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