3}}{3}})|_{x^{2}}^{9}=\int _{0}^{1}{\mathsf {d}}x\;[(x^{2}\cdot 9+{\frac {9^{3}}{3}})-(x^{2}\cdot x^{2}+{\frac {(x^{2})^{3}}{3}})]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5f4e26fad0546b94e7db8e44b99fd77b5692133)
![{\displaystyle =\int _{0}^{3}[9x^{2}+{\frac {729}{3}}-x^{4}-{\frac {x^{6}}{3}}]{\mathsf {d}}x=\left({\frac {9x^{3}}{3}}+{\frac {729x}{3}}-{\frac {x^{5}}{5}}-{\frac {x^{7}}{3\cdot 7}}\right)|_{0}^{3}=\left({\frac {9\cdot 3^{3}}{3}}+{\frac {729\cdot 3}{3}}-{\frac {3^{5}}{5}}-{\frac {3^{7}}{21}}\right)-\left({\frac {9\cdot 0^{3}}{3}}+{\frac {729\cdot 0}{3}}-{\frac {0^{5}}{5}}-{\frac {0^{7}}{21}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af22a950d0311b4e2a2bb755dd2a888f7d4ecff9)
![{\displaystyle =\int _{0}^{9}{\mathsf {d}}y\int _{0}^{\sqrt {y}}(x^{2}+y^{2}){\mathsf {d}}x=\int _{0}^{9}{\mathsf {d}}y\;({\frac {x^{3}}{3}}+y^{2}x)|_{0}^{\sqrt {y}}=\int _{0}^{9}{\mathsf {d}}y\;[({\frac {({\sqrt {y}})^{3}}{3}}+y^{2}{\sqrt {y}})-({\frac {0^{3}}{3}}+y^{2}\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdbfccde61b3a30ac2d0f2ab93b23d81719684c6)
![{\displaystyle =\int _{0}^{9}[{\frac {y^{3 \over 2}}{3}}+y^{5 \over 2}]{\mathsf {d}}y=\left({\frac {y^{{\frac {3}{2}}+1}}{3({\frac {3}{2}}+1)}}+{\frac {y^{{\frac {5}{2}}+1}}{{\frac {5}{2}}+1}}\right)|_{0}^{9}=\left({\frac {y^{\frac {5}{2}}}{3\cdot {\frac {5}{2}}}}+{\frac {y^{\frac {7}{2}}}{\frac {5+2}{2}}}\right)|_{0}^{9}=\left({\frac {2y^{\frac {5}{2}}}{15}}+{\frac {2y^{\frac {7}{2}}}{7}}\right)|_{0}^{9}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd8fd75ae4b6450d6981b1488f9e145163ab89f1)
![{\displaystyle V=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}{\mathsf {d}}y\int _{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}{\mathsf {d}}z=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}{\mathsf {d}}y\;z|_{{\frac {x^{2}+y^{2}}{4}}-2}^{4-x^{2}-y^{2}}=\int _{0}^{1}{\mathsf {d}}x\int _{0}^{1}[(4-x^{2}-y^{2})-({\frac {x^{2}+y^{2}}{4}}-2)]{\mathsf {d}}y=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a13d4af4941dc815ba375dac01be51f287846b8)
![{\displaystyle =4\int _{0}^{\pi \over 2}(-{{\sqrt {(6-\rho ^{2})^{3}}} \over 3}-{\rho ^{4} \over 4})|_{0}^{\sqrt {2}}d\phi =-4\int _{0}^{\pi \over 2}[({{\sqrt {(6-2)^{3}}} \over 3}+{4 \over 4})-({{\sqrt {(6-0)^{3}}} \over 3}+{0 \over 4})]d\phi =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a755d44090de6ffb573d37ea737f6b5925cd49c2)
![{\displaystyle =-4\int _{0}^{\pi \over 2}[({{\sqrt {64}} \over 3}+1)-{{\sqrt {216}} \over 3}]d\phi =-4\int _{0}^{\pi \over 2}[({8 \over 3}+1)-{6{\sqrt {6}} \over 3}]d\phi ={4 \over 3}(6{\sqrt {6}}-11)\int _{0}^{\pi \over 2}d\phi ={4\pi \over 6}(6{\sqrt {6}}-11).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6b1b10f9bb53a55986ebf4998d6df9e6f9bd1cc)
![{\displaystyle =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(\int _{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho {\mathsf {d}}z\right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(z|_{0}^{\sqrt {a^{2}-\rho ^{2}}}\rho \right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\left(({\sqrt {a^{2}-\rho ^{2}}}-0)\rho \right){\mathsf {d}}\rho \right]{\mathsf {d}}\phi =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d2109fc6e97c8d93f716fc278e579d670058a7b)
