Past exam of the mathematical methods II course of the University of Oxford / 2014 / 10

The divergence theorem states ${\oint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS={\int }_V\nabla \cdot \mathbf{F}dV$ with outward orientation. For $\mathbf{F}=(x,y,0)$, its divergence is $1+1=2$. The enclosed region has cylindrical bounds $0\le r\le 2$, $0\le z\le 4-r^2$, $0\le \varphi \le 2\pi$, so its volume isThereforeThe planar base contributes zero because its normal is vertical and $F_z=0$; the curved part accounts for the entire outward flux.

The original integration region is the triangle with vertices $(0,0)$, $(0,1)$ and $(1/2,1/2)$. Under $r=x+y$, $s=y-x$, its inverse map is $x=(r-s)/2$, $y=(r+s)/2$, with absolute inverse Jacobian determinant $1/2$. The boundaries $x=0$, $y=x$, $x+y=1$ become $s=r$, $s=0$, $r=1$. Thus the transformed triangle is $0\le s\le r\le 1$, andThe endpoint singularities are integrable after the inner integral: $r=0$ is only a collapsed vertex, while $r=1$ has the usual inverse-square-root behavior. The diagrams indicate the complete triangles, rather than rectangular bounds that would include points absent from the original region.