Changing order of integration in spherical coordinates

  1. 1. The problem statement, all variables and given/known data

    Let D be the region bounded below by the plane z=0, above by the sphere x^2+y^2+z^2=4, and on the sides by the cylinder x^2+y^2=1. Set up the triple integral in spherical coordinates that gives the volume of D using the order of integration dφdρdθ.


    2. Relevant equations

    The solution says that D is:

    [​IMG]

    3. The attempt at a solution

    I thought that the solution was:

    [​IMG]

    Could you please tell me where I’m going wrong? Many thanks!
     
  2. jcsd
  3. vela

    vela 12,728
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    Your answer is correct. The solution is wrong.
     
  4. Many thanks!
     
  5. LCKurtz

    LCKurtz 8,386
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    I'm afraid I don't agree with vela. The answer given is correct. The problem is the the value of φ on the cylinder depends on ρ. The equation of the cylinder is r = 1 where r is the polar radius. This gives r = ρsin(φ) = 1. If you are at the top of the cylinder φ = π/6 and ρ = 2 while at the bottom of the cylinder φ = π/2 and ρ = 1. φ is the function of ρ given by φ = sin-1(1/ρ). In the second integral, if you start at z axis and move in the φ direction, you start at φ = π/6 and how far you move depends on the value of ρ, and that value is sin-1(1/ρ) That gives the inner integral limits. Then ρ goes from 1 to 2 etc.

    [Edit] Nope, I take it back. vela is right as he explains below.
     
    Last edited: Jun 27, 2010
  6. vela

    vela 12,728
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    You're describing the volume bounded by the cylinder and the sphere ρ=1, but the volume of D also include the region between ρ=0 and ρ=1 where φ can run unrestricted from 0 to π/2. That's what the OP's third integral corresponds to.

    It's easy to write the volume in terms of cylindrical coordinates:

    [tex]V=\int_0^1\int_0^{2\pi}\int_0^{\sqrt{4-r^2}} r\,dz\,d\theta\,dr[/tex]

    This integral evaluates to the same result as the OP's answer, but not the answer from the solution.
     
  7. LCKurtz

    LCKurtz 8,386
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    Ahhh yes, you're right. That's easy to miss; puts me in good company with the author :redface: And just emphasizes the point that you should choose your coordinate system and order of integration carefully, trying not to make the problem trickier than it is.
     
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