Thread: Double integral in polar coordinates

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.

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

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

$\displaystyle r=\pm\sqrt{2}$

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

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