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Homework Help: Spherical coordinates find volume

  1. Jul 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Use spherical coordinates to find the volume of the solid that lies above the cone z= sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = z.


    3. The attempt at a solution

    I'm having trouble solving for rho (p). I know it starts from 0, and it reaches to the sphere and to the cone. But I don't know where to go from there.
     
  2. jcsd
  3. Jul 18, 2008 #2
    Assuming you meant [tex]\rho_o[/tex], the question becomes one of finding [tex]\phi[/tex].
    In this case, you have to look at your cone. Essentially, it says [tex]z = r[/tex], where r is in cylindrical coordinates. Now use the conversions to spherical, [tex]z=\rho cos(\phi)[/tex] and [tex] r = \rho sin (\phi)[/tex]. Equating those gives you [tex]cos(\phi) = sin(\phi)[/tex], or [tex] \phi = \pi/4[/tex]. So [tex]0<\phi<\pi/4[/tex].
     
    Last edited: Jul 18, 2008
  4. Jul 18, 2008 #3
    It is a sphere, though. If you move the z over to the left side of the equation, complete the square, you get a sphere with center (0,0,1/2). Also, I need to find rho, not phi.
     
  5. Jul 18, 2008 #4
    Sorry.
    [tex] x^2 + y^2 +z^2 = \rho^2[/tex]

    [tex] z = \rho cos(\phi)[/tex]

    [tex]x^2 + y^2 +z^2 = z[/tex] implies [tex]\rho^2 = \rho cos(\phi)[/tex]

    or [tex]\rho = cos(\phi)[/tex]

    so [tex]0 < \rho < cos(\phi)[/tex]
     
  6. Jul 18, 2008 #5
    Why are we just looking at the sphere though and not the cone boundary?
     
  7. Jul 18, 2008 #6
    Because you said "below the sphere x^2 + y^2 + z^2 = z" which implies the sphere is the upper bound, so its value for [tex]\rho[/tex] is the maximum value that [tex]\rho[/tex] can take on, thus establishing the upper limit of integration. Hope that made sense :)
     
  8. Jul 18, 2008 #7
    That makes sense...but we are also looking at the cone too. And the value of p for the cone doesn't depend on theta...it is constant. How do we know if we need to divide the triple integral expression into 2 separate ones, one for each different p?
     
  9. Jul 18, 2008 #8
    Okay, this took me a while to find, but here's an example of a time when you need to split the expression for different values of [tex]\rho[/tex].

    There are several evident differences between your problem and this one. Most importantly, notice that your solid is "above the cone z= sqrt(x^2 + y^2)" and the example I show is enclosed by the planes. There is a uniquely different value of [tex]\rho[/tex] that the function takes on for different values of [tex]\phi[/tex] namely because the solid that is being integrated must satisfy either being inside a cone or inside a sphere.

    Essentially, in the example I give, there are uniquely different values of [tex]\rho[/tex] for different values of [tex]\phi[/tex], whereas in your case, no such thing occurs. If you try to find a different [tex]\rho[/tex], you get what I did in an earlier post:
    Notice that the [tex]\rho[/tex] cancels, thus showing that there is no other unique value to make different bounds of integration.



    http://img528.imageshack.us/my.php?image=11fm5.jpg
     
    Last edited: Jul 18, 2008
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