Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$
My work
Changing to cylindrical coordinates
For Paraboloid
$$x^2+y^2=az\;\;\;\implies r^2=az\;\;\;\implies z=\frac{r^2}{a}$$
For the cylinder
$$x^2+y^2=2ay\;\;\;\implies r=2a \sin \theta$$
Since this volume lies only in first two quadrants $\theta$ goes from $0$ to $\pi$
Volume=$\int_{\theta=0}^{\pi}\int_{r=0}^{2a\sin \theta}\int_{z=0}^{r^2/a}r\ dr\ d\theta \ dz$
Best Answer
Using cylinder coordinate is the best way to solve this problem : $$\left\{(r\cos\theta ,r\sin\theta +a ,z)\mid \theta \in [0,2\pi], r\in[0,a], z\in \left[0,\frac{2r^2+2ar\sin \theta+a^2}{a}\right]\right\}.$$