1. ## Divergence theorem question

Calculate the flux of F across S.

Let $F = \frac{r}{|r|}$ where $r=x\vec{i}+y\vec{j}+z\vec{k}$, S consists of the hemisphere $z=\sqrt{1-x^2-y^2}$ and the disk $x^2+y^2 \leq 1$ in the xy plane

so for F I got:

$\frac{x\vec{i}+y\vec{j}+z\vec{k}}{\sqrt{x^2+y^2+z^ 2}}$

therefore my divergence will be:

$\frac{\partial}{\partial x} = \frac{y^2+z^2}{(x^2+y^2+z^2)^{3/2}}$

$\frac{\partial}{\partial y} = \frac{x^2+z^2}{(x^2+y^2+z^2)^{3/2}}$

$\frac{\partial}{\partial z} = \frac{x^2+y^2}{(x^2+y^2+z^2)^{3/2}}$

now putting everything together I get:

$\frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{3/2}}= \frac{2}{\sqrt{x^2+y^2+z^2}}$

now converting into cylindrical coordinates I get:

$\int_{\theta=0}^{\theta=2\pi} \int_{\phi=0}^{\phi=\pi/2} \int_{\rho=0}^{\rho=1} = \frac{2\rho^2 \sin(\phi)}{\rho} \ d\rho \ d\phi \ d\theta = 2\pi$

now I don't know if this is correct since $x^2+y^2 \leq 1$ is throwing me off.

2. Originally Posted by lllll
Calculate the flux of F across S.

Let $F = \frac{r}{|r|}$ where $r=x\vec{i}+y\vec{j}+z\vec{k}$, S consists of the hemisphere $z=\sqrt{1-x^2-y^2}$ and the disk $x^2+y^2 \leq 1$ in the xy plane

so for F I got:

$\frac{x\vec{i}+y\vec{j}+z\vec{k}}{\sqrt{x^2+y^2+z^ 2}}$

therefore my divergence will be:

$\frac{\partial}{\partial x} = \frac{y^2+z^2}{(x^2+y^2+z^2)^{3/2}}$

$\frac{\partial}{\partial y} = \frac{x^2+z^2}{(x^2+y^2+z^2)^{3/2}}$

$\frac{\partial}{\partial z} = \frac{x^2+y^2}{(x^2+y^2+z^2)^{3/2}}$

now putting everything together I get:

$\frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{3/2}}= \frac{2}{\sqrt{x^2+y^2+z^2}}$

now converting into cylindrical coordinates I get:

$\int_{\theta=0}^{\theta=2\pi} \int_{\phi=0}^{\phi=\pi/2} \int_{\rho=0}^{\rho=1} = \frac{2\rho^2 \sin(\phi)}{\rho} \ d\rho \ d\phi \ d\theta = 2\pi$

now I don't know if this is correct since $x^2+y^2 \leq 1$ is throwing me off.
Looks fine to me. The disc (the bottom of the hemisphere) is described by $x^2+y^2 \le 1$ but since your using the divergence thm, you really don't consider this part of the surface but instead the volume as a whole.