Another Polar Double Integral

  1. G01

    G01 2,698
    Homework Helper
    Gold Member

    Hey everyone, My task this time is to derive the volume of a sphere using a polar double integral.

    The sphere has radius [tex] a[/tex] we know that r goes from 0 to a in this integral.

    The equation for a sphere is:

    [tex] x^2 + y^2 +z^2 = r^2[/tex]
    or [tex] f(x,y) = \sqrt{r^2 -x^2 -y^2}[/tex]
    and it intersects the x-y plane in a circle:

    [tex] x^2 + y^2 = r^2 [/tex]


    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 [tex]\theta[/tex] goes from 0 to[tex]2\pi[/tex].

    So we get:

    [tex]2 \int_0^{2\pi} \int_0^a f( r\cos\theta , r\sin\theta ) r dr d\theta [/tex]

    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. jcsd
  3. Office_Shredder

    Office_Shredder 4,499
    Staff Emeritus
    Science Advisor
    Gold Member

    Because you're confusing your r's. One of them is a variable, the other is the radius of the sphere
     
  4. G01

    G01 2,698
    Homework Helper
    Gold Member

    wow! i must be tired!! Thanks alot shredder
     
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