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Finding the volume of a sphere with a double integral

  1. Jun 5, 2013 #1
    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. jcsd
  3. Jun 5, 2013 #2
    And yes, everything in this proof refers to a sphere with radius 1.
     
  4. Jun 5, 2013 #3

    LCKurtz

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    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.
     
  5. Jun 5, 2013 #4
    Oh hey, I see where I went wrong. I took the derivative when I should have taken the integral. Thank you!
     
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