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Vol of spherical cap?

  1. Aug 16, 2006 #1
    finished

    ok i think ive got it. thanks
     
    Last edited: Aug 16, 2006
  2. jcsd
  3. Aug 16, 2006 #2

    siddharth

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    You need to show your work before you get help. What are your thoughts/ideas on this problem?
     
  4. Aug 16, 2006 #3

    Integral

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    I do not think that there is any difference between a "spherical cap" and a hemisphere.

    Finding a volume using calculus generally involves an integral, there are several methods which can be used to solve such a problem. Do you recall seeing something having to do with "disks"?
     
  5. Apr 15, 2011 #4
    Let F defined on [0,r]x[0,2\pi]x[0,\pi] by
    F(R, S, T)=(R\cos(S)\sin(T), R\sin(S)\sin(T), R\cos(T))
    (the change into spherical coordinates).
    The absolute value of its Jacobian is R^2\sin(T).

    Your domain of integration (cap) is defined by
    {(x, y, z) : x^2 + y^2 + z^2 \leq r^2 , r-h \leq z \leq r}
    and its pre-image through F is
    D=[0,r]x[0,2\pi]x[0,T_0],
    where T_0 satisfies \cos(T_0)=(r-h)/r=1-h/r.

    Volume = \int_{D}R^2\sin(T)dRdSdT
    = (\int_0^rR^2dR).(\int_0^{2\pi}dS).(\int_…
    = 1/3r^3.2\pi.(1-\cos(T_0)) = 2\pi.h.r^2/3.
    QED.
     
  6. Apr 15, 2011 #5

    HallsofIvy

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    Please do not erase a post after it has been resolved! Other people are sure to have the same question and can still learn from thread.
     
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