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

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

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  • $\begingroup$ 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

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