Past exam of the mathematical methods II course of the University of Oxford / 2021 / 10

Let $\mathbf{F}=(F_x,F_y,F_z)$ be continuously differentiable near $P$. Take a small axis-aligned rectangular volume around $P$, with side lengths $\Delta x,\Delta y,\Delta z$ shrinking with bounded aspect ratios. The outward flux through its two $x$ faces is, to leading order,where $\Delta V=\Delta x\Delta y\Delta z$. The analogous face pairs give $\partial F_y/\partial y$ and $\partial F_z/\partial z$. Dividing total flux by $\Delta V$ and taking the limit givesThe divergence is thus the local outward flux per unit volume; the symbols on the right are Cartesian partial derivatives.

For a bounded volume $V$ with piecewise smooth closed boundary $S$, subdivide into small cells. Each internal face occurs twice with opposite normal vectors, so its flux cancels. The remaining sum is the exterior surface integral; the cell sums of $(\nabla \cdot \mathbf{F})\Delta V$ converge to the volume integral. Assuming continuous first partial derivatives on a neighborhood of the closure of $V$, this proves the divergence theorem:The unit normal vector $\mathbf{n}$ points out of $V$. Multiple boundary components are allowed; at a cavity wall it points into the cavity.

Assume $a>0$ and choose the outward orientation. On the sphere, $\mathbf{n}=(x,y,z)/a$, so $\mathbf{F}\cdot \mathbf{n}=(x^2+y^2+2z^2)/a=(a^2+z^2)/a$. In spherical coordinates, $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\varphi$. The direct surface integral isIndependently, $\nabla \cdot \mathbf{F}=1+1+2=4$. Its volume integral over the ball isThe two calculations agree exactly with the divergence theorem. The expression requested is a scalar flux of a vector field, namely the dot product with the directed surface area element.

For this vector field,The divergence theorem and spherical coordinates therefore give the outward fluxThe $a^5$ dependence also follows by combining the cubic field magnitude with the quadratic scaling of surface area.