Formulė Gauso - Ostrogradskio yra analogas Gryno formulės .
Tada kai formulė Gryno - Ostrogradskio suriša kreivinį integralą antrojo tipo uždara kreive su dvilypiu integralu plokščia sritimi, apribota šia kreive, tai formulė Gauso - Ostragradskio nustato ryšį tarp paviršinio integralo (antrojjo tipo) uždaru paviršiumi ir trilypiu integralu palei erdvinę sritį, apribotą šiuo paviršiumi.
Formulė Gauso - Ostrogradskio :
∭
V
(
∂
P
∂
x
+
∂
Q
∂
y
+
∂
R
∂
z
)
d
x
d
y
d
z
=
∬
S
P
d
y
d
z
+
Q
d
z
d
x
+
R
d
x
d
y
.
{\displaystyle \iiint _{V}\left({\frac {\partial P}{\partial x}}+{\frac {\partial Q}{\partial y}}+{\frac {\partial R}{\partial z}}\right)dxdydz=\iint _{S}Pdydz+Qdzdx+Rdxdy.}
Ji išreiškia paviršinį integralą bendro pavidalo palei išorinę pusę uždaro paviršiaus S per trilypį integralą palei trimatę sritį V , apribotą šiuo paviršiumi.
Formulę Gauso - Ostrogradksio galimą naudoti apskaičiavimui paviršinių integralų uždaru paviršiumi.
Vienam iš pritaikymų formulės Gauso - Ostrogradskio, paimkime uždavinį apie apskaičiavimą kūno tūrio su paviršiniu integralu, palei išorinę pusę paviršiaus, apribojantį šitą kūną.
Tikrai, jei sritis V turi nurodyta aukščiau formą, tai
P
(
x
;
y
;
z
)
=
x
,
Q
(
x
;
y
;
z
)
=
0
,
R
(
x
;
y
;
z
)
=
0
,
{\displaystyle P(x;\;y;\;z)=x,\quad Q(x;\;y;\;z)=0,\quad R(x;\;y;\;z)=0,}
pagal formulę Gauso - Ostrogradksio randame:
∬
S
x
d
y
d
z
=
∭
V
d
x
d
y
d
z
=
v
.
{\displaystyle \iint _{S}x\;dy\;dz=\iiint _{V}dx\;dy\;dz=v.}
Keisdami rolėmis x , y , z , gausime taip pat, kad
∬
S
y
d
z
d
x
=
v
,
∬
S
z
d
x
d
y
=
v
.
{\displaystyle \iint _{S}y\;dz\;dx=v,\quad \iint _{S}z\;dx\;dy=v.}
tokiu budu yra formulės:
∬
S
x
d
y
d
z
=
v
,
∬
S
y
d
z
d
x
=
v
,
∬
S
z
d
x
d
y
=
v
,
{\displaystyle \iint _{S}x\;dy\;dz=v,\quad \iint _{S}y\;dz\;dx=v,\quad \iint _{S}z\;dx\;dy=v,}
išreiškiančios kūno tūrį v per integralą palei išorinę pusę jo paviršiaus.
Paėmę funkciją
P
(
x
;
y
;
z
)
=
1
3
⋅
x
,
Q
(
x
;
y
;
z
)
=
1
3
⋅
y
,
R
(
x
;
y
;
z
)
=
1
3
⋅
z
,
{\displaystyle P(x;\;y;\;z)={\frac {1}{3}}\cdot x,\quad Q(x;\;y;\;z)={\frac {1}{3}}\cdot y,\quad R(x;\;y;\;z)={\frac {1}{3}}\cdot z,}
gausime formulę, išreiškiančią kūno tūrį per paviršinį integralą bendro pavidalo:
v
=
1
3
∬
S
x
d
y
d
z
+
y
d
z
d
x
+
z
d
x
d
y
.
{\displaystyle v={\frac {1}{3}}\iint _{S}x\;dy\;dz+y\;dz\;dx+z\;dx\;dy.}
Nes tada
∂
P
∂
x
+
∂
Q
∂
y
+
∂
R
∂
z
=
1
3
+
1
3
+
1
3
=
1
,
{\displaystyle {\frac {\partial P}{\partial x}}+{\frac {\partial Q}{\partial y}}+{\frac {\partial R}{\partial z}}={\frac {1}{3}}+{\frac {1}{3}}+{\frac {1}{3}}=1,}
P
=
x
3
,
Q
=
y
3
,
R
=
z
3
{\displaystyle P={\frac {x}{3}},\quad Q={\frac {y}{3}},\quad R={\frac {z}{3}}}
∭
V
(
∂
P
∂
x
+
∂
Q
∂
y
+
∂
R
∂
z
)
d
x
d
y
d
z
=
∭
V
d
x
d
y
d
z
=
v
.
{\displaystyle \iiint _{V}\left({\frac {\partial P}{\partial x}}+{\frac {\partial Q}{\partial y}}+{\frac {\partial R}{\partial z}}\right)dxdydz=\iiint _{V}dxdydz=v.}
Todėl kūno V , apriboto paviršiumi S tūris v lygūs:
v
=
∭
V
d
x
d
y
d
z
=
1
3
∬
S
x
d
y
d
z
+
y
d
z
d
x
+
z
d
x
d
y
.
