I'm having a very hard time with the following problem:
Consider an area $D$ given by: \begin{cases} \frac{x^2}{4} + y^2 \leq 1\\ 0 \leq z \leq -x \end{cases}
Let $\vec{F} = (y+e^y)\vec{j} + z(1-e^y)\vec{k}$. Check the Divergence Theorem for $\vec{F}$ on $D$ by calculating both sides of the theorem:
\begin{equation} \iiint_{D}div\vec{F}dV = \oint\oint_{S} \vec{F} \cdot \vec{N}dS \quad \text{(\oiint doesn't seem to exist here)} \end{equation}
For the divergence I've got \begin{equation} div\vec{F} = 0 + (1+e^y) + (1-e^y) = 2 \end{equation} So I'd get \begin{equation} \iiint_{D} 2dxdydz \end{equation}
But from this point on I can't seem to figure out what to do. I've tried using the integration limits \begin{cases} -2 \leq x \leq 0 \\ -1 \leq y \leq 1 \\ 0 \leq z \leq -x \end{cases} (I understand now after writing that these integration limits would only work when $D$ was rectangular instead of having a ellipse shape)
I chose these according to this sketch I made of $D$, which seemed right considering that $\frac{x^2}{4} + y^2 \leq 1$ should give $D$ an ellipse shaped base:
But when integrating over these limits I got \begin{equation} \iiint_{D} 2dxdydz = 0 \end{equation}
After this I tried a parametrization for ellipses, which I've never used or seen before and am not very confident in being right. I got: \begin{cases} x = 2u\cos(v)\\ y = u\sin(v)\\ z = w \end{cases} With the Jacobian of this transformation being $J = 2u$ Now for the limits for these new variables I chose \begin{cases} 0 \leq u \leq 1\\ 0 \leq v \leq \pi \\ 0 \leq w \leq 2u\cos(v) \end{cases}
When integrating this I get
\begin{equation} \int_{0}^{1}\int_{0}^{\pi}\int_{0}^{2cos(v)}(2 \cdot 2u) dudvdw\\ \int_{0}^{1}\int_{0}^{\pi}4u \cdot 2u\cos(v) dudv \end{equation} But integrating this from $0$ to $\pi$ would again result in $0$.
I can't figure out what I should do and my lecture notes and example problems don't seem to help me further. Can someone give me a hint on how to continue?
The limits of integration for $D$ are wrong. The LHS should be $$\begin{align}\iiint_{D} 2dxdydz &=\int_{y=-1}^{1}(\int_{x=-2\sqrt{1-y^2}}^0(\int_{z=0}^{-x}2dz)dx)dy \\&=-\int_{y=-1}^{1}(\int_{x=-2\sqrt{1-y^2}}^02xdx)dy\\ &=4\int_{y=-1}^{1}(1-y^2)dy=\frac{16}{3}. \end{align}$$ The RHS is the sum of the fluxes through the three surfaces given by the boundary of $D$: $$\iint_{S_1} \vec{F} \cdot \vec{N}dS+ \iint_{S_2} \vec{F} \cdot \vec{N}dS+\iint_{S_3} \vec{F} \cdot \vec{N}dS$$ where $$S_1=\{(x,y,0):y\in[-1,1], -2\sqrt{1-y^2}\leq x \leq 0\},$$ $$S_2=\{(x,y,-x):y\in[-1,1], -2\sqrt{1-y^2}\leq x \leq 0\},$$ $$S_3=\{(2\cos(t),\sin(t),z):t\in[\pi/2,3\pi/2], 0\leq z\leq -2\cos(t)\}.$$ The orientation is outward. Try to evaluate the three fluxes and verify the equality.