Given a vector field $$\text{F}=(xz^2)\vec i+(x^2y-z^3)\vec j+(2xy+y^2z)\vec k$$
- How to describe $\Sigma_1$ in cylindrical coordinates?
- How to generally find $\vec n$?
And $S$ is the sphere half sphere $z= \sqrt {4-x^2-y^2}$ and $z=0$, then compute $$\iint_ S \text{F}\cdot\vec n\; \text{d}S$$
From Divergence theorem I know that the double integral is indeed:
$$\begin{align} \iiint_V \nabla \cdot \text{F} \;\text{d} V &= \iiint_V x^2+y^2+z^2 \;\text{d} V\\[1ex] &= \int_{0}^{2 \pi}\int_{0}^{2}\int_{0}^{\sqrt {4- \rho^2 }}(\rho^2+z^2)\rho \;\text{d}z\text{d}\rho\text{d}\theta\\[1ex] &=\frac{64\pi}{5}\end{align}$$
If I was going to compute $\iint_ S \text{F}\cdot\vec n\; \text{d}S$ directly without using Divergence theorem, then how I could proceed?
The surface $S$ can be written as $$S= \Sigma_1 \cup \Sigma_2$$
Where $$\Sigma_1= \left\{(\rho \cos\theta,\rho \sin\theta ,z): 0\le \theta \le 2\pi , 0 \le z \le \sqrt{4-\rho^2} \right\}$$ $$\Sigma_2= \left\{(2 \cos\theta,2 \sin\theta,0): 0\le \theta \le 2\pi \right\}$$ And $$\vec n_1=\frac{\partial \Sigma_1}{\partial \theta}× \frac{\partial \Sigma_1}{\partial z}=\langle \rho\cos\theta,\rho\sin\theta,0 \rangle $$
So:$$\iint_ {\Sigma_1} \text{F}\cdot\vec n_1\; \text{d}S$$$$=\int_{0}^{2 \pi}\int_{0}^{\sqrt{4-\rho^2} } z^{2}r^{2}\cos^{2}\theta+\left(r^{2}\cos^{2}\theta\sin\theta-z^{3}\right)r\sin\theta\;\text{d}z\text{d}\theta\tag{I} $$
And $$\vec n_2=\frac{\partial \Sigma_2}{\partial \theta}\times \frac{\partial \Sigma_2}{\partial z}=\langle 0,0,0 \rangle $$
So:$$\iint_ {\Sigma_2} \text{F}\cdot\vec n_2\; \text{d}S=0 \tag{II}$$
Now we need to add $(\text{I})$ and $(\text{II})$
$$\iint_ S \text{F}\cdot\vec n\; \text{d}S=\iint_ {\Sigma_1} \text{F}\cdot\vec n_1\; \text{d}S+\iint_ {\Sigma_2} \text{F}\cdot\vec n_2\; \text{d}S$$
However I doubt if $\Sigma_1$ describes the half sphere correctly.
First, let's work on $\Sigma_1$, we consider the parametrization,
$$x=r\cos \theta, y=r\sin \theta, z=\sqrt{4-r^2}$$
$$\begin{bmatrix} \cos \theta \\ \sin \theta \\ \frac{-r}{\sqrt{4-r^2}}\end{bmatrix} \times \begin{bmatrix} -r\sin \theta \\ r \cos \theta \\ 0 \end{bmatrix}=\begin{bmatrix} \frac{r^2\cos \theta}{\sqrt{4-r^2}} \\ \frac{r^2 \sin \theta}{\sqrt{4-r^2}}\\ r\end{bmatrix}$$
$$F(r, \theta) =\begin{bmatrix} r(4-r^2)\cos \theta \\ r^3\cos^2 \theta \sin \theta - (4-r^2)^\frac32\\ 2r^2\sin \theta \cos \theta + r^2 \sin ^2 \theta\sqrt{4-r^2}\end{bmatrix}$$
\begin{align} &\int \int_{\Sigma_1} F \cdot \vec{n}_1 dS\\ &=\int_0^2 r^3\sqrt{4-r^2}\, dr \int_0^{2\pi}\cos^2 \theta \, d\theta + \int_0^2 \frac{r^5}{\sqrt{4-r^2}}\, dr \int_0^{2\pi}\sin^2 \theta \cos^2 \theta \, d\theta + \int_0^2r^3 \sqrt{4-r^2}\, dr \int_0^{2\pi} \sin^2 \theta \, d\theta \\ &= 2\pi \int_0^{2}r^3\sqrt{4-r^2}\, dr + \frac{\pi}4 \int_0^2 \frac{r^5}{\sqrt{4-r^2}}\, dr\\ &= 2\pi \left(\frac{64}{15} \right) + \frac{\pi}{4} \left( \frac{256}{15}\right)\\ &=3\pi \left(\frac{64}{15} \right)\\ &= \frac{64\pi}{5} \end{align}
Now, let's work on $\Sigma_2$,
$$x=r\cos \theta, y=r\sin \theta, z=0$$
$$ \begin{bmatrix} -r\sin \theta \\ r\cos\theta \\ 0\end{bmatrix} \times \begin{bmatrix} \cos \theta \\ \sin \theta \\ 0\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ -r\end{bmatrix}$$
\begin{align} &\int\int_{\Sigma_2} F \cdot \vec{n}_2 dS\\ &= \int_0^2-r^3 \, dr \int_0^{2\pi} \sin \theta \cos \theta\, d\theta \\&=0 \end{align}