![{\displaystyle =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\rho {\sqrt {a^{2}-\rho ^{2}}}\,{\mathsf {d}}\rho \right]{\mathsf {d}}\phi =\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }\rho {\sqrt {a^{2}-\rho ^{2}}}\,{\frac {{\mathsf {d}}(a^{2}-\rho ^{2})}{-2\rho }}\right]{\mathsf {d}}\phi =-{\frac {1}{2}}\int _{0}^{\pi \over 2}\left[\int _{0}^{a\cos \phi }{\sqrt {a^{2}-\rho ^{2}}}\,{\mathsf {d}}(a^{2}-\rho ^{2})\right]{\mathsf {d}}\phi =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93fadb2c316b55bf765d057be50ab2fde547a0a5)
![{\displaystyle =-{\frac {1}{2}}\int _{0}^{\pi \over 2}\left[{\frac {(a^{2}-\rho ^{2})^{3 \over 2}}{3 \over 2}}|_{0}^{a\cos \phi }\right]{\mathsf {d}}\phi =-{\frac {1}{2}}\cdot {\frac {2}{3}}\cdot \int _{0}^{\pi \over 2}\left[{\sqrt {(a^{2}-(a\cos \phi )^{2})^{3}}}-{\sqrt {(a^{2}-0^{2})^{3}}}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[{\sqrt {(a^{2}(1-\cos ^{2}\phi ))^{3}}}-{\sqrt {a^{6}}}\right]{\mathsf {d}}\phi =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6840c5a03e39f5ca2fd46f10d78a417487ac8e69)
![{\displaystyle =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[{\sqrt {a^{6}(1-\cos ^{2}\phi )^{3}}}-a^{3}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[a^{3}{\sqrt {(\sin ^{2}\phi )^{3}}}-a^{3}\right]{\mathsf {d}}\phi =-{\frac {1}{3}}\int _{0}^{\pi \over 2}\left[a^{3}\sin ^{3}\phi -a^{3}\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba567a82976721f4b1222d79b0ea9d04fecee539)
![{\displaystyle =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[{\frac {1}{4}}(3\sin \phi -\sin(3\phi ))-1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{12}}\int _{0}^{\pi \over 2}\left[3\sin \phi -\sin(3\phi )-4\right]{\mathsf {d}}\phi =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b3228c62a8d8cf8f6892e1d3ec9a4c5fb194050)
![{\displaystyle =-{\frac {a^{3}}{12}}\left[-3\cos \phi +{\frac {\cos(3\phi )}{3}}-4\phi \right]|_{0}^{\pi \over 2}=-{\frac {a^{3}}{12}}\left[-3\cos {\frac {\pi }{2}}+{\frac {\cos(3\cdot {\frac {\pi }{2}})}{3}}-4\cdot {\frac {\pi }{2}}-(-3\cos 0+{\frac {\cos(3\cdot 0)}{3}}-4\cdot 0)\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a4321811132e63045aa54d0bea4cec6211e0001)
![{\displaystyle =-{\frac {a^{3}}{12}}\left[-3\cdot 0+{\frac {0}{3}}-2\pi -(-3\cdot 1+{\frac {1}{3}})\right]=-{\frac {a^{3}}{12}}\left[-2\pi +3-{\frac {1}{3}}\right]=-{\frac {a^{3}}{12}}\cdot {\frac {-6\pi +9-1}{3}}=-{\frac {a^{3}}{12}}\cdot {\frac {-6\pi +8}{3}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69e75e3f1f69e5b66e5127bc7e4a264602d784bb)
![{\displaystyle V=-{\frac {a^{3}}{3}}\int _{0}^{\pi \over 2}\left[\sin ^{3}\phi -1\right]{\mathsf {d}}\phi =-{\frac {a^{3}}{3}}(\int _{0}^{\pi \over 2}\sin ^{3}\phi \;{\mathsf {d}}\phi -\int _{0}^{\pi \over 2}{\mathsf {d}}\phi )=-{\frac {a^{3}}{3}}({\frac {(3-1)!!}{3!!}}-({\pi \over 2}-0))=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0da1319d78b0809a79eabe00dffa7798e8e753d8)
![{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba35c893d1ed55762e5511ba861517c434d77b7a)
![{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fc8c550ec38ada598f8b45967f7ee5f6c66f1e5)
![{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {1}{32}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{64}}=0.147262155.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a921f377da60951f5a9c45fe60b54a2b7e92c30e)
![{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})+2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)+2\sin(2\cdot 0)+3\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8ad731f0cf21f6bb0848346cc046f0afa47ccc0)
