Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$

Question

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=$\int_{\theta=0}^{\pi}\int_{r=0}^{2a\sin \theta}\int_{z=0}^{r^2/a}r\ dr\ d\theta \ dz$

Answer 1

Using cylinder coordinate is the best way to solve this problem : $$\left\{(r\cos\theta ,r\sin\theta +a ,z)\mid \theta \in [0,2\pi], r\in[0,a], z\in \left[0,\frac{2r^2+2ar\sin \theta+a^2}{a}\right]\right\}.$$

Answer 2

You may wish to re-order your integration. You can also exploit the symmetry of the region of integration about the $ \ yz- $ plane by integrating over $ \theta $ from $ \ 0 \ $ to $ \ \frac{\pi}{2} \ $ . Thus,

$$ V \ = \ 2 \ \int_{ 0}^{ \pi / 2}\int_{0}^{2a\sin \theta}\int_{0}^{r^2/a} \ \ \ dz \ \ r \ dr \ \ d\theta \ \ = \ \ 2 \ \int_{ 0}^{ \pi / 2}\int_{0}^{2a\sin \theta} \ \ \frac{r^2}{a} · r \ dr \ \ d\theta $$

$$ = \ \ \frac{2}{a} \ \int_{ 0}^{ \pi / 2} \ \ \frac{1}{4} \ (2a \sin \theta)^4 \ \ d\theta \ \ = \ \ \frac{2·2^4·a^4}{4a} \ \int_{ 0}^{ \pi / 2} \sin^4 \theta \ \ d\theta \ \ = \ \ 8a^3 \ · \ \frac{3 \pi}{16} $$

$$ = \ \ \frac{3 \pi}{2}a^3 \ \ , $$

the last definite integral being of a kind $ \left( \ \int_{ 0}^{ \pi / 2} \sin^n \theta \ \ d\theta \ = \ \int_{ 0}^{ \pi / 2} \cos^n \theta \ \ d\theta \ \right) $ for which a formula to evaluate it is available.