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Setting up limits of integration for multiple integral

  1. Apr 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to find the volume of the region bounded by

    [tex] (x-1)^2 + y^2 =1 \ \ \text{and} \ \ x^2+y^2+z^2=4 \ .[/tex]
    But I only need help setting up the limits of integration.

    2. Relevant equations

    The typical cylindrical change of variables.

    3. The attempt at a solution

    I have [itex] 0 \leq r \leq 2\cos\theta, \ -\sqrt{4-r^2} \leq z \leq \sqrt{4-r^2}, \ \text{and} \ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}. [/itex] Then the volume is given by
    \int\limits_{-\pi/2}^{\pi/2}\int\limits_0^{(2\cos\theta)}\int\limits_{(-\sqrt{4-r^2})}^{(\sqrt{4-r^2})} dz\,(r\,dr)\,d\theta \ .
  2. jcsd
  3. Apr 19, 2013 #2


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    Pls define your cylindrical coordinate frame.
  4. Apr 19, 2013 #3


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    Your integral is over a cylinder with center at (0, 0). The cylinder of the problem has center at (1, 0).
  5. Apr 19, 2013 #4
    @haruspex: Sorry, but I do not know what you mean by cylindrical coordinate frame.

    @HallsofIvy: I thought that taking [itex] 0 \leq r \leq 2\cos\theta [/itex] with [itex] \theta\in(-\pi/2, \pi/2) [/itex] made it so that I would be integrating over the projection of the cylinder onto the [itex] x,y [/itex] plane as a circle of radius 1 centered at (1,0). The region is bound by [itex] \pm\sqrt{4-r^2} [/itex] on [itex] z [/itex]. Can I get some insight into how I can fix my limits of integration?
  6. Apr 19, 2013 #5


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    That integral looks fine to me.

    What's your question?
  7. Apr 19, 2013 #6


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    Agreed. Not clear what the fuss is about.
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