# Another Polar Double Integral

by G01
Tags: double, integral, polar
 HW Helper P: 2,688 Hey everyone, My task this time is to derive the volume of a sphere using a polar double integral. The sphere has radius $$a$$ we know that r goes from 0 to a in this integral. The equation for a sphere is: $$x^2 + y^2 +z^2 = r^2$$ or $$f(x,y) = \sqrt{r^2 -x^2 -y^2}$$ and it intersects the x-y plane in a circle: $$x^2 + y^2 = r^2$$ So if we find the volume over this circle and under the positive half of the sphere and double it we should get the volume we want. r goes from 0 to a, and $$\theta$$ goes from 0 to$$2\pi$$. So we get: $$2 \int_0^{2\pi} \int_0^a f( r\cos\theta , r\sin\theta ) r dr d\theta$$ My problem is that whenever I fill in the polar conversions for x and y my sphere equation becomes z=0. I know I'm missing something simple but I can't figure it out. Any help would be appreciated. Thank you.