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