Let $S$ be part of the ellipsoid $3x^2+2y^2+z^2=28$ that is above the plane $z=1$ and $F=\langle yz^2,4xz, zx^2\rangle$ a vector field. Calculate $\iint_S F\cdot ndS$.
I thought of first "closing" the surface $S$ with the plane $z=1$ and then calculating the integral with Gauss theorem (let the closed surface be $B$) and then calculating separately $\iint F\cdot ndS$ over the plane.
So: $$ \iint_S F\cdot ndS=\iint_B F\cdot ndS-\iint_{z=1} F\cdot ndS $$ Calculating $\iint_{z=1} F\cdot ndS$ is relatively easy. The projection (let is be $P$) of $S$ onto the plane is $3x^2+2y^2+1^2=28\iff x^2+y^2=9$, then: $$ 0\le r\le 3\\ 0\le \theta\le2\pi\\ n=\langle 0,0,1\rangle\text{ (the normal vector to the projection is essentially the negative axis z)} $$ Then: $$ \iint_{P} F\cdot ndS=\iint_P zx^2 dA\stackrel{z=1}{=}\iint_P x^2 dA\stackrel{\text{in polar coord.}}{=}\int_0^{2\pi}\int_0^3 r^3\cos^2\theta=\frac{81}{4}\pi. $$ Link to Wolfram Alpha calculation.
Calculation of $\iint_B F\cdot ndS$ is much more problematic. According to Gauss theorem: $$ \iint_B F\cdot ndS=\iiint_G\operatorname{div}F \,dv\\ \operatorname{div}F=x^2\implies \iiint_G x^2 \,dv $$ I thought that we can do the calculation in polar coordinates again: $$ 0\le r\le 3\\ 0\le \theta\le2\pi\\ 1\le z\le\sqrt{28-3(x^2+y^2)} $$ Thus: $$ \iiint_{1}^{\sqrt{28-3r^2}} r^2\cos^2\theta\cdot r $$ Very quickly the integral turns ugly but manageable: $$ \int \cos^2\theta\int \bigg[\frac{r^4}{4}\bigg]^{\sqrt{28-3r^2}}_1=\frac{1}{4}\int \cos^2\theta\int (28-3r^2)^2-1=\\\frac{1}{4}\int \cos^2\theta\int 2862-168r^2+r^4-1=\big(462+\frac{243}{20}-\frac{3}{16}\big)\pi. $$ So the final answer is: $$ \iint_S F\cdot ndS=\big(462+\frac{243}{20}-\frac{3}{16}\big)\pi-\frac{81}{4}\pi. $$ Am I in the right direction?
Note that $3x^2+2y^2+1=28$ is not equivalent to $x^2+y^2=9$.
I suggest to compute the surface integral. To this end let $a$, $b$, $c$ be the semiaxes of the ellipsoid. Then you have a parametric representation of the form $${\bf r}(\phi,\theta)=\bigl(a\cos\theta\cos\phi,b\cos\theta\sin\phi,c\sin\theta\bigr)\ ,$$ whereby $\phi$ and $\theta$ are GPS coordinates on $S^2$. Now it so happens that $F_1$ is odd in $y$, and $F_2$ is odd in $x$, whereas the normal vector ${\bf r}_\phi\times{\bf r}_\theta$ is even in both $x$ and $y$, by symmetry of the ellipsoid. This implies that from $F\cdot({\bf r}_\phi\times{\bf r}_\theta)$ we only have to compute $$F_3\>({\bf r}_\phi\times{\bf r}_\theta)_3=a^2c\cos^2\theta\sin\theta\cos^2\phi\cdot ab\cos\theta\sin\theta\ .$$ Now you have to integrate this over $0\leq\phi\leq2\pi$ and over $\theta_1\leq\theta\leq{\pi\over2}$, whereby $\theta_1$ is defined by the condition $c\sin\theta_1=1$. Since $\int_0^{2\pi}\cos^2\phi\>d\phi=\pi$ we obtain in this way $$\Phi=\pi a^3bc \int_{\theta_1}^{\pi/2}\cos^3\theta\sin^2\theta\>d\theta\ .$$ Here the last integral can be written as $$\int_{\theta_1}^{\pi/2}(\sin^2\theta-\sin^4\theta)\>\cos\theta\>d\theta=\int_{\sin \theta_1}^1(u^2-u^4)\>du\ ,$$ which simplifies matters considerably.