# Evaluate this triple integral (volume)

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.

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?
• Could you explain why It is from 0 to $\pi$? – NPLS Jan 16 '19 at 7:56
• 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). – Robert Z Jan 16 '19 at 8:00
• yup $10\pi -32/3$ – NPLS Jan 16 '19 at 8:55