# Finding the volume of a sphere with a double integral

1. Jun 5, 2013

### Sleepycoaster

1. The problem statement, all variables and given/known data

I know how to find the volume of a sphere just by adding the areas of circles, so I decided to do a double integral to find the same volume, just for fun.

Here's what I've set up. I put 8 out front and designed the integrals to find an eighth of a sphere that has its center at the origin. This piece of the sphere is what you would get if you cut a sphere in half three times, one from each of the three dimensions.

8∫010√(1-y2)√(1-x2-y2)dxdy

√(1-x2-y2) is the equation I used to measure the z-dimension of a point on the sphere given values for x and y.

2. Relevant equations

3. The attempt at a solution

Here's the integral I got for the inner integration, the one with respect to x:

0√(1-y2)√(1-x2-y2)dx

=[-2x/(2√(1-x2-y2))] with 0 on the bottom and √(1-y2) on the top.

As you can see, if I plug in √(1-y2) for x, the values cancel out so that there is only a zero in the denominator.

Is this simply a limitation in using multiple integration, or did I do something wrong? Any help is appreciated.

Last edited: Jun 5, 2013
2. Jun 5, 2013

### Sleepycoaster

And yes, everything in this proof refers to a sphere with radius 1.

3. Jun 5, 2013

### LCKurtz

That antiderivative doesn't look correct. Try changing your dxdy integral to polar coordinates. It will be much easier. Otherwise you will need to do a trig substitution.

4. Jun 5, 2013

### Sleepycoaster

Oh hey, I see where I went wrong. I took the derivative when I should have taken the integral. Thank you!