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Find volume of the solid which is the intersection of the solid sphere $x^2+y^2+z^2 \leq 9$ and the solid cylinder $x^2+y^2\leq 1$

I have used cylindrical coordinates and set up the following triple integral

$$V=2\int_{0}^{2 \pi} \int_{0}^{1} \int_{0}^{9-r^2} \; r \; dz \; dr \;d \theta$$

Is this integral correct ?

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Almost.

$$V=2\int_{0}^{2 \pi} \int_{0}^{1} \int_{0}^{\color{red}{\sqrt{9-r^2}}} \; r \; dz \; dr \;d \theta$$

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    $\begingroup$ There is a $2$ outside. $\endgroup$ – Siong Thye Goh Dec 13 '17 at 20:33
  • $\begingroup$ Not sure where the $2$ came from? If its standard cylindrical coordinates shouldn't be just $r$? $\endgroup$ – K Split X Dec 13 '17 at 21:07
  • $\begingroup$ If we just convert directly, the lower limit for $z$ should be $-\sqrt{9-r^2}$ but we can take advantage of symmetry and set the lower limit to $0$ and times $2$ outside. $\endgroup$ – Siong Thye Goh Dec 13 '17 at 21:12

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