Suppose the surface $S$ is made up of the half-ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1, \quad z\geqslant 0,$$ together with its base in the $xy$-plane, i.e., $x^2/a^2 + y^2/b^2\leqslant 1$.
Goal: To find $I=\int_S(y,x,z+c)\cdot\vec{\mathbf n}\,dS$, where $\vec{\mathbf n}$ is the outward pointing normal of $S$.
My idea was to split up the integral as $S_1\cup S_2$ where $S_1$ is the half-ellipsoid and $S_2$ is the base. Over $S_2$, we have $\vec{\mathbf n} = -\mathbf k$, $$dS_2 = \frac{dx\,dy}{\vec{\mathbf n}\cdot\mathbf k}=-dx\,dy,$$ so \begin{align*} \int_{S_2}(y,x,z+c)\cdot\vec{\mathbf n}\,dS_2&=- \iint_{S_2}(0+c)\,dx\,dy\\[4pt] &=-c\iint_{S_2}\,dx\,dy\\[4pt] &= -\pi abc. \end{align*} Now I'm not sure how to proceed with the integral over $S_1$, I suspect that some kind of change of coordinates would be helpful, but I'm not sure how to go about it.
I appreciate any assistance on how to proceed.
The question asks you to find flux over closed surface, which is half ellipsoid with its base. So the easiest is to apply divergence theorem. For a closed surface and a vector field defined over the entire closed region,
$ \displaystyle \iint_S \vec F \cdot\hat{n}\,dS = \iiint_V \text{div} \vec F \, dV$
Here, $~ \vec F = (y,x,z+c)$
$\nabla \cdot \vec F = 0 + 0 + 1 = 1$
As the divergence is $1$, the flux is equal to the volume of half ellipsoid, which is nothing but $\frac{2 \pi a b c}{3}$.
$ \displaystyle I = \iiint_V 1 \, dV = \frac{2\pi abc}{3} $