# Divergence Theorem limits

• May 2nd 2011, 06:26 AM
Divergence Theorem limits
F = 2xz i + yz j + ${z}^{ 2}$ k over the upper half of the sphere ${x}^{2 } + {y}^{2 } + {z}^{2 } = {a}^{ 2}$

I thought it was dz d\theta z between a and 0 and \theta between \pi and 0 , but im wrong (Crying)
• May 2nd 2011, 06:40 AM
TheEmptySet
Quote:

F = 2xz i + yz j + ${z}^{ 2}$ k over the upper half of the sphere ${x}^{2 } + {y}^{2 } + {z}^{2 } = {a}^{ 2}$

I thought it was dz d\theta z between a and 0 and \theta between \pi and 0 , but im wrong (Crying)

I guess I am not sure what you are asking.

Do you want to compute the flux out of the top half of the sphere using the divergence theorem?

$\iint_{S}\mathbf{F}\cdot d\mathbf{S}=\iiint_{V}\nabla \cdot \mathbf{F}dV$

If so you end up with the integral

$5\iiint_{V}zdV$

If you convert to spherical coordinates you get

$5 \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{a} (\rho \cos(\theta))\rho^2\sin(\theta)d\rho d\theta d\phi$

Note that phi is the plane angle and theta is the angle measured from the positive z axis
• May 8th 2011, 04:29 AM
Ant
Change to spherical polar coordintes would be my best bet. Then integrate wrt theta, phi and r with limits:

theta [0, pi/2]
phi [0, 2pi]
r [0, a]

with the Jacobian being r^2 sin(theta)

that's just how I'd try it though!!
• May 8th 2011, 04:31 AM
Quote:

Originally Posted by TheEmptySet
$(\rho \cos(\theta))\rho^2\sin(\theta)d\rho d\theta d\phi$

How do you get this, i have got the 5z?
• May 8th 2011, 04:55 AM
Ant
Polar coords are:

$x = rsin(\theta)cos(\phi)$
$y = rsin(\theta)sin(\phi)$
$z = rcos(\theta)$

The Jacobian is

$r^2 sin(\theta)$

The Jacobian is used when changing from x, y, z into polar coordinates.

So

$5x dx dy dz$

becomes

$5 rcos(\theta) r^2 sin(\theta)$
• May 8th 2011, 05:01 AM
Quote:

Originally Posted by Ant
The Jacobian is

$r^2 sin(\theta)$

Is the Jacobian always that?
• May 8th 2011, 05:54 AM
Quote:

Is the Jacobian always that?

Can i use this for a cylinder?
• May 8th 2011, 06:34 AM
TheEmptySet
Quote:

$r dr d\phi dz$
$x = r\cos(\phi) \quad y = r\sin(\phi) \quad z= z$
$\begin{vmatrix} \cos(\phi) & \sin(\phi) & 0 \\ -r\sin(\phi) & r \cos(\phi) & 0 \\ 0 & 0 & 1 \end{vmatrix} = r\cos^2(\phi)+r\sin^2(\phi)=r$