Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 6

Let $V\subset {\mathbb{R}}^3$ be a bounded region with a piecewise smooth closed boundary $S=\partial V$, and let $\mathbf{F}$ have continuous first partial derivatives on a neighborhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. The right side is the outward flux; reversing the orientation reverses its sign.

On the radius-$a$ sphere, the outward unit normal vector is $(x,y,z)/a$, so $\mathbf{F}\cdot \mathbf{n}=(x^2+2y^2+3z^2)/a$. For a direct surface integral, put $x=a\sin \theta \cos \phi$, $y=a\sin \theta \sin \phi$, $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\phi$. The integral isThe azimuthal integrals are $\pi ,\pi ,2\pi$, while ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. ThusThe flux of a linear field through a sphere also follows from the equality of the three spherical second moments, without using the divergence theorem.

The divergence is $\nabla \cdot \mathbf{F}=1+2+3=6$. Since the volume enclosed by a sphere is $4\pi a^3/3$,This agrees with the independently computed outward surface integral, as the divergence theorem requires.