{\displaystyle v=\iiint _{V}dx\;dy\;dz={\frac {1}{3}}\iint _{S}x\;dydz+y\;dzdx+z\;dxdy.}
Pavyzdis . Apskaičiuoti integralą
∬
S
x
3
d
y
d
z
+
y
3
d
z
d
x
+
z
3
d
x
d
y
{\displaystyle \iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy}
x
2
+
y
2
+
z
2
=
R
2
.
{\displaystyle x^{2}+y^{2}+z^{2}=R^{2}.}
Taikydami formulę Gauso - Ostragradksio, gauname:
∬
S
x
3
d
y
d
z
+
y
3
d
z
d
x
+
z
3
d
x
d
y
=
∭
V
(
3
x
2
+
3
y
2
+
3
z
2
)
d
x
d
y
d
z
=
3
∫
0
2
π
d
ϕ
∫
−
π
2
π
2
cos
θ
d
θ
∫
0
R
ρ
4
d
ρ
=
{\displaystyle \iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy=\iiint _{V}(3x^{2}+3y^{2}+3z^{2})dxdydz=3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta \int _{0}^{R}\rho ^{4}d\rho =}
=
3
∫
0
2
π
d
ϕ
∫
−
π
2
π
2
cos
θ
d
θ
ρ
5
5
|
0
R
=
3
⋅
R
5
5
∫
0
2
π
d
ϕ
∫
−
π
2
π
2
cos
θ
d
θ
=
3
R
5
5
∫
0
2
π
d
ϕ
sin
θ
|
−
π
2
π
2
=
3
R
5
5
∫
0
2
π
(
sin
π
2
−
sin
−
π
2
)
d
ϕ
=
{\displaystyle =3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta {\frac {\rho ^{5}}{5}}|_{0}^{R}=3\cdot {\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta ={\frac {3R^{5}}{5}}\int _{0}^{2\pi }d\phi \;\sin \theta |_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(\sin {\frac {\pi }{2}}-\sin {\frac {-\pi }{2}})d\phi =}
=
3
R
5
5
∫
0
2
π
(
1
−
(
−
1
)
)
d
ϕ
=
6
R
5
5
ϕ
|
0
2
π
=
6
R
5
5
⋅
2
π
=
12
π
R
5
5
.
{\displaystyle ={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(1-(-1))d\phi ={\frac {6R^{5}}{5}}\phi |_{0}^{2\pi }={\frac {6R^{5}}{5}}\cdot 2\pi ={\frac {12\pi R^{5}}{5}}.}
Teisingiau skaičiuoti taip (
x
2
+
y
2
+
z
2
=
ρ
2
{\displaystyle x^{2}+y^{2}+z^{2}=\rho ^{2}}
∬
S
x
3
d
y
d
z
+
y
3
d
z
d
x
+
z
3
d
x
d
y
=
∭
V
(
3
x
2
+
3
y
2
+
3
z
2
)
d
x
d
y
d
z
=
3
∫
0
2
π
d
ϕ
∫
0
π
sin
θ
d
θ
∫
0
R
ρ
2
⋅
ρ
2
d
ρ
=
{\displaystyle \iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy=\iiint _{V}(3x^{2}+3y^{2}+3z^{2})dxdydz=3\int _{0}^{2\pi }d\phi \int _{0}^{\pi }\sin \theta d\theta \int _{0}^{R}\rho ^{2}\cdot \rho ^{2}d\rho =}
=
3
∫
0
2
π
d
ϕ
∫
0
π
sin
θ
d
θ
ρ
5
5
|
0
R
=
3
R
5
5
∫
0
2
π
d
ϕ
∫
0
π
sin
θ
d
θ
=
3
R
5
5
∫
0
2
π
d
ϕ
(
−
cos
θ
)
|
0
π
=
{\displaystyle =3\int _{0}^{2\pi }d\phi \int _{0}^{\pi }\sin \theta d\theta {\frac {\rho ^{5}}{5}}|_{0}^{R}=3{\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi \int _{0}^{\pi }\sin \theta d\theta =3{\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi (-\cos \theta )|_{0}^{\pi }=}
=
3
R
5
5
∫
0
2
π
d
ϕ
(
−
[
−
1
−
1
]
)
=
6
R
5
5
∫
0
2
π
d
ϕ
=
6
R
5
5
⋅
(
2
π
−
0
)
=
12
π
R
5
5
.
{\displaystyle =3{\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi (-[-1-1])={\frac {6R^{5}}{5}}\int _{0}^{2\pi }d\phi ={\frac {6R^{5}}{5}}\cdot (2\pi -0)={\frac {12\pi R^{5}}{5}}.}
Patikrinsime apskaičiuodami
∬
S
x
3
d
y
d
z
,
∬
S
y
3
d
z
d
x
{\displaystyle \iint _{S}x^{3}\mathbf {d} y\mathbf {d} z,\;\iint _{S}y^{3}\mathbf {d} z\mathbf {d} x}
∬
S
z
3
d
x
d
y
{\displaystyle \iint _{S}z^{3}\mathbf {d} x\mathbf {d} y}
x
2
=
R
2
−
y
2
−
z
2
,
{\displaystyle x^{2}=R^{2}-y^{2}-z^{2},}
x
=
R
2
−
y
2
−
z
2
;
{\displaystyle x={\sqrt {R^{2}-y^{2}-z^{2}}};}
∂
x
∂
y
=
∂
(
R
2
−
y
2
−
z
2
)
∂
y
=
−
2
y
2
R
2
−
y
2
−
z
2
=
−
y
R
2
−
y
2
−
z
2
;
{\displaystyle {\frac {\partial x}{\partial y}}={\frac {\partial ({\sqrt {R^{2}-y^{2}-z^{2}}})}{\partial y}}={\frac {-2y}{2{\sqrt {R^{2}-y^{2}-z^{2}}}}}={\frac {-y}{\sqrt {R^{2}-y^{2}-z^{2}}}};}
∂
x
∂
z
=
∂
(
R
2
−
y
2
−
z
2
)
∂
z
=
−
2
z
2
R
2
−
y
2
−
z
2
=
−
z
R
2
−
y
2
−
z
2
.
