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I need to calculate this volume of $$D=\{x^2+y^2-2y\le 0,0\le z\le 10-3\sqrt{x^2+y^2} \}.$$

My attempt. So the first one is a shifted cylinder $x^2+(y-1)^2\le 1$ , and the second one is an upside-down cone with vertex $z=10$.

By using cylindrical coordinates I get:

$$\int_{0}^{2\pi}\int_{0}^{\sqrt{2\sin{\theta}}}\int_{0}^{10-3r}rdzdrd\theta$$

Is this the correct way to approach a shifted region? $x^2+y^2=2y$ implies $r^2=2\sin{\theta}$.

I think I got the integral wrong.

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You are almost correct. The angle $\theta$ should be in the interval $[0,\pi]$, so $0\leq r\leq 2\sin\theta\geq 0$ (note that $r^2\leq 2r\sin\theta$). Thus your triple integral should be $$\int_{0}^{\pi}\int_{0}^{2\sin{\theta}}\int_{0}^{10-3r}rdzdrd\theta.$$ So what is the final result?

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  • $\begingroup$ Could you explain why It is from 0 to $\pi$? $\endgroup$ – NPLS Jan 16 '19 at 7:56
  • $\begingroup$ Draw your circle: it is centered in $(0,1)$ with radius $1$. It is completely contained in the upper half-plane and therefore the angle $\theta$ varies in $[0,\pi]$ (the ray starts in the origin in cylindrical coordinates). $\endgroup$ – Robert Z Jan 16 '19 at 8:00
  • $\begingroup$ Thanks, you are magic! $\endgroup$ – NPLS Jan 16 '19 at 8:07
  • $\begingroup$ @NPLS Fine! Let me know if you obtain the correct result. $\endgroup$ – Robert Z Jan 16 '19 at 8:12
  • $\begingroup$ yup $10\pi -32/3$ $\endgroup$ – NPLS Jan 16 '19 at 8:55

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