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
"\\begin{aligned}\nx=& r \\cos \\theta, \\quad y=r \\cos \\theta, \\quad z=z, \\quad d V=r d z d r d \\theta \\\\\n\\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 \\\\\n&=\\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 \\\\\n&=\\int_{0}^{2 \\pi} \\frac{32 \\sqrt{2}}{3} d \\theta=\\frac{64 \\pi \\sqrt{2}}{3}\n\\end{aligned}"
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