{\displaystyle {\frac {\partial x}{\partial z}}={\frac {\partial ({\sqrt {R^{2}-y^{2}-z^{2}}})}{\partial z}}={\frac {-2z}{2{\sqrt {R^{2}-y^{2}-z^{2}}}}}={\frac {-z}{\sqrt {R^{2}-y^{2}-z^{2}}}}.}
V
x
=
∬
S
x
3
1
+
(
∂
x
∂
y
)
2
+
(
∂
x
∂
z
)
2
d
y
d
z
=
{\displaystyle V_{x}=\iint _{S}x^{3}{\sqrt {1+\left({\frac {\partial x}{\partial y}}\right)^{2}+\left({\frac {\partial x}{\partial z}}\right)^{2}}}\mathbf {d} y\mathbf {d} z=}
=
∬
S
x
3
1
+
(
−
y
R
2
−
y
2
−
z
2
)
2
+
(
−
z
R
2
−
y
2
−
z
2
)
2
d
y
d
z
=
{\displaystyle =\iint _{S}x^{3}{\sqrt {1+\left({\frac {-y}{\sqrt {R^{2}-y^{2}-z^{2}}}}\right)^{2}+\left({\frac {-z}{\sqrt {R^{2}-y^{2}-z^{2}}}}\right)^{2}}}\mathbf {d} y\mathbf {d} z=}
=
∬
S
x
3
1
+
y
2
R
2
−
y
2
−
z
2
+
z
2
R
2
−
y
2
−
z
2
d
y
d
z
=
{\displaystyle =\iint _{S}x^{3}{\sqrt {1+{\frac {y^{2}}{R^{2}-y^{2}-z^{2}}}+{\frac {z^{2}}{R^{2}-y^{2}-z^{2}}}}}\mathbf {d} y\mathbf {d} z=}
=
∬
S
x
3
R
2
−
y
2
−
z
2
+
y
2
+
z
2
R
2
−
y
2
−
z
2
d
y
d
z
=
{\displaystyle =\iint _{S}x^{3}{\sqrt {\frac {R^{2}-y^{2}-z^{2}+y^{2}+z^{2}}{R^{2}-y^{2}-z^{2}}}}\mathbf {d} y\mathbf {d} z=}
=
∬
S
x
3
R
2
R
2
−
y
2
−
z
2
d
y
d
z
=
∬
S
(
R
2
−
y
2
−
z
2
)
3
R
2
R
2
−
y
2
−
z
2
d
y
d
z
=
∬
S
(
R
2
−
y
2
−
z
2
)
R
d
y
d
z
=
{\displaystyle =\iint _{S}x^{3}{\sqrt {\frac {R^{2}}{R^{2}-y^{2}-z^{2}}}}\mathbf {d} y\mathbf {d} z=\iint _{S}{\sqrt {(R^{2}-y^{2}-z^{2})^{3}}}{\sqrt {\frac {R^{2}}{R^{2}-y^{2}-z^{2}}}}\mathbf {d} y\mathbf {d} z=\iint _{S}(R^{2}-y^{2}-z^{2})R\mathbf {d} y\mathbf {d} z=}
=
∬
S
(
R
2
−
ρ
2
)
R
ρ
d
ρ
d
ϕ
=
∫
0
2
π
(
∫
0
R
(
R
3
ρ
−
R
ρ
3
)
d
ρ
)
d
ϕ
=
{\displaystyle =\iint _{S}(R^{2}-\rho ^{2})R\rho \mathbf {d} \rho \mathbf {d} \phi =\int _{0}^{2\pi }\left(\int _{0}^{R}(R^{3}\rho -R\rho ^{3})\mathbf {d} \rho \right)\mathbf {d} \phi =}
=
∫
0
2
π
(
R
3
ρ
2
2
−
R
ρ
4
4
)
|
0
R
d
ϕ
=
∫
0
2
π
(
R
3
R
2
2
−
R
R
4
4
)
d
ϕ
=
{\displaystyle =\int _{0}^{2\pi }\left(R^{3}{\frac {\rho ^{2}}{2}}-R{\frac {\rho ^{4}}{4}}\right)|_{0}^{R}\mathbf {d} \phi =\int _{0}^{2\pi }\left(R^{3}{\frac {R^{2}}{2}}-R{\frac {R^{4}}{4}}\right)\mathbf {d} \phi =}
=
∫
0
2
π
(
R
5
2
−
R
5
4
)
d
ϕ
=
∫
0
2
π
R
5
4
d
ϕ
=
R
5
4
⋅
2
π
=
π
R
5
2
.
