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

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.