Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another Polar Double Integral

  1. Nov 24, 2006 #1


    User Avatar
    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. Nov 24, 2006 #2


    User Avatar
    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. Nov 24, 2006 #3


    User Avatar
    Homework Helper
    Gold Member

    wow! i must be tired!! Thanks alot shredder
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook