Volume using triple integrals

  • Thread starter PhysDrew
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Homework Statement


Use a triple integral to calculate the volume of the solid enclosed by the sphere
x^2 + y^2 + z^2=4a^2 and the planes z=0 and z=a


Homework Equations


Transform to spherical coordinates (including the Jacobian)


The Attempt at a Solution


I'm stuck, as the radius of the sphere is not constant through the area of integration due to the plane z=a. It looks like I should split this integral up, but I'm just not sure on how to do it. It looks like the angle rho is (pi)/3 when the radius of the sphere(2a, from the origin) hits the z=a plane. Please help!
 

Answers and Replies

  • #2
HallsofIvy
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Yes, for [itex]\phi< \pi/3[/itex] the upper boundary is the z= a plane. So, going up from the origin to z= a, then over to the angle [itex]\phi[/itex], you have a right triangle with angle [itex]\phi[/itex] and near side of length a. [itex]\rho[/itex] is the length of the hypotenuse of that right triangle. [itex]\cos(\phi)= a/\rho[/itex] so
[tex]\rho= \frac{a}{\cos(\phi)}= a \sec(\phi)[/tex].
 
  • #3
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Yes, for [itex]\phi< \pi/3[/itex] the upper boundary is the z= a plane. So, going up from the origin to z= a, then over to the angle [itex]\phi[/itex], you have a right triangle with angle [itex]\phi[/itex] and near side of length a. [itex]\rho[/itex] is the length of the hypotenuse of that right triangle. [itex]\cos(\phi)= a/\rho[/itex] so
[tex]\rho= \frac{a}{\cos(\phi)}= a \sec(\phi)[/tex].
Oh good so I was right, thanks. So should I integrate with rho being between 0 and 2a, with phi being between pi/2 and pi/3, and then....well I get stuck there. I've got that little cone bit in the middle to go and I don't know how to get him (or her). Sorry I'm just not getting this one
 

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