![{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )+2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)+2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/759292ba969fa0c2a7b20969a24dfe001bf207f2)
![{\displaystyle ={\frac {1}{32}}[({\frac {1}{4}}\cdot 0+2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0+2\cdot 0)]={\frac {1}{32}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{64}}=0.147262155.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0ae53a8033dd98c6596f1358b918361be9593a6)
![{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47255130527c1110b7b8259f22f1722f8e76ef1f)
![{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d4fb37879c9a9c22dacad998121ceed00b2b405)
![{\displaystyle ={\frac {1}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {1}{2}}\cdot {\frac {3\pi }{2}}={\frac {3\pi }{4}}=2.35619449.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6df193e5b15b6023a5634980ed95d10fcb013963)
![{\displaystyle =128[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ec0b7e8eb2363b8d6104b31b9a95dd15041cec)
![{\displaystyle =128[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/809bd815c14f6cda721d58c7ecc29fcab5a8ece6)
![{\displaystyle =128[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]=128\cdot {\frac {3\pi }{2}}=192\pi =603.1857895.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41cb21c770036967b5b2784ff9bb6f3ecfe985fb)
![{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e630ef850a92c0e5ebeb05784bf7c341cf12674)
![{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/807971678ffc6461c920a8118d291cdcdd0bbc75)
![{\displaystyle ={\frac {6561}{32}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {6561}{32}}\cdot {\frac {3\pi }{2}}={\frac {19683\pi }{64}}=307.546875\pi =966.1870031.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/352af184e028fcd00033114a4a3a40404658dc92)
![{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot \sin(4\cdot {\frac {\pi }{2}})-2\sin(2\cdot {\frac {\pi }{2}})+3\cdot {\frac {\pi }{2}})-({\frac {1}{4}}\cdot \sin(4\cdot 0)-2\sin(2\cdot 0)+3\cdot 0)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f696ea4e23845c8f363bf50cf2b4ee5dffce7c67)
![{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot \sin(2\pi )-2\sin(\pi )+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot \sin(0)-2\sin(0))]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25e2fbb029790f7e7b6354bdd91d0c7f950429e3)
![{\displaystyle ={\frac {625}{2}}[({\frac {1}{4}}\cdot 0-2\cdot 0+{\frac {3\pi }{2}})-({\frac {1}{4}}\cdot 0-2\cdot 0)]={\frac {625}{2}}\cdot {\frac {3\pi }{2}}={\frac {1875\pi }{4}}=468.75\pi =1472.621556.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a666f8fb9fb67288468b9f91718767f70b311051)
![{\displaystyle =\int _{0}^{2\pi }d\phi \int _{0}^{2}[(4\rho ^{3}-\rho ^{5})\sin ^{2}\phi +{64 \over 9}\rho -{\rho ^{7} \over 3}+{8\rho ^{5} \over 3}-{64 \over 9}\rho ^{3}]d\rho =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93f718baa43ac711bccd57a856064a656073cbe0)
![{\displaystyle =\int _{0}^{2\pi }[(\rho ^{4}-{\rho ^{6} \over 6})\sin ^{2}\phi +{32 \over 9}\rho ^{2}-{\rho ^{8} \over 24}+{4\rho ^{6} \over 9}-{16 \over 9}\rho ^{4}]|_{0}^{2}d\phi =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54d57385083f8fa995982c7e6efdc6282e1976fd)
![{\displaystyle \iiint _{V}f(x,y,z)dxdydz=\iiint _{T}f[\rho \sin \theta \cos \phi ,\;\rho \sin \theta \sin \phi ,\;\rho \cos \theta ]\rho ^{2}\sin \theta d\rho d\phi d\theta .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5beb5ddd6e14ed537ee700e880138f19d2a0afa)
