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Math Help - help with triple integral spherical coordinates

  1. #1
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    help with triple integral spherical coordinates

    Set up a triple integral that represents the volume of part of the ball x^2 + (y-1)^2 + z^2 </= 1 that is outside the cone rho = pi/6 and inside the cone rho = pi/3

    In the answer, the dTheta part of the triple integral goes from 0 to pi. My question is how do you find that? Normally theta goes from 0 to 2pi.
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  2. #2
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    Quote Originally Posted by messianic View Post
    Set up a triple integral that represents the volume of part of the ball x^2 + (y-1)^2 + z^2 </= 1 that is outside the cone rho = pi/6 and inside the cone rho = pi/3

    In the answer, the dTheta part of the triple integral goes from 0 to pi. My question is how do you find that? Normally theta goes from 0 to 2pi.
    If you look at the sphere from the top it's a circle of radius 1 and centered at (0,1). The values of \theta for this is [0,\pi]. If it was centerd at (0,0), then it would be [0,2 \pi] as you say.
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  3. #3
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    Perhaps I'm missing something, but cones in spherical coordinates are usually given by \phi=c and not \rho. If the question was find the volume between the two cones \phi=\pi/6, \phi=\pi/3 and the sphere x^2+(y-1)^2+z^2=1, then that looks pretty tough to me. It's the part of the sphere in the plot below between the two cones. Note a small part of the sphere goes through the smaller cone so would have to take that into account. Maybe though this isn't what you want.
    Attached Thumbnails Attached Thumbnails help with triple integral spherical coordinates-sphere-cones.jpg  
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