Evaluating \int\int_{\sigma} F.n ds with Divergence Theorem

[boneill3]

Homework Statement

Use the divergence theorem to evaluate

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

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

Homework Equations

\int\int_{s}F . dA = \int\int\int_{R}divF dV

The Attempt at a Solution

I've worked out divF to be 6.

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


To calulate this using spherical co-ordinates.

I would need to calculate a triple integral

I know there's

\int\int\int p^2 sin(\theta) dp d\theta d\phi

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

I guess the limits would be 0<p<30<\phi<2pi and 0<\theta<\phi

Any help greatly appreciated

 

[HallsofIvy]

Homework Statement

Use the divergence theorem to evaluate

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

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

Homework Equations

\int\int_{s}F . dA = \int\int\int_{R}divF dV

The Attempt at a Solution

I've worked out divF to be 6.

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


To calulate this using spherical co-ordinates.

I would need to calculate a triple integral

I know there's

\int\int\int p^2 sin(\theta) dp d\theta d\phi

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

I guess the limits would be 0<p<30<\phi<2pi and 0<\theta<\phi

No. Your limits on \rho and \phi are correct but \theta goes from 0 to \pi.

Any help greatly appreciated

 

[Hootenanny]

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.

 

[boneill3]

When I calculate the integral I'm getting 36\pi
I'm not sure where I'm going wrong.

So I compute.


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

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

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



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

=\int_{0}^{2\pi}18 d\phi

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

=36\pi

regards

 

[HallsofIvy]

When I calculate the integral I'm getting 36\pi
I'm not sure where I'm going wrong.

So I compute.


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

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

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



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

=\int_{0}^{2\pi}18 d\phi

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

=36\pi

regards

Yes, 36\pi is the volume of that sphere. Now multiply by 6: 6(36\pi)= 216\pi.

 

[boneill3]

Thanks for your help