{\displaystyle =\int _{0}^{2\pi }\left({\frac {R^{5}}{2}}-{\frac {R^{5}}{4}}\right)\mathbf {d} \phi =\int _{0}^{2\pi }{\frac {R^{5}}{4}}\mathbf {d} \phi ={\frac {R^{5}}{4}}\cdot 2\pi ={\frac {\pi R^{5}}{2}}.}
Kadangi ir teigiama ir neigiama kryptimi reikia apskaičiuoti, tai
V
X
=
2
V
x
=
2
⋅
π
R
5
2
=
π
R
5
.
{\displaystyle V_{X}=2V_{x}=2\cdot {\frac {\pi R^{5}}{2}}=\pi R^{5}.}
x
3
{\displaystyle x^{3}}
y
3
{\displaystyle y^{3}}
z
3
{\displaystyle z^{3}}
M
=
3
V
X
=
3
π
R
5
.
{\displaystyle M=3V_{X}=3\pi R^{5}.}
Gauta masė (skardinės sferos tankis priklauso nuo x reikšmės trečiame laipsnyje) nesutampa su Gauso formulės logika.
Update . Pagal oficialią teoriją taip ir neturi sutapti atsakymai ir kas ką tik buvo apskaičiuota (
M
=
3
V
X
=
3
π
R
5
{\displaystyle M=3V_{X}=3\pi R^{5}}
Pastaba. Rutulio paviršiaus plotą įmanoma apskaičiuoti cilindinėse koordinatėse. Bandydami, gauname:
R
2
=
x
2
+
y
2
+
z
2
,
{\displaystyle R^{2}=x^{2}+y^{2}+z^{2},}
z
=
R
2
−
x
2
−
y
2
;
{\displaystyle z={\sqrt {R^{2}-x^{2}-y^{2}}};}
z
x
′
=
−
2
x
2
R
2
−
x
2
−
y
2
=
−
x
R
2
−
x
2
−
y
2
;
{\displaystyle z_{x}'={\frac {-2x}{2{\sqrt {R^{2}-x^{2}-y^{2}}}}}={\frac {-x}{\sqrt {R^{2}-x^{2}-y^{2}}}};}
z
y
′
=
−
2
y
2
R
2
−
x
2
−
y
2
=
−
y
R
2
−
x
2
−
y
2
;
{\displaystyle z_{y}'={\frac {-2y}{2{\sqrt {R^{2}-x^{2}-y^{2}}}}}={\frac {-y}{\sqrt {R^{2}-x^{2}-y^{2}}}};}
1
+
(
z
x
′
)
2
+
(
z
y
′
)
2
=
1
+
(
−
x
R
2
−
x
2
−
y
2
)
2
+
(
−
y
R
2
−
x
2
−
y
2
)
2
=
{\displaystyle {\sqrt {1+(z_{x}')^{2}+(z_{y}')^{2}}}={\sqrt {1+\left({\frac {-x}{\sqrt {R^{2}-x^{2}-y^{2}}}}\right)^{2}+\left({\frac {-y}{\sqrt {R^{2}-x^{2}-y^{2}}}}\right)^{2}}}=}
=
1
+
x
2
R
2
−
x
2
−
y
2
+
y
2
R
2
−
x
2
−
y
2
=
R
2
−
x
2
−
y
2
+
x
2
+
y
2
R
2
−
x
2
−
y
2
=
R
2
R
2
−
x
2
−
y
2
;
{\displaystyle ={\sqrt {1+{\frac {x^{2}}{R^{2}-x^{2}-y^{2}}}+{\frac {y^{2}}{R^{2}-x^{2}-y^{2}}}}}={\sqrt {\frac {R^{2}-x^{2}-y^{2}+x^{2}+y^{2}}{R^{2}-x^{2}-y^{2}}}}={\sqrt {\frac {R^{2}}{R^{2}-x^{2}-y^{2}}}};}
x
2
+
y
2
=
ρ
2
{\displaystyle x^{2}+y^{2}=\rho ^{2}}
Iš internetinio integratoriaus
∫
0
R
ρ
R
2
R
2
−
ρ
2
d
ρ
=
(
ρ
2
−
R
2
)
R
2
R
2
−
ρ
2
|
0
R
=
−
R
R
2
−
ρ
2
|
0
R
;
{\displaystyle \int _{0}^{R}\rho {\sqrt {\frac {R^{2}}{R^{2}-\rho ^{2}}}}\;d\rho =(\rho ^{2}-R^{2}){\sqrt {\frac {R^{2}}{R^{2}-\rho ^{2}}}}|_{0}^{R}=-R{\sqrt {R^{2}-\rho ^{2}}}|_{0}^{R};}
S
p
a
v
.
