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Find the volume of material cut from the solid sphere $x^2 + y^2 + z^2 \leq 9$ by the cylinder $r = 3\sin(\theta)$ using cylindrical coordinates.

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  • $\begingroup$ Im getting the answer as 18pi which is the volume of the hemisphere and hence obviously wrong $\endgroup$ – user34304 Nov 19 '15 at 15:07
  • $\begingroup$ How did you set up your triple integral? Please include this in your question. $\endgroup$ – user147263 Nov 19 '15 at 15:33
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In cylindrical coordinates the equation of the sphere is: $r^2+z^2=9$ so, since $r\le 3 \sin \theta$ the volume is limited by the inequalities:

$$ 0\le \theta \le 2\pi \qquad 0\le r \le 3 \sin \theta \qquad-\sqrt{9-r^2}\le z \le \sqrt{9-r^2} $$

so, given the symmetry, the volume is: $$ V=4 \int_0^{\frac{\pi}{2}}\int_0^{3\sin \theta}\int_0^{\sqrt{9-r^2}}rdzdrd\theta $$ where $rdzdrd\theta$ is the volume element in cylindrical coordinates.

so we have:

$$ V=4\int_0^{\frac{\pi}{2}}\int_0^{3\sin \theta}r\sqrt{9-r^2}drd\theta= 4\int_0^{\frac{\pi}{2}}\sqrt{(9-9\sin^2 \theta)^3}drd\theta=-36\int_0^{\frac{\pi}{2}}\left(\cos^3 \theta-1 \right)d\theta $$

can you do from this?

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