 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. HenceThis 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 $\pi a^{3}b/4$; the absolute-value form respects the original allowance of arbitrary nonzero signs.