=
∬
s
1
+
(
z
x
′
)
2
+
(
z
y
′
)
2
d
x
d
y
=
∬
s
R
2
R
2
−
x
2
−
y
2
d
x
d
y
=
∫
0
2
π
d
ϕ
∫
0
R
ρ
R
2
R
2
−
ρ
2
d
ρ
=
{\displaystyle S_{pav.}=\iint _{s}{\sqrt {1+(z_{x}')^{2}+(z_{y}')^{2}}}\;dx\;dy=\iint _{s}{\sqrt {\frac {R^{2}}{R^{2}-x^{2}-y^{2}}}}\;dx\;dy=\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho {\sqrt {\frac {R^{2}}{R^{2}-\rho ^{2}}}}\;d\rho =}
=
∫
0
2
π
−
R
R
2
−
ρ
2
|
0
R
d
ϕ
=
−
R
∫
0
2
π
(
R
2
−
R
2
−
R
2
−
0
2
)
d
ϕ
=
−
R
∫
0
2
π
(
0
−
R
2
)
d
ϕ
=
R
2
∫
0
2
π
d
ϕ
=
2
π
R
2
.
{\displaystyle =\int _{0}^{2\pi }-R{\sqrt {R^{2}-\rho ^{2}}}|_{0}^{R}d\phi =-R\int _{0}^{2\pi }({\sqrt {R^{2}-R^{2}}}-{\sqrt {R^{2}-0^{2}}})d\phi =-R\int _{0}^{2\pi }(0-{\sqrt {R^{2}}})d\phi =R^{2}\int _{0}^{2\pi }d\phi =2\pi R^{2}.}
Taigi, visas rutulio paviršiaus plotas susideda iš dviejų pusrutulių, todėl
S
v
i
s
a
s
=
2
⋅
S
p
a
v
.
=
2
⋅
2
π
R
2
=
4
π
R
2
.
{\displaystyle S_{visas}=2\cdot S_{pav.}=2\cdot 2\pi R^{2}=4\pi R^{2}.}
Kitaip patikrinsime apskaičiuodami
∬
S
x
3
d
y
d
z
,
∬
S
y
3
d
z
d
x
{\displaystyle \iint _{S}x^{3}\mathbf {d} y\mathbf {d} z,\;\iint _{S}y^{3}\mathbf {d} z\mathbf {d} x}
∬
S
z
3
d
x
d
y
{\displaystyle \iint _{S}z^{3}\mathbf {d} x\mathbf {d} y}
x
2
=
R
2
−
y
2
−
z
2
,
{\displaystyle x^{2}=R^{2}-y^{2}-z^{2},}
x
=
R
2
−
y
2
−
z
2
.
{\displaystyle x={\sqrt {R^{2}-y^{2}-z^{2}}}.}
V
x
=
∬
S
x
3
d
y
d
z
=
∬
S
(
R
2
−
y
2
−
z
2
)
3
d
y
d
z
=
∬
S
(
R
2
−
ρ
2
)
3
ρ
d
ρ
d
ϕ
=
{\displaystyle V_{x}=\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-y^{2}-z^{2}}}\right)^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-\rho ^{2}}}\right)^{3}\rho \mathbf {d} \rho \mathbf {d} \phi =}
=
∫
0
2
π
(
∫
0
R
ρ
(
R
2
−
ρ
2
)
3
d
ρ
)
d
ϕ
.
{\displaystyle =\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi .}
Pasinaudodami internetiniu integratoriumi , gauname, kad
∫
0
R
ρ
(
R
2
−
ρ
2
)
3
d
ρ
=
−
1
5
(
(
R
−
ρ
)
(
R
+
ρ
)
)
5
/
2
|
0
R
=
{\displaystyle \int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho =-{\frac {1}{5}}((R-\rho )(R+\rho ))^{5/2}|_{0}^{R}=}
=
−
1
5
(
(
R
−
R
)
(
R
+
R
)
)
5
/
2
−
(
−
1
5
(
(
R
−
0
)
(
R
+
0
)
)
5
/
2
)
=
{\displaystyle =-{\frac {1}{5}}((R-R)(R+R))^{5/2}-\left(-{\frac {1}{5}}((R-0)(R+0))^{5/2}\right)=}
=
0
+
1
5
(
(
R
−
0
)
(
R
+
0
)
)
5
/
2
=
1
5
(
R
2
)
5
/
2
=
R
5
5
;
{\displaystyle =0+{\frac {1}{5}}((R-0)(R+0))^{5/2}={\frac {1}{5}}(R^{2})^{5/2}={\frac {R^{5}}{5}};}
V
x
=
∫
0
2
π
(
∫
0
R
ρ
(
R
2
−
ρ
2
)
3
d
ρ
)
d
ϕ
=
∫
0
2
π
R
5
5
d
ϕ
=
2
π
R
5
5
.
{\displaystyle V_{x}=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi =\int _{0}^{2\pi }{\frac {R^{5}}{5}}\mathbf {d} \phi ={\frac {2\pi R^{5}}{5}}.}
Kadangi reikia dviejų rutulio pusrutulių (teigiama ir neigiama Ox kryptimi), tai
V
X
=
2
V
x
=
2
⋅
2
π
R
5
5
=
4
π
R
5
5
.
{\displaystyle V_{X}=2V_{x}=2\cdot {\frac {2\pi R^{5}}{5}}={\frac {4\pi R^{5}}{5}}.}
Kadangi, pagal sąlyga bus
V
X
=
V
Y
=
V
Z
,
{\displaystyle V_{X}=V_{Y}=V_{Z},}
V
=
3
V
X
=
3
⋅
4
π
R
5
5
=
12
π
R
5
5
.
