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Polar Coordinates, intersection of a cylinder with a spher

  1. Aug 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the Volume of the solid that the cylinder ##r = acos\theta## cuts out of the sphere of radius a centered at the origin.


    2. Relevant equations



    3. The attempt at a solution

    I have defined the polar region as follows,
    $$D = \{ (r,\theta) | -\pi/2 ≤ \theta ≤ \pi/2 , 0 ≤ r ≤acos\theta \} $$

    I will take f(x,y) as follows,
    $$ f(x,y) = \sqrt{a^2 - x^2 - y^2} $$

    Because of symmetry, I should be able to take the removed volume as double the double integral of f(x,y) over the region D.

    $$ V = 2 *\int_{-\pi/2}^{\pi/2} \int_{0}^{acos\theta} r\sqrt{a^2 - r^2} \mathrm{d} r \mathrm{d} \theta $$

    After solving this double integral, I keep ariving at ## {2\pi a^3}/3 ## Which I don't think makes sense because that would be half the volume of the sphere. What am I doing wrong???
     
    Last edited: Aug 13, 2014
  2. jcsd
  3. Aug 13, 2014 #2

    Ray Vickson

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    Show your work after the above; we cannot tell where you went wrong if you don't show us what you did---in detail, step-by-step.
     
  4. Aug 13, 2014 #3
    ## V = -2/3 * \int_{0}^{2\pi} (a^2 - a^2 cos^2(\theta))^{3/2} - a^{3/2} d\theta ##
    ## \Rightarrow V = -2/3 * \int_{-\pi/2}^{\pi/2} a^3 (1- cos^2(\theta))^{3/2} - a^3 d\theta ##
    ## \Rightarrow V = -(2*a^3)/3 * \int_{-\pi/2}^{\pi/2} sin^3(\theta) - 1 d\theta ##
    ## \Rightarrow V = (2*a^3*\pi)/3 ##

    Hope that's enough, any help with be very appreciated.
     
  5. Aug 13, 2014 #4

    Ray Vickson

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    One obvious problem is the fact that ##\int_{-\pi/2}^{\pi/2} \sin^3 (\theta) \, d \theta = 0## because ##\sin## is an odd function. This, in turn, arises from the mistake of taking ##(1-\cos^2)^{3/2} = \sin^3##. The fact is that you started off with positive square roots, then got a 3/2 power via integration. In fact, ##\sqrt{1-\cos^2} = |\sin|##. You could have avoided all that by using symmetry to just integrate from 0 to ##\pi/2##, then multiply the result by 2.
     
  6. Aug 13, 2014 #5
    Thank you very much, I see where I went wrong. After using your method, I arrived at a final answer of
    ##V = 4/3 a^3 *((3\pi - 4)/6) ##
    which is about 29 percent of the sphere's volume. Makes more sense. Thanks again for your help!
     
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