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Divergence Theorem

  • Thread starter boneill3
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  • #1
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Homework Statement



Use the divergence theorem to evaluate

[itex]\int\int_{\sigma}F . n ds[/itex]
Where n is the outer unit normal to [itex]\sigma[/itex]

we have
[itex]F(x,y,z)=2x i + 2y j +2z k [/itex] and [itex]\sigma[/itex] is the sphere [itex]x^2 + y^2 +z^2=9[/itex]

Homework Equations



[itex]\int\int_{s}F . dA = \int\int\int_{R}divF dV[/itex]


The Attempt at a Solution



I've worked out [itex]divF[/itex] to be 6.

so I multyiply that by the Volume of a sphere [itex]6\times\frac{4}{3}\pi r^3 = 216\pi[/itex]


To calulate this using spherical co-ordinates.

I would need to calculate a triple integral

I know theres

[itex]\int\int\int p^2 sin(\theta) dp d\theta d\phi[/itex]

I know that p = 3 but what would the values of [itex]\theta [/itex] and [itex]\phi [/itex] be

I guess the limits would be [itex]0<p<3[/itex][itex] 0<\phi<2pi[/itex] and [itex]0<\theta<\phi[/itex]

Any help greatly appreciated

Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,792
920

Homework Statement



Use the divergence theorem to evaluate

[itex]\int\int_{\sigma}F . n ds[/itex]
Where n is the outer unit normal to [itex]\sigma[/itex]

we have
[itex]F(x,y,z)=2x i + 2y j +2z k [/itex] and [itex]\sigma[/itex] is the sphere [itex]x^2 + y^2 +z^2=9[/itex]

Homework Equations



[itex]\int\int_{s}F . dA = \int\int\int_{R}divF dV[/itex]


The Attempt at a Solution



I've worked out [itex]divF[/itex] to be 6.

so I multyiply that by the Volume of a sphere [itex]6\times\frac{4}{3}\pi r^3 = 216\pi[/itex]


To calulate this using spherical co-ordinates.

I would need to calculate a triple integral

I know theres

[itex]\int\int\int p^2 sin(\theta) dp d\theta d\phi[/itex]

I know that p = 3 but what would the values of [itex]\theta [/itex] and [itex]\phi [/itex] be

I guess the limits would be [itex]0<p<3[/itex][itex] 0<\phi<2pi[/itex] and [itex]0<\theta<\phi[/itex]
No. Your limits on [itex]\rho[/itex] and [itex]\phi[/itex] are correct but [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex].

Any help greatly appreciated

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #3
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
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Your limits are correct (assuming of course theta runs from 0 to pi, rather than phi).

All you need to so now is explicitly compute the integral, which is straightforward.
 
  • #4
127
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When I calculate the integral I'm getting 36[itex]\pi[/itex]
I'm not sure where I'm going wrong.

So I compute.


[itex]\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3} p^2 sin(\theta) dp d\theta d\phi[/itex]

[itex]=\int_{0}^{2\pi}\int_{0}^{\pi}\left[\frac{ p^3 sin(\theta)}{3} \right]_{0}^{3} d\theta d\phi[/itex]

[itex]=\int_{0}^{2\pi}\int_{0}^{\pi}9sin(\theta) d\theta d\phi[/itex]



[itex]=\int_{0}^{2\pi}\left[-9cos(\theta) \right]_{0}^{\pi} d\phi[/itex]

[itex]=\int_{0}^{2\pi}18 d\phi[/itex]

[itex]=\left[18\phi\right]_{0}^{2\pi[/itex]

[itex]=36\pi[/itex]

regards
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,792
920
When I calculate the integral I'm getting 36[itex]\pi[/itex]
I'm not sure where I'm going wrong.

So I compute.


[itex]\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3} p^2 sin(\theta) dp d\theta d\phi[/itex]

[itex]=\int_{0}^{2\pi}\int_{0}^{\pi}\left[\frac{ p^3 sin(\theta)}{3} \right]_{0}^{3} d\theta d\phi[/itex]

[itex]=\int_{0}^{2\pi}\int_{0}^{\pi}9sin(\theta) d\theta d\phi[/itex]



[itex]=\int_{0}^{2\pi}\left[-9cos(\theta) \right]_{0}^{\pi} d\phi[/itex]

[itex]=\int_{0}^{2\pi}18 d\phi[/itex]

[itex]=\left[18\phi\right]_{0}^{2\pi[/itex]

[itex]=36\pi[/itex]

regards
Yes, [itex]36\pi[/itex] is the volume of that sphere. Now multiply by 6: [itex]6(36\pi)= 216\pi[/itex].
 
  • #6
127
0
Thanks for your help
 

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