{\displaystyle V=3V_{X}=3\cdot {\frac {4\pi R^{5}}{5}}={\frac {12\pi R^{5}}{5}}.}
Ką norėta rasti ir kas rasta . Norėta apskaičiuoti (kaip supranta redaguotojas) sferos iš skardos masę. Skardos tankis vienu skaičiavimu kinta tik priklausomai nuo Ox koordinatės pagal funkcija
γ
(
x
)
=
x
3
.
{\displaystyle \gamma (x)=x^{3}.}
y koordinatės pagal funkciją
γ
(
y
)
=
y
3
.
{\displaystyle \gamma (y)=y^{3}.}
z koordinatės pagal funkciją
γ
(
z
)
=
z
3
.
{\displaystyle \gamma (z)=z^{3}.}
∬
S
z
3
d
x
d
y
{\displaystyle \iint _{S}z^{3}\mathbf {d} x\mathbf {d} y}
M
=
6
π
R
3
.
{\displaystyle M=6\pi R^{3}.}
M
=
6
π
R
3
{\displaystyle M=6\pi R^{3}}
M
=
∬
S
x
3
d
y
d
z
+
∬
S
y
3
d
z
d
x
+
∬
S
z
3
d
x
d
y
.
{\displaystyle M=\iint _{S}x^{3}dydz+\iint _{S}y^{3}dzdx+\iint _{S}z^{3}dxdy.}
M
=
12
π
R
5
5
.
{\displaystyle M={\frac {12\pi R^{5}}{5}}.}
integravimas paviršiumi , nes tik su iškraipyta prasme integravimas paviršiumi Gauso formulė gali egzistuoti kaip teisinga formulė.Gauto rezultato
∬
S
z
3
d
x
d
y
=
1
3
⋅
12
π
R
5
5
=
4
π
R
5
5
{\displaystyle \iint _{S}z^{3}\mathbf {d} x\mathbf {d} y={\frac {1}{3}}\cdot {\frac {12\pi R^{5}}{5}}={\frac {4\pi R^{5}}{5}}}
R
=
10
{\displaystyle R=10}
xOy ; sferos centras yra taškas O , sferos projekcija į xOy plokštumą yra skritulys, kurio centras koordinačių pradžios taškas O ; skritulio formulė yra
x
2
+
y
2
≤
R
2
;
{\displaystyle x^{2}+y^{2}\leq R^{2};}
x
2
+
y
2
=
R
2
;
{\displaystyle x^{2}+y^{2}=R^{2};}
xOy spindulys
r
=
10
{\displaystyle r=10}
π
r
2
=
π
⋅
10
2
=
100
π
=
314.1592653589793
{\displaystyle \pi r^{2}=\pi \cdot 10^{2}=100\pi =314.1592653589793}
Oz ašiai ir atstumas tarp strypų ant xOy plokštumos yra vienodas; kiekvienas strypas susikerta su sferos paviršiumi ir kiekvieno strypo ilgis yra nuo plokštumos xOy iki susikirtimo su sferos paviršiumi (mes skaičiuojame tik vienam pusrutuliui, tik teigiama Oz ašimi); trumpiausias strypo ilgis yra 0, o ilgiausias strypo ilgis yra iš centro O ir lygus R=10; dabar kiekvieną strypo ilgį reikia pakelti kubu, nes
z
3
;
{\displaystyle z^{3};}
O (sutampančiam su ašimi Oz ) ilgis yra
z
3
=
R
3
=
10
3
=
1000
;
{\displaystyle z^{3}=R^{3}=10^{3}=1000;}
4
π
R
5
5
=
4
π
10
5
5
=
80000
π
=
251327.412287
{\displaystyle {\frac {4\pi R^{5}}{5}}={\frac {4\pi 10^{5}}{5}}=80000\pi =251327.412287}
xOy ir lygiagrečių ašiai Oz ilgių suma (dviejų pusrutulių).
Pavyzdis . Apskaičiuoti integralą
∬
S
x
2
d
y
d
z
+
0
d
z
d
x
+
0
d
x
d
y
{\displaystyle \iint _{S}x^{2}\;dydz+0\;dzdx+0\;dxdy}
x
2
+
y
2
+
z
2
=
R
2
.
{\displaystyle x^{2}+y^{2}+z^{2}=R^{2}.}
Taikydami formulę Gauso, gauname:
∬
S
x
2
d
y
d
z
=
∭
V
(
2
x
+
0
+
0
)
d
x
d
y
d
z
=
2
∫
0
2
π
d
ϕ
∫
−
π
2
π
2
cos
θ
d
θ
∫
0
R
ρ
3
d
ρ
=
{\displaystyle \iint _{S}x^{2}dydz=\iiint _{V}(2x+0+0)dxdydz=2\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta \int _{0}^{R}\rho ^{3}d\rho =}
=
2
∫
0
2
π
d
ϕ
∫
−
π
2
π
2
cos
θ
d
θ
ρ
4
4
|
0
R
=
2
⋅
R
4
4
∫
0
2
π
d
ϕ
∫
−
π
2
π
2
cos
θ
d
θ
=
R
4
2
∫
0
2
π
d
ϕ
sin
θ
|
−
π
2
π
2
=
R
4
2
∫
0
2
π
(
sin
π
2
−
sin
−
π
2
)
d
ϕ
=
{\displaystyle =2\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta {\frac {\rho ^{4}}{4}}|_{0}^{R}=2\cdot {\frac {R^{4}}{4}}\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta ={\frac {R^{4}}{2}}\int _{0}^{2\pi }d\phi \;\sin \theta |_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}={\frac {R^{4}}{2}}\int _{0}^{2\pi }(\sin {\frac {\pi }{2}}-\sin {\frac {-\pi }{2}})d\phi =}
=
R
4
2
∫
0
2
π
(
1
−
(
−
1
)
)
d
ϕ
=
R
4
ϕ
|
0
2
π
=
R
4
⋅
2
π
=
2
π
R
4
.
