= Solution
Since $\mathbf F=\nabla\Omega$, its <divergence> is the <Laplacian> of $\Omega$:
$$
\nabla\cdot\mathbf F=\nabla^2\Omega=6x+4z.
$$
The <divergence theorem> changes the outward <flux> through the whole closed boundary into $I=\int_V(6x+4z)dV$. Rotational symmetry makes the $x$ term vanish. With $z=(u^2-v^2)/2$ and the <Jacobian determinant> from part (c),
$$
\begin{aligned}
I&=2\int_{-\pi}^{\pi}\int_0^1\int_0^v(u^2-v^2)uv(u^2+v^2)\,du\,dv\,d\phi\\
&=4\pi\int_0^1v\left[\frac{u^6}{6}-\frac{v^4u^2}{2}\right]_{u=0}^{u=v}dv\\
&=-\frac{4\pi}{3}\int_0^1v^7\,dv
=\boxed{-\frac\pi6}.
\end{aligned}
$$
Both the curved face and the top disk are included in this <surface integral> through the <divergence theorem>.
Back to article page
