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Need help setting up triple integral in spherical coordinates

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Use spherical coordinates to find the volume of the solid bounded above by the sphere with radius 4 and below by the cone z=(x^2 + y^2)^(1/2).

    2. Relevant equations
    All general spherical conversions
    Cone should be [tex]\phi[/tex]=[tex]\pi[/tex]/4

    3. The attempt at a solution
    So far I think the triple integral setup is

    My question is, for dV, do I need anything more than ([tex]\rho[/tex]^2)sin[tex]\phi[/tex]d[tex]\rho[/tex]d[tex]\theta[/tex]d[tex]\phi[/tex]? Or do I need to figure out the intersection and volume that describes the area bounded above by the sphere and below the cone? Or do I already have that with my limits and standard dV question? (if I am correct so far). Any help would be great. Thanks.
  2. jcsd
  3. Nov 23, 2008 #2


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    The boundaries are given by the limits to the integration. How did you arrive at the limits? No, they are not wrong, if the center of the sphere is at the point where all the cartesian coordinates are zero.
  4. Nov 23, 2008 #3
    Yup. Sphere is x^2 + y^2 + z^2 = 16. So with the limits for the three variables and dV converted I am good to integrate?
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