Double integral in polar coordinates

**Yan** · May 24th 2008, 04:01 PM

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

Bounded by the paraboloid z=10-3x^2-3y^2 and the plane z=4.

I don't know how to defined the r going from where to where and what should be the equation for the double integral.

**galactus** · May 24th 2008, 04:33 PM

You can convert $10-3(x^{2}+y^{2})=10-3r^{2}$

$10-3r^{2}=4$

$r=\pm\sqrt{2}$

Don't forget about the extra r in polar. That gives us $10r-3r^{3}$ .

$\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}(10r-3r^{3})drd{\theta}$

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