Let $S$ be the ellipse $\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 + \left(\frac{z}{c}\right)^2=1,$ with $\vec{n}$ oriented outwards. Compute $\int\!\!\!\int_S \vec{F}\cdot \vec{n}\,dA$ for $\vec{F}=(-a^2y, b^2x, z^2).$
So far I let $R = \{(x,y,z)\in \mathbb{R}^3 : \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 + \left(\frac{z}{c}\right)^2\leq1\}$ which means that $S = \partial R.$ So, by the Divergence Theorem $$\int\!\!\!\int_{\partial R} \vec{F}\cdot\vec{n} \, dA = \int\!\!\!\int\!\!\!\int_{R} \text{div}\vec{F}\, dV.$$ Since div$\vec{F}=2z$ we have $$\displaystyle\int_{-a}^a\int_{-b\sqrt{1-\left(\frac{x}{a}\right)^2}}^{b\sqrt{1-\left(\frac{y}{b}\right)^2}}\int_{-c\sqrt{1-\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2}}^{c\sqrt{1-\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2}} 2z \,\,dzdydx.$$ I chose to use the change of variable $(x,y,z)=(ar\cos\theta, br\sin\theta, z)=g(r,\theta,z)$ where $|J|=abr$ giving me the new integral $$\int_0^1\int_{0}^{2\pi}\int_{-c\sqrt{1-r^2}}^{c\sqrt{1-r^2}}2z \,\,dzdrd\theta.$$
When I go to compute this integral I have a problem since the inner most integral equals zero. I think I may be missing something conceptually.
There's nothing wrong. Since the divergence is a constant multiple of $z$, by symmetry it will integrate to $0$.