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Homework Help: Triple integral (spherical coordinate)

  1. Jul 8, 2011 #1
    find the volume of the solid D that lies above the cone z = (x^2 + y^2)^1/2
    and below the sphere z = (x^2 + y^2 + z^2)

    i've done the integration until i need to substitute cos phi = u..
    however.. i dont know to change the range..

    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 8, 2011 #2


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    [Edit] Never mind the below paragraph. Your work is correct; I misread the equation at first. All you have to do is put in the correct u limit in your last equation. Since u = cos(φ) and φ goes from 0 to pi/4, your upper u limit is 1/sqrt(2). Be careful with your lower limit for u, it isn't 0. Just finish your integration.

    I don't think your integral is set up correctly. I see a sphere centered at (0,0,1/2) with radius 1/2 intersecting the cone at z = 1/2. So you apparently want the volume under the top half of the sphere but above the cone. Are you working this in two parts, the hemisphere and the cone with a flat base? If so, for the cone, ρ goes from 0 to z=1/2 which is ρcos(φ)=1/2 or ρ = (1/2)sec(φ). And you have to add the hemisphere.
    Last edited: Jul 8, 2011
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