Elliptical Shell (Impossible)
This is actually impossible to evaluate
Parameterizing the Surface
We use modified spherical coordinates, which we have to express in Cartesian coordiantes.
\[\b{r}(\theta, \phi) = a \sin \theta \; \cos \phi \; \u{x} + b \sin \theta \sin \phi \; \u{y} + c \cos \theta \; \u{z}\]Therefore
\[\frac{\partial \b{r}}{\partial \theta} = a \cos \theta \; \cos \phi \; \u{x} + b \cos \theta \sin \phi \; \u{y} - c \sin \theta \; \u{z} \qquad\qquad \frac{\partial \b{r}}{\partial \phi} = a \sin \theta \; \sin \phi \; \u{x} - b \sin \theta \cos \phi \; \u{y}\] \[\begin{align} d \b{A} &= \left ( \frac{\partial \b{r}}{\partial \theta} d\theta \right ) \times \left ( \frac{\partial \b{r}}{\partial \phi} d\phi \right ) \\[10pt] &= \left [ (bc \; \sin^2 \theta \; \cos \phi) \; \u{x} + (- ac \; \sin^2 \theta \; \sin \phi) \; \u{y} + (- ab \sin \theta \; \cos \theta \; \cos^2 \phi - ab \sin \theta \; \cos \theta \; \sin^2 \phi) \; \u{z} \right ] \; d\theta \; d\phi \end{align}\]Now we compute.
\[\begin{align} dA &= \abs{d \b{A}} \\[10pt] &= \sqrt{ (bc \; \sin^2 \theta \; \cos \phi)^2 + (ac \; \sin^2 \theta \; \sin \phi)^2 + (ab \sin \theta \; \cos \theta \; \cos^2 \phi + ab \sin \theta \; \cos \theta \; \sin^2 \phi)^2 } \; d\theta \; d\phi \\[10pt] &= \sqrt{ (bc \; \sin^2 \theta \; \cos \phi)^2 + (ac \; \sin^2 \theta \; \sin \phi)^2 + (ab \sin \theta \; \cos \theta)^2 } \; d\theta \; d\phi \\[10pt] &= \sqrt{ (bc \; \sin \theta \; \cos \phi)^2 + (ac \; \sin \theta \; \sin \phi)^2 + (ab \cos \theta)^2 } \; \sin \theta \; d\theta \; d\phi \\[10pt] \end{align}\]And here is where we get stuck. There’s really nothing else we can do to further simplify this. Obviously there is no way we are going to be able to integrate this, so we can’t evaluate any of the subsequent integrals.