= Solution
In <cylindrical coordinates>, $\rho=(x^2+y^2)^{1/2}$ and $\mathbf F=(f(\rho)x/\rho,f(\rho)y/\rho,0)$. The <chain rule> gives $\partial_x\rho=x/\rho$ and $\partial_y\rho=y/\rho$. Therefore
$$
\boxed{\begin{aligned}
\frac{\partial F_1}{\partial x}&=f'(\rho)\cos^2\phi+\frac{f(\rho)}\rho\sin^2\phi,\\
\frac{\partial F_2}{\partial y}&=f'(\rho)\sin^2\phi+\frac{f(\rho)}\rho\cos^2\phi,\\
\frac{\partial F_3}{\partial z}&=0,\qquad
\nabla\cdot\mathbf F=f'(\rho)+\frac{f(\rho)}\rho.
\end{aligned}}
$$
This is <cylindrically radial divergence>, valid for $\rho>0$.
The bounded domain is outside the unit cylinder, below the inverse-square surface and above the lower plane:
$$
V=\left\{\frac14\le z\le1,\quad 1\le\rho\le z^{-1/2},\quad0\le\phi<2\pi\right\}
=\left\{1\le\rho\le2,\quad\frac14\le z\le\rho^{-2}\right\}.
$$
The upper plane touches only the rim $\rho=1,z=1$; it is not an extra positive-area top face. Revolving the shaded meridional region below around the $z$ axis gives the three-dimensional domain.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-3-domain.png]
{title=Meridional section of the annular volume between the unit cylinder, inverse-square surface and lower plane}
With the <Jacobian determinant> $dV=\rho\,d\rho\,d\phi\,dz$ of <cylindrical coordinates>,
$$
\boxed{\operatorname{Vol}(V)=\pi\int_{1/4}^1(z^{-1}-1)\,dz=\pi\left(\log4-\frac34\right).}
$$
Zero <divergence> gives $(\rho f)'=0$, hence the most general solution on the positive radial interval is
$$
\boxed{f(\rho)=\frac A\rho,\qquad A\in\mathbb R.}
$$
Although this <vector field> is singular on the axis if $A\ne0$, the entire closure of $V$ lies at $\rho\ge1$, so it is smooth on a neighbourhood of $V$ and the <divergence theorem> applies.
Its volume integral of <divergence> is zero. Compute the outward <surface integral> directly. On the inner cylinder, $\mathbf n=-\mathbf e_\rho$, $dS=d\phi\,dz$, and the flux is $-2\pi A(1-1/4)=-3\pi A/2$. On the curved roof $z=g(\rho)=\rho^{-2}$, the outward <vector surface element of a graph> is
$$
d\mathbf S=(-g'(\rho)\mathbf e_\rho+\mathbf e_z)\rho\,d\rho\,d\phi
=\left(\frac2{\rho^2}\mathbf e_\rho+\rho\mathbf e_z\right)d\rho\,d\phi.
$$
Therefore its flux is
$$
\int_0^{2\pi}\int_1^2\frac{2A}{\rho^3}\,d\rho\,d\phi=\frac{3\pi A}{2}.
$$
The horizontal bottom has zero flux and the top rim has zero surface area. Summing gives
$$
\boxed{\int_{\partial V}\mathbf F\cdot d\mathbf S=0=\int_V\nabla\cdot\mathbf F\,dV,}
$$
which verifies the <divergence theorem> with every boundary contribution accounted for.
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