{\displaystyle ={\frac {R^{4}}{2}}\int _{0}^{2\pi }(1-(-1))d\phi =R^{4}\phi |_{0}^{2\pi }=R^{4}\cdot 2\pi =2\pi R^{4}.}
Neteisingai skaičiuota, nes x polinėse ir sferinėse koordinatėse nėra
ρ
,
{\displaystyle \rho ,}
x
=
R
2
−
y
2
−
z
2
{\displaystyle x={\sqrt {R^{2}-y^{2}-z^{2}}}}
ρ
{\displaystyle \rho }
ρ
2
=
R
2
=
x
2
+
y
2
+
z
2
,
{\displaystyle \rho ^{2}=R^{2}=x^{2}+y^{2}+z^{2},}
x išreikšti galima taip:
x
=
R
sin
θ
cos
ϕ
{\displaystyle x=R\sin \theta \cos \phi }
x
=
ρ
sin
θ
cos
ϕ
{\displaystyle x=\rho \sin \theta \cos \phi }
x pakeisti
R
2
−
y
2
−
z
2
=
R
2
−
ρ
2
{\displaystyle {\sqrt {R^{2}-y^{2}-z^{2}}}={\sqrt {R^{2}-\rho ^{2}}}}
Kitaip patikrinsime apskaičiuodami
∬
S
x
2
d
y
d
z
.
{\displaystyle \iint _{S}x^{2}\mathbf {d} y\mathbf {d} z.}
x
2
=
R
2
−
y
2
−
z
2
,
{\displaystyle x^{2}=R^{2}-y^{2}-z^{2},}
x
=
R
2
−
y
2
−
z
2
.
{\displaystyle x={\sqrt {R^{2}-y^{2}-z^{2}}}.}
V
x
=
∬
S
x
2
d
y
d
z
=
∬
S
(
R
2
−
y
2
−
z
2
)
2
d
y
d
z
=
∬
S
(
R
2
−
ρ
2
)
2
ρ
d
ρ
d
ϕ
=
{\displaystyle V_{x}=\iint _{S}x^{2}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-y^{2}-z^{2}}}\right)^{2}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-\rho ^{2}}}\right)^{2}\rho \mathbf {d} \rho \mathbf {d} \phi =}
=
∫
0
2
π
(
∫
0
R
ρ
(
R
2
−
ρ
2
)
d
ρ
)
d
ϕ
.
{\displaystyle =\int _{0}^{2\pi }\left(\int _{0}^{R}\rho (R^{2}-\rho ^{2})\mathbf {d} \rho \right)\mathbf {d} \phi .}
Pasinaudodami internetiniu integratoriumi , gauname, kad
∫
0
R
ρ
(
R
2
−
ρ
2
)
d
ρ
=
∫
0
R
(
ρ
R
2
−
ρ
3
)
d
ρ
=
(
ρ
2
R
2
2
−
ρ
4
4
)
|
0
R
=
{\displaystyle \int _{0}^{R}\rho (R^{2}-\rho ^{2})\mathbf {d} \rho =\int _{0}^{R}(\rho R^{2}-\rho ^{3})\mathbf {d} \rho =({\frac {\rho ^{2}R^{2}}{2}}-{\frac {\rho ^{4}}{4}})|_{0}^{R}=}
=
R
2
⋅
R
2
2
−
R
4
4
=
R
4
4
.
{\displaystyle ={\frac {R^{2}\cdot R^{2}}{2}}-{\frac {R^{4}}{4}}={\frac {R^{4}}{4}}.}
V
x
=
∫
0
2
π
(
∫
0
R
ρ
(
R
2
−
ρ
2
)
2
d
ρ
)
d
ϕ
=
∫
0
2
π
R
4
4
d
ϕ
=
2
π
R
4
4
=
π
R
4
2
.
{\displaystyle V_{x}=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{2}}}\mathbf {d} \rho \right)\mathbf {d} \phi =\int _{0}^{2\pi }{\frac {R^{4}}{4}}\mathbf {d} \phi ={\frac {2\pi R^{4}}{4}}={\frac {\pi R^{4}}{2}}.}
Kadangi reikia dviejų rutulio pusrutulių (teigiama ir neigiama Ox kryptimi), tai
V
X
=
2
V
x
=
2
⋅
π
R
4
2
=
π
R
4
.
