Double integral in polar coordinates 2

**Yan** · May 24th 2008, 04:48 PM

Use polar coordinates to find the volume of the given solid.

Inside the sphere x^2+y^2+z^2=16 and outside the cylinder x^2+y^2=4

**galactus** · May 24th 2008, 06:07 PM

On thing you could do is find the volume of the sphere and then the volume of the cylinder, then subtract them. That will be the volume between the sphere and the cylinder.

$V_{s}=\int_{0}^{2\pi}\int_{0}^{\sqrt{16-r^{2}}}\int_{0}^{4}\int_{0}^{2\pi}rdzdrd{\theta}$

This can be checked by just using the formula for the volume of a sphere of radius 4.

$r^{2}=4, \;\ z^{2}=\sqrt{16-r^{2}}$

$z=\pm\sqrt{12}$

The cylinder has height $\sqrt{12}$ and radius 2

It's volume is $V_{c}=\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{\sqrt{1 2}}rdzdrd{\theta}$

This can be checked by using the formula for the volume of a cylinder with height $\sqrt{12}$ and radius 2.

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