Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 19

The divergence isFor any closed piecewise smooth surface bounding a volume, use the divergence theorem with the outward orientation:Reversing the orientation still gives zero. The field is smooth everywhere, so there is no hidden singular source inside the surface.

Choose the plane normal to be $+\hat{y}$. Only the second field component contributes, and on $y=0$ it is $e^{-(x^2+z^2)}$. Thus the surface integral factors into two Gaussian integrals:The original question does not specify an orientation; choosing $-\hat{y}$ instead gives $-\pi$.

The printed first component $\sin y/y$ is undefined on the entire integration plane. Its removable continuous extension there is $1$, since ${\lim }_{y\to 0}\sin y/y=1$. The stated flux uses this natural extension; in fact the normal component alone is already well-defined and integrable on the plane, so the first component does not affect either oriented value.

Take the sphere's outward normal. Since $F=k(x,y,z)$ with constant $k$, its divergence is $3k$. The divergence theorem givesThe translated centre does not enter the answer, because the divergence is constant. Equivalently the constant field produced by translating coordinates has zero net flux through a closed sphere. The inward orientation would negate the result.

Choose the upward orientation of the paraboloid cap. Its boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise as viewed from above. By Stokes theorem,On the boundary, $dz=0$ and $A\cdot dr=ydx-3xydy$. Parametrize by $x=2\cos \theta$, $y=2\sin \theta$, $0\le \theta \le 2\pi$. ThenThe second term integrates to zero over a full period, while the first contributes $-4\pi$. Equivalently, replace the cap by the upward disk at $z=1$: there $(\nabla \times A{)}_z=-3y-1$, whose integral is $-4\pi$ by symmetry.

An orientation is not stated in the original question. The displayed result uses the upward normal, also the outward normal on the curved top of the enclosed cap; the opposite orientation gives $4\pi$.