Finding the volume of a sphere with a double integral

[Sleepycoaster]

Homework Statement

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∫₀¹∫₀^√(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.

Homework Equations

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.

 

[Sleepycoaster]

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

 

[LCKurtz]

Homework Statement

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, from each of the three dimensions.

8∫₀¹∫₀^√(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.

Homework Equations

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.

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.

 

[Sleepycoaster]

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