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?
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.