F = 2xz i + yz j + $\displaystyle {z}^{ 2}$ k over the upper half of the sphere $\displaystyle {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)

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- May 2nd 2011, 06:26 AMadam_leedsDivergence Theorem limits
F = 2xz i + yz j + $\displaystyle {z}^{ 2}$ k over the upper half of the sphere $\displaystyle {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 AMTheEmptySet
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?

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

If so you end up with the integral

$\displaystyle 5\iiint_{V}zdV$

If you convert to spherical coordinates you get

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

Note that phi is the plane angle and theta is the angle measured from the positive z axis - May 8th 2011, 04:29 AMAnt
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 AMadam_leeds
- May 8th 2011, 04:55 AMAnt
Polar coords are:

$\displaystyle x = rsin(\theta)cos(\phi) $

$\displaystyle y = rsin(\theta)sin(\phi) $

$\displaystyle z = rcos(\theta) $

The Jacobian is

$\displaystyle r^2 sin(\theta) $

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

So

$\displaystyle 5x dx dy dz $

becomes

$\displaystyle 5 rcos(\theta) r^2 sin(\theta) $ - May 8th 2011, 05:01 AMadam_leeds
- May 8th 2011, 05:54 AMadam_leeds
- May 8th 2011, 06:34 AMTheEmptySet
The Jacobian in cylindrical coordinates

$\displaystyle r dr d\phi dz$

This can be calculated

$\displaystyle x = r\cos(\phi) \quad y = r\sin(\phi) \quad z= z$

This gives

$\displaystyle \begin{vmatrix} \cos(\phi) & \sin(\phi) & 0 \\ -r\sin(\phi) & r \cos(\phi) & 0 \\ 0 & 0 & 1 \end{vmatrix} = r\cos^2(\phi)+r\sin^2(\phi)=r$