I have a problem such as:
Find the net charge contained in a solid hemisphere $x^2 + y^2 + z^2 \le 64$, where $z \ge 0$ if the electric field is $E=[5x, 5y, 5z]$
I know from my sources that this is a surface integral problem and that I should set up Gauss's Law problems in the following way:
$$Q = \epsilon_0 \int\int_S \vec{E} \cdot d\vec{S}$$
In this case $\epsilon_0$ is just a constant referred to as the permittivity of free space. Therefore, it can be ignored.
My main question is, what exactly would my limits of integration be? I thought to say maybe polar coordinates, we would have a radius of $8$. The integrand would be $5x+5y+5z$ and the angle would be from $0 \ \ ... \ \ 2\pi$
The easy way is $\vec{\nabla}\cdot\vec{E}=\frac{\rho}{\epsilon_0}=15$, so $\rho=15\epsilon_0$, a constant, so $$Q=\int\int\int\rho d^3V=\rho V=15\epsilon_0\left(\frac12\right)\frac{4\pi}3(8)^3=5120\pi\epsilon_0$$ To set up the surface integral in terms of $r$ and $\theta$, we have $x=r\cos\theta$, $y=r\sin\theta$, $z=\sqrt{64-x^2-y^2}=\sqrt{64-r^2}$. So $$\vec r=\left\langle x,y,z\right\rangle=\left\langle r\cos\theta,r\sin\theta,\sqrt{64-r^2}\right\rangle$$ $$d\vec r=\left\langle \cos\theta,\sin\theta,\frac{-r}{\sqrt{64-r^2}}\right\rangle dr+\left\langle -r\sin\theta,r\cos\theta,0\right\rangle d\theta$$ $$\begin{align}d^2\vec A & =\pm\left\langle \cos\theta,\sin\theta,\frac{-r}{\sqrt{64-r^2}}\right\rangle dr\times\left\langle -r\sin\theta,r\cos\theta,0\right\rangle d\theta \\ & =\pm\left\langle\frac{r^2\cos\theta}{\sqrt{6-r^2}},\frac{r^2\sin\theta}{\sqrt{6-r^2}},r\right\rangle drd\theta \\ & =\left\langle\frac{r^2\cos\theta}{\sqrt{6-r^2}},\frac{r^2\sin\theta}{\sqrt{6-r^2}},r\right\rangle drd\theta\end{align}$$ Where we have taken the plus sign because outward normal is up on top. Then on the upper surface, $\vec E=\left\langle 5x,5y,5z\right\rangle=\left\langle 5r\cos\theta,5r\sin\theta,5\sqrt{64-r^2}\right\rangle$, so $$\begin{align}\int\int\vec E\cdot d^2\vec A & =\int_0^{2\pi}\int_0^8\left(\frac{5r^3}{\sqrt{64-r^2}}+5r\sqrt{64-r^2}\right)drd\theta \\ & =2\pi\int_0^8\left(\frac{320r}{\sqrt{64-r^2}}\right)dr=2\pi(-320)\sqrt{64-r^2}|_0^{2\pi}=5120\pi\end{align}$$ On the bottom, $z=0$, so $\vec r=\langle r\cos\theta,r\sin\theta,0\rangle$, $d\vec r=\langle\cos\theta,\sin\theta,0\rangle dr+\langle-r\cos\theta,r\sin\theta,0\rangle d\theta$, and $$\begin{align}d^2\vec A & =\pm\langle\cos\theta,\sin\theta,0\rangle dr\times\langle-r\sin\theta,r\cos\theta,0\rangle d\theta \\ & =\pm\langle0,0,r\rangle=-\langle0,0,r\rangle\end{align}$$ Where we have taken the minus sign because outward normal points down on the bottom. Also $\vec E=\langle x,y,z\rangle=\langle r\cos\theta,r\sin\theta,0\rangle$, so $$\int\int\vec E \cdot d^2\vec A=0$$ Because $\vec E$ has no downward component. Then $$Q=\epsilon_0\int\int\vec E \cdot d^2\vec A=5120\epsilon_0\pi+0=5120\epsilon_0\pi$$ This is the same as we got from the volume integral. You choose which approach you prefer.