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Triple integrals in spherical & cylindrical coordinates

  1. Jul 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Set up triple integrals for the volume of the sphere rho = 2 in (a) spherical, (b) cylindrical, and (c) rectangular coordinates.

    2. Relevant equations
    Volume in cylindrical coordinates: Triple integral of dz r dr d(theta) over region D.
    Volume in spherical coordinates: Triple integral of [p^2 sin(phi) dp d(phi) d(theta)] over region D.

    p = rho
    3. The attempt at a solution
    Trying to find limits of integration:
    For (a), I looked at the limits starting with theta, then phi and then p. Because the region is a sphere, starting from the positive x-axis I thought theta would range from 0 to 2pi (make one complete revolution).

    phi I thought should range from the positive z-axis to the negative z-axis, so from 0 to pi.

    And finally p should range from 0 to 2.

    However, the answers are the following:
    8 ||| p^2 sin(phi) dp d(phi) d(theta) (note: using "|||" for triple integral)

    with these limits of integration:
    0 <= theta <= pi/2
    0 <= phi <= pi/2
    0 <= p <= 2

    Of these, I only got the limit for p correct. Can someone explain with theta and phi are only from 0 to pi/2? Also, where did the 8 come from before the triple integral?

    (b) Using a similar approach, I thought these should be limits of integration: 0 <= theta <= 2pi, 0 <= r <= 2, 0 <= z <= sqrt(4-r^2)

    The correct answers are:
    8 ||| r dz dr d(theta)

    with these limits of integration:
    0 <= theta <= pi/2
    0 <= r <= 2
    0 <= z <= sqrt(4-r^2)

    Once again I have no idea how to get those correct limits for theta and z. Since the graph is a sphere, I thought theta has to range from 0 to 2pi. And again I don't know where the 8 is coming from.

    I figure if I can understand these then rectangular coordinates shouldn't be much harder. Can anyone explain these to me? Thanks
  2. jcsd
  3. Jul 1, 2008 #2


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    They are integrating over one octant of the sphere and then multiplying by 8. I don't know exactly why, but I guess they like to do it that way. Your solution is also correct, without the 8 factor. Except I would change the z range in cylindrical coordinates to -sqrt to +sqrt.
  4. Jul 2, 2008 #3
    Oh I see, yeah that is weird, but thanks for the response!

    Another quick question though if you don't mind answering:

    Find the volume of the smaller region cut from the solid sphere p <= 2 by the plane z = 1.

    So theta should still range from 0 to 2pi, right? And z should be from -sqrt(4-r^2) to sqrt(4-r^2) like in the problem I originally posted, right? But how do you get phi? I know it should start at 0, but this problem only asks for the smaller region (which is above the plane z = 1), so where does phi end? It would make sense to be somewhere between 0 and pi/2, but I'm not sure how to find where exactly is the top limit of phi.

    Thanks again!
  5. Jul 2, 2008 #4


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    Draw a picture. If it's above z=1, then 1<=z<=sqrt(4-r^2), yes? You only need to worry about phi if you are in spherical coordinates. Then you use trig to find the angular value where z=1 and r=2. Don't get confused by the coordinate systems. r in cylindrical coordinates is a different animal from r in spherical coordinates.
  6. Jul 2, 2008 #5
    What do you mean by use trig to find the angular value where z=1 and r=2? Is this for finding theta? Isn't theta the same for both cylindrical and spherical coordinates, so it would be 0 to 2pi? I'm a little unsure about r. r is the distance from the origin to the point, so would r range from 1 to 2?
  7. Jul 2, 2008 #6


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    If you notice, Dick is talking about phi not theta.
    The solid is a section cut from a solid sphere, what is the only condition you are given on r (aka [itex]\rho[/itex])?
  8. Jul 2, 2008 #7


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    I should probably have said that the phi limit is defined by the intersection of the plane z=1 and the sphere p=2 (and not mixed up r and p). But notice within the integration range of phi, the lower limit of p depends on phi. A picture would really help. Did you sketch one?
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