Past exam of the mathematical methods II course of the University of Oxford / 2015 resit / 6

The divergence theorem states that, for a sufficiently smooth vector field $\mathbf{F}$ on a volume $\Omega$ with piecewise smooth closed boundary $\partial \Omega$,Here $\hat{\mathbf{n}}$ is the outward unit normal. The left-hand side is the total outward flux; the right-hand side adds the local divergence throughout the volume. The orientation and the requirement to include the complete closed boundary are essential.

Read the given vector field as the three-dimensional field $\mathbf{F}=(x,y,0)$. Its divergence is the sum of the three matching-coordinate partial derivatives:The absence of a $z$ component does not make the three-dimensional flux vanish: the field still flows outward through surface normals with horizontal components.

The enclosed volume lies above $z=0$ and below $z=4-x^2-y^2$. In cylindrical coordinates, its limits are $0\le r\le 2$, $0\le \theta \le 2\pi$, $0\le z\le 4-r^2$, with volume element $rdzdrd\theta$. The divergence theorem givesThus the volume is $8\pi$, and the outward flux is $\boxed{16\pi }$.

A direct check separates the boundary into its cap and base. On the base, $\mathbf{F}\cdot (-\hat{\mathbf{z}})=0$. On the graph $z=4-x^2-y^2$, the upward vector area element is $(2x,2y,1)dxdy$, so the cap flux is ${\iint }_{x^2+y^2\le 4}2(x^2+y^2)dxdy=16\pi$, in agreement with the volume calculation.