Question

Use double integrals and polar coordinates to find the volume of the solid that lies inside the graph of both x²+y²=1 and x²+y²+z²=9

Answer 1

$\begin{aligned} x=& r \cos \theta, \quad y=r \cos \theta, \quad z=z, \quad d V=r d z d r d \theta \\ \operatorname{volume}(E) &=\iiint_{E} 1 d V=\int_{0}^{2 \pi} \int_{1}^{3} \int_{-\sqrt{9-r^{2}}}^{\sqrt{9-r^{2}}} r d z d r d \theta \\ &=\int_{0}^{2 p i} \int_{1}^{e} 2 r \sqrt{9-r^{2}} d r d \theta=\left.\int_{0}^{2 \pi} \frac{-2}{3}\left(9-r^{2}\right)^{3 / 2}\right|_{r=1} ^{r=3} d \theta \\ &=\int_{0}^{2 \pi} \frac{32 \sqrt{2}}{3} d \theta=\frac{64 \pi \sqrt{2}}{3} \end{aligned}$