Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 16

Each of the three surface pieces is invariant under $y\mapsto -y$, while $y^3$ is an odd function under that reflection. Its surface integral therefore vanishes on each piece separately. Only $z$ contributes.

On the cylindrical side, use cylindrical coordinates $x=a\cos \varphi$, $y=a\sin \varphi$, with area element $dS=ad\varphi dz$:The upper disc has constant $z=b$ and area $\pi a^2$, giving $I_{\mathrm{top}}=\pi a^{2}b$. On the hemispherical indentation, use spherical coordinates with $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\varphi$:Adding the three contributions yieldsThis is a scalar surface integral, so the indentation contributes positively through $dS$; an inward or outward orientation does not change its sign.

Write $q(x)=1/[x(1+(\ln x{)}^2)]$. The vector field is undefined at $x=0$ because $q$ diverges there, but this component is tangential to that face. The face fluxes have finite improper integral limits. In particular, the substitution $s=\ln x$ givesUsing the outward unit normal on each face, the flux contributions areHenceFor a precise justification at the singular face, first use the truncated cube $\epsilon \le x\le 1$. Its opposite $y$-face integrals cancel, the $x$-faces contribute $e^{-1}-e^{-\epsilon }$, and the upper $z$-face contributes $1-\epsilon$. The limit as $\epsilon \downarrow 0$ is $e^{-1}$. On each truncated cube the divergence theorem gives the same answer, since $\nabla \cdot \mathbf{F}=1-e^{-x}$. Thus the result is a flux of a vector field with an integrable tangential singularity, rather than an application of the classical smooth-field theorem directly at $x=0$.