# Another Polar Double Integral

1. Nov 24, 2006

### G01

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.

2. Nov 24, 2006

### Office_Shredder

Staff Emeritus
Because you're confusing your r's. One of them is a variable, the other is the radius of the sphere

3. Nov 24, 2006

### G01

wow! i must be tired!! Thanks alot shredder