Need help setting up triple integral in spherical coordinates

  • Thread starter Outlaw747
  • Start date
  • #1
10
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Homework Statement


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).


Homework Equations


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


The Attempt at a Solution


So far I think the triple integral setup is
0[tex]\leq[/tex][tex]\rho[/tex][tex]\leq[/tex]4
0[tex]\leq[/tex][tex]\theta[/tex][tex]\leq[/tex]2[tex]\pi[/tex]
0<[tex]\phi[/tex][tex]\leq[/tex][tex]\pi[/tex]/4

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.
 

Answers and Replies

  • #2
ptr
28
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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.
 
  • #3
10
0
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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