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<3$$0<\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<3$$0<\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