= Solution
First compute the <curl> directly:
$$
\nabla\times\mathbf F=(0,1,2x).
$$
Reflection symmetry in $x$ makes the last component integrate to zero. The middle component gives the <hemisphere>'s volume, so
$$
\int_V\nabla\times\mathbf F\,dV
=\left(0,\frac{2\pi R^3}{3},0\right).
$$
For an independent <surface integral>, the closed boundary consists of the curved <hemisphere> and its flat base. The right-hand side in part (b) is $\int_S\mathbf n\times\mathbf F\,dS$.
On the curved part use <spherical coordinates> with $0\leq\theta\leq\pi/2$ and $0\leq\phi<2\pi$. There,
$$
\begin{aligned}
\mathbf n&=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\\
\mathbf F&=(R\cos\theta,R^2\sin^2\theta,0),\qquad dS=R^2\sin\theta\,d\theta\,d\phi,\\
\mathbf n\times\mathbf F
&=(-R^2\cos\theta\sin^2\theta,\ R\cos^2\theta,R^2\sin^3\theta\cos\phi-R\sin\theta\cos\theta\sin\phi).
\end{aligned}
$$
The last component has zero angular integral. The first two components give
$$
\int_{S_{\mathrm{curved}}}\mathbf n\times\mathbf F\,dS
=\left(-2\pi R^4\int_0^{\pi/2}\cos\theta\sin^3\theta\,d\theta,
2\pi R^3\int_0^{\pi/2}\cos^2\theta\sin\theta\,d\theta,0\right)
=\left(-\frac{\pi R^4}{2},\frac{2\pi R^3}{3},0\right).
$$
On the base, $z=0$ and the outward <normal vector> is $-\mathbf e_z$. Thus $\mathbf n\times\mathbf F=(x^2+y^2,0,0)$. In <polar coordinates> its integral is
$$
\int_{S_{\mathrm{base}}}\mathbf n\times\mathbf F\,dS
=\left(2\pi\int_0^R\rho^3\,d\rho,0,0\right)
=\left(\frac{\pi R^4}{2},0,0\right).
$$
The first components cancel, and therefore both required sides equal
$$
\boxed{\left(0,\frac{2\pi R^3}{3},0\right),\qquad \alpha=\frac{2\pi R^3}{3}}.
$$
Back to article page
