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

  • $\begingroup$ Well, $y\in[0..2a], x\in[-\sqrt{2ay-y^2}..\sqrt{2ay-y^2}], z\in[0..(x^2+y^2)/a]$ but a change of variables might be better. $\endgroup$ – Graham Kemp Aug 19 '17 at 9:25

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

  • $\begingroup$ Yes, it does look much easier. But what would be the limits in cartesian coords? $\endgroup$ – user467745 Aug 19 '17 at 9:21
  • $\begingroup$ @user467745 : Something as $y\in [0,2a]$, $x\in [-\sqrt{2a-(y-1)^2},\sqrt{2a-(y-1)^2}]$ and $z\in [0, \frac{x^2}{a}+\frac{y^2}{a}]$. $\endgroup$ – Surb Aug 19 '17 at 9:25
  • $\begingroup$ Just to cross check-the answer would be $\pi a^3$, right? $\endgroup$ – user467745 Aug 19 '17 at 10:27
  • $\begingroup$ @user467745: I'm not a computer ;-) $\endgroup$ – Surb Aug 19 '17 at 10:35
  • $\begingroup$ Did I imply that? I am sorry. Thanks for your time $\endgroup$ – user467745 Aug 19 '17 at 10:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.