Verifiy the Divergence Theorem The vector field $\vec{F}(x,y,z)=(z,x,y+z^2)$, on the limited region $x^2+y^2=4$, $z=x+1$, plane $XY$. Verify the Divergence Theorem:
$\textbf{Answer:}$
I know that the divergence theorem is:
${\displaystyle \iint _{S}\mathbf {F} \cdot d\mathbf {S} =\iiint _{\Omega}\nabla \cdot \mathbf {F} \;dV} $
$i)$ I have first calculated $\iiint_{\Omega}\nabla \cdot \mathbf {F} \;dV $
$$\iiint_{\Omega}\nabla \cdot \mathbf {F} \;dV = \iiint_{\Omega} 2zdxdydz$$
$x=rcos\theta, y=rsin\theta, z=z, r\in[0,2], \theta \in [0,2\pi], z \in [0, rcos\theta+1]$
$$ \rightarrow \iiint_{\Omega}\nabla \cdot \mathbf {F} \;dV = \iiint_{\Omega} 2zdxdydz = \int_{0}^{2\pi} d\theta \int_{0}^{2} dr \int_{0}^{rcos\theta+1} 2zr dz=8\pi $$
$ii)$ I then calculated $ \iint _{S}\mathbf {F} \cdot d\mathbf {S} $
At this point my doubt arises, do I have to evaluate the surface integral in 3 regions, the upper part, the surface and the lower part? How could you parameterize to find the normal vectors in each region?
 A: First of all the given equations do not clearly define a region. I will come to that in a while but to answer your question, yes you need to find surface integral over cylindrical surface $x^2 + y^2 = 4$ and planar surfaces $z = 0$ and $z = x + 1$. Sum of them should be equal to the volume integral but the bounds of volume integral are not correct in your work.
The given surface is $\{S: S_1 \cup S_2 \cup S_3\}$ where $\{S_1: x^2 + y^2 = 4, S_2: z = x + 1, S_3: z = 0\}$. However the question misses to state whether we consider $S_1$ and $S_2$ above $z = 0$ or below. Let's consider $z \geq 0$ to verify divergence theorem.
In polar coordinates, $S_1: \{x = 2 \cos\theta, y = 2 \sin\theta\}, 0 \leq \theta \leq 2\pi$
At intersection of $S_2$ and $S_3$, $x = - 1$ and at its intersection with $S_1$, $x = 2 \cos\theta = - 1 \implies \theta = - \frac{2\pi}{3}, \frac{2 \pi}{3}$.
Using the symmetry about xz plane, let's consider region above xz plane and multiply the volume integral by $2$. For $0 \leq \theta \leq 2\pi/3, 0 \leq r \leq 2$ and for $2 \pi / 3 \leq \theta \leq \pi, 0 \leq r \leq - \sec\theta$
So the volume integral is,
$\displaystyle 2 \int_0^{2 \pi / 3} \int_0^2 \int_0^{1 + r \cos\theta} 2 z r ~dz ~dr ~d\theta + $
$$ \displaystyle 2 \int_{2 \pi / 3}^{\pi} \int_0^{-\sec\theta} \int_0^{1 + r \cos\theta} 2z r ~dz ~dr ~d\theta$$
The outward flux through closed surface $S$ is $~ \displaystyle  \frac{9 \sqrt3}{2} + \frac{16 \pi}{3}$.
Now verify this using surface integral.
