= Solution
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> $x=a\cos\phi$, $y=a\sin\phi$, with <area element> $dS=a\,d\phi\,dz$:
$$
I_{\mathrm{side}}=\int_0^{2\pi}\int_0^b az\,dz\,d\phi=\pi ab^2.
$$
The upper disc has constant $z=b$ and area $\pi a^2$, giving $I_{\mathrm{top}}=\pi a^2b$. On the hemispherical indentation, use <spherical coordinates> with $z=a\cos\theta$ and $dS=a^2\sin\theta\,d\theta\,d\phi$:
$$
I_{\mathrm{hemisphere}}=2\pi a^3\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=\pi a^3.
$$
Adding the three contributions yields
$$
\boxed{\iint_S(z+y^3)\,dS=\pi a(b^2+ab+a^2)}.
$$
This is a scalar <surface integral>, so the indentation contributes positively through $dS$; an inward or outward orientation does not change its sign.
Back to article page