{\displaystyle V_{X}=2V_{x}=2\cdot {\frac {\pi R^{4}}{2}}=\pi R^{4}.}
Atsakymai
2
π
R
4
{\displaystyle 2\pi R^{4}}
π
R
4
{\displaystyle \pi R^{4}}
π
R
4
2
{\displaystyle {\frac {\pi R^{4}}{2}}}
Šiaip, atsakymas
π
R
4
{\displaystyle \pi R^{4}}
Teisingas skaičiavimas pagal tūrio formulę (skaičiavimas sferinėje koordinačių sistemoje). Mums prireiks šitos trigonometrinės formulės
cos
(
2
θ
)
=
1
−
2
sin
2
θ
,
{\displaystyle \cos(2\theta )=1-2\sin ^{2}\theta ,\;}
sin
2
θ
=
1
−
cos
(
2
θ
)
2
.
{\displaystyle \sin ^{2}\theta ={\frac {1-\cos(2\theta )}{2}}.}
x
=
ρ
sin
θ
cos
ϕ
{\displaystyle x=\rho \sin \theta \cos \phi }
∬
S
x
2
d
y
d
z
=
∭
V
(
2
x
+
0
+
0
)
d
x
d
y
d
z
=
2
∫
0
2
π
cos
ϕ
d
ϕ
∫
0
π
sin
2
θ
d
θ
∫
0
R
ρ
⋅
ρ
2
d
ρ
=
{\displaystyle \iint _{S}x^{2}dydz=\iiint _{V}(2x+0+0)dxdydz=2\int _{0}^{2\pi }\cos \phi \;d\phi \int _{0}^{\pi }\sin ^{2}\theta \;d\theta \int _{0}^{R}\rho \cdot \rho ^{2}\;d\rho =}
=
2
∫
0
2
π
cos
ϕ
d
ϕ
∫
0
π
sin
2
θ
d
θ
ρ
4
4
|
0
R
=
2
⋅
R
4
4
∫
0
2
π
cos
ϕ
d
ϕ
∫
0
π
sin
2
θ
d
θ
=
{\displaystyle =2\int _{0}^{2\pi }\cos \phi \;d\phi \int _{0}^{\pi }\sin ^{2}\theta \;d\theta \;{\frac {\rho ^{4}}{4}}|_{0}^{R}=2\cdot {\frac {R^{4}}{4}}\int _{0}^{2\pi }\cos \phi \;d\phi \int _{0}^{\pi }\sin ^{2}\theta \;d\theta =}
=
R
4
2
⋅
1
2
∫
0
2
π
cos
ϕ
d
ϕ
∫
0
π
(
1
−
cos
(
2
θ
)
)
d
θ
=
R
4
2
⋅
1
2
∫
0
2
π
cos
ϕ
d
ϕ
(
θ
−
sin
(
2
θ
)
2
)
|
0
π
=
{\displaystyle ={\frac {R^{4}}{2}}\cdot {\frac {1}{2}}\int _{0}^{2\pi }\cos \phi \;d\phi \int _{0}^{\pi }(1-\cos(2\theta ))\;d\theta ={\frac {R^{4}}{2}}\cdot {\frac {1}{2}}\int _{0}^{2\pi }\cos \phi \;d\phi (\theta -{\frac {\sin(2\theta )}{2}})|_{0}^{\pi }=}
=
R
4
4
∫
0
2
π
cos
ϕ
d
ϕ
(
[
π
−
0
]
−
[
0
−
0
]
)
=
π
R
4
4
∫
0
2
π
cos
ϕ
d
ϕ
=
π
R
4
4
sin
ϕ
|
0
2
π
=
π
R
4
4
(
sin
(
2
π
)
−
sin
0
)
=
0.
{\displaystyle ={\frac {R^{4}}{4}}\int _{0}^{2\pi }\cos \phi \;d\phi ([\pi -0]-[0-0])={\frac {\pi R^{4}}{4}}\int _{0}^{2\pi }\cos \phi \;d\phi ={\frac {\pi R^{4}}{4}}\sin \phi |_{0}^{2\pi }={\frac {\pi R^{4}}{4}}(\sin(2\pi )-\sin 0)=0.}
Kažkas nesiintegruoja. Galima pabandyti integruoti per
ϕ
{\displaystyle \phi }
π
/
2
{\displaystyle \pi /2}
4
π
R
4
4
sin
ϕ
|
0
π
/
2
=
4
⋅
π
R
4
4
(
sin
(
π
/
2
)
−
sin
0
)
=
π
R
4
(
1
−
0
)
=
π
R
4
.
{\displaystyle 4{\frac {\pi R^{4}}{4}}\sin \phi |_{0}^{\pi /2}=4\cdot {\frac {\pi R^{4}}{4}}(\sin(\pi /2)-\sin 0)=\pi R^{4}(1-0)=\pi R^{4}.}
![{\displaystyle =3{\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi (-[-1-1])={\frac {6R^{5}}{5}}\int _{0}^{2\pi }d\phi ={\frac {6R^{5}}{5}}\cdot (2\pi -0)={\frac {12\pi R^{5}}{5}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9f84f056d2175f517ae728fece23c48b5260701)
![{\displaystyle ={\frac {R^{4}}{4}}\int _{0}^{2\pi }\cos \phi \;d\phi ([\pi -0]-[0-0])={\frac {\pi R^{4}}{4}}\int _{0}^{2\pi }\cos \phi \;d\phi ={\frac {\pi R^{4}}{4}}\sin \phi |_{0}^{2\pi }={\frac {\pi R^{4}}{4}}(\sin(2\pi )-\sin 0)=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/806f76241c374f742216c30afe4633feb60a6104)
