= Solution
The modified upper surface has exactly the same <ellipse> at $z=0$ and the same <boundary orientation>. The field retains the global <polynomial> <vector potential> from part (a). By the <Stokes theorem>, both surface fluxes equal the same boundary <line integral>; the cap's height and interior shape do not enter. Hence
$$
\boxed{A=\frac\pi4a^2|ab|.}
$$
This is an instance of the fact that <solenoidal flux depends on the oriented boundary>, with the domain condition automatically satisfied by the smooth <polynomial> field on all of $\mathbb R^3$. When $a,b$ are chosen positive, the formula simplifies to $\pi a^3b/4$; the absolute-value form respects the original allowance of arbitrary nonzero signs.
Back to article page
