Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V / 3

The curl is $\nabla \times \mathbf{F}=(x,y,-2z)$. For the upward orientation of the paraboloid, the vector surface area element is $(2x,2y,1)dxdy$. Therefore Stokes theorem givesThe upward normal gives the prescribed anticlockwise boundary orientation viewed from above.

Take the boundary pieces in the orderOn the circular arc $C_1$, use $(x,y,z)=(2\cos \theta ,2\sin \theta ,0)$ for $0\le \theta \le \pi /2$. Then $\mathbf{F}=(0,0,1)$ and $dz=0$, so its line integral is zero.

On $C_2$, use $x=0$, $z=4-y^2$, with $y$ decreasing from two to zero. Since $dx=0$ and $F_y=0$, $\mathbf{F}\cdot d\mathbf{r}=dz$, giving four. On $C_3$, use $y=0$, $z=4-x^2$, with $x$ increasing from zero to two. Again $\mathbf{F}\cdot d\mathbf{r}=dz$, giving minus four. The total is $\boxed{0+4-4=0}$, agreeing with Stokes theorem.

Write $r=\sqrt{x^2+y^2}>0$, so $\mathbf{F}=(f(z)x/r,f(z)y/r,0)$. The Cartesian curl isThe last component is zero because both differentiated terms are $-f(z)xy/r^3$. ThusThe condition $r>0$ excludes the cylindrical-axis coordinate singularity.