= Solution
In <cylindrical coordinates> $r=\sqrt{x^2+y^2}$, the bounded solid enclosed by the two conical sheets and the <circular cylinder> is
$$
0\le r\le1,\qquad -r\le z\le r,\qquad 0\le\theta<2\pi.
$$
Its meridional section consists of two triangular regions; revolving them about the $z$-axis gives the solid. The two conical sheets meet at the origin and meet the <circular cylinder> in its circles at $z=\pm1$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-3-cone-cylinder.png]
{title=The bounded solid between two cones and a unit cylinder, with its meridional section}
Since $\mathbf F=(x,y,z)$, its <divergence> is $3$. Integrate directly with the <cylindrical coordinates> volume element $r\,dz\,dr\,d\theta$:
$$
\int_V\nabla\cdot\mathbf F\,dV
=\int_0^{2\pi}\int_0^1\int_{-r}^r3r\,dz\,dr\,d\theta
=12\pi\int_0^1r^2\,dr
=\boxed{4\pi}.
$$
The same calculation gives $\operatorname{Vol}(V)=4\pi/3$.
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