Find volume bounded by two cylinders using triple integration

Question

Given $x^2+y^2=2ay$ and $z^2=2ay$. I tried to use cylindrical corordinated but unable to solve the integral

Answer 1

I will assume that $a>0$. Then $x^2+y^2=2ay\iff x^2+(y-a)^2=a^2$ and therefore $y\in[0,2a]$. In particular, $y\geqslant0$.

In cylindrical coordinates, you have $r^2=z^2=2ar\sin\theta$. Besides $\theta\in[0,\pi]$, since $y\geqslant0$. So, one needs to compute the triple integral$$\int_0^\pi\int_0^{2a\sin\theta}\int_{-\sqrt{2ar\sin\theta}}^{\sqrt{2ar\sin\theta}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$It turns out that\begin{align}\int_0^\pi\int_0^{2a\sin\theta}\int_{-\sqrt{2ar\sin\theta}}^{\sqrt{2ar\sin\theta}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta&=\int_0^\pi\int_0^{2a\sin\theta}2\sqrt2r^{\frac32}\sqrt{a\sin\theta}\,\mathrm dr\,\mathrm d\theta\\&=\int_0^\pi\frac{32}5a^3\sin^3\theta\,\mathrm d\theta\\&=\frac{128}{15}a^3.\end{align}