= Solution
The integrand field is exactly the <curl> computed in part (a), so use its <polynomial> <vector potential> $\mathbf G=(z^3,x^3,y^3)/3$. The upper cap meets the plane $z=0$ in an <ellipse>. Since the parameters are only assumed nonzero, set $a_0=|a|$ and $b_0=|b|$ to use positive semiaxes. Its correctly oriented boundary is
$$
\mathbf r(t)=(a_0\cos t,b_0\sin t,0),\qquad0\leq t\leq2\pi.
$$
On that boundary, $G_x=0$ and $dz=0$, so only $G_y\,dy=x^3dy/3$ contributes. The <Stokes theorem> therefore gives
$$
\begin{aligned}
A&=\oint_C\mathbf G\cdot d\mathbf r
=\frac13\int_0^{2\pi}(a_0\cos t)^3(b_0\cos t)dt\\
&=\frac{a_0^3b_0}{3}\int_0^{2\pi}\cos^4t\,dt
=\frac{a_0^3b_0}{3}\frac{3\pi}{4}.
\end{aligned}
$$
The last integral follows by expanding $\cos^4t=(3+4\cos2t+\cos4t)/8$. Thus
$$
\boxed{A=\frac\pi4|a|^3|b|=\frac\pi4a^2|ab|.}
$$
The result is independent of $c$. A sign check uses the <divergence theorem>: closing the cap with its base disk adds outward normal $-\widehat{\mathbf z}$ there. Since $\boldsymbol\nabla\cdot\mathbf F=0$, the cap <flux> equals $\int_{\rm disk}x^2\,dx\,dy>0$, consistent with the boxed expression and its absolute values.
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