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Surface of a sphere intersecting two cylinders

  1. Aug 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the surface area of the surface [tex] x^2 + y^2 + z^2 = a^2 [/tex], that is outside the two cylinders [tex] x^2+y^2=\pm ax[/tex].

    2. Relevant equations

    [tex]ds= \sqrt{1+(Z_x')^2+(Z_y')^2}dxdy[/tex]

    3. The attempt at a solution

    The intersections are clearly four ellipses. I will try to find a area of one of these ellipses in order to multiply it by four and subtract that from the known area of a whole sphere.
    I have tried two different attitudes to solve it. First, I used the equation mentioned above, to convert this integral to a double integral.
    [tex]S = \iint_D \sqrt{\frac{a^2}{a^2-x^2-y^2}}dxdy[/tex] when D is a circle given by [tex] \left(x-\frac{a}{2}\right)^2 + y^2 \le \left(\frac{a}{2}\right)^2 [/tex]
    Alas, I haven't succeeded in solving this one (not when staying cartesian, nor with polar co-ordinates).
    The second idea was to use spheric coordinates in the first place, but then the limits for theta and phi turn out to be [tex]0<\phi < \frac{\pi}{2} , -a\cos\phi<\sin^2\theta < a\cos\phi[/tex] Which ain't nice either.
    ANY help would be appreciated, especially before my exam on Sunday
  2. jcsd
  3. Aug 15, 2008 #2


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    Welcome to PF!

    Hi Jaron! Welcome to PF! :smile:
    Nooo … it's not even a plane figure … an ellipse is a plane figure, and any plane intersects a sphere in a circle, but these cylinders in ellipses.
    hmm … clever isn't always the easiest. :wink:

    have you tried the dumb approach, just using horizontal slices of thickness dy?

    (and didn't Archimedes have a theorem about areas on spheres and cylinders?)
  4. Aug 17, 2008 #3
    Re: Welcome to PF!

    Thanks tiny Tim!

    Yep.. You were right.
    About Archimedes, I only find his ideas over a cylinder of the same radius as the sphere.

    Anyway, thanks for you help
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