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Homework Help: Cylindrical Coordinates

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Use Cylindrical Coordinates.
    Find the volume of the solid that the cylinder [tex]r=acos\theta[/tex] cuts out of the sphere of radius a centered at the origin.


    2. Relevant equations
    Sphere = x2+y2+z2=a3


    3. The attempt at a solution
    I think that the limits are from -pi/2 to positive pi/2 for theta, 0 to acos(theta) for r, and negative (a3-r2)1/2 to positive (a3-r2)1/2. This gives me the equation:
    [tex]\int^{\pi/2}_{-\pi/2}\int^{acos\theta}_0\int^{\sqrt{a^3-r^2}}_{-\sqrt{a^3-r^2}} dzrdrd\theta[/tex]
    Solving this, I get a volume of [tex]\frac{4\pi}{3}a^{9/2}+\frac{8}{9}a^3[/tex]
    But is this right?
     
  2. jcsd
  3. Nov 17, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi squeeky! Welcome to PF! :smile:

    erm … you need oiling! :biggrin:

    i know it's three-dimensional, but still …

    it should be x2+y2+z2=a2 :redface:
     
  4. Nov 17, 2008 #3

    HallsofIvy

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    No, it's not. The the graph of the equation [itex]r= acos(\theta)[/itex], in polar coordinates is a circle with center at (0, a/2) and radius a/2. Since it lies only in the upper half plane, [itex]\theta[/itex] ranges from 0 to [itex]\pi[/itex], not [itex]-\pi/2[/itex] to [itex]\pi/2[/itex].
     
  5. Nov 17, 2008 #4

    tiny-tim

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    Hi HallsofIvy! :smile:

    No, that would be r = a sintheta. :wink:
     
  6. Nov 17, 2008 #5

    HallsofIvy

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    Oops! Well, I'll just go back and edit it so it looks like I never made that mistake!
     
  7. Nov 18, 2008 #6
    Re: Welcome to PF!

    Ah! That's right, I don't know how I got that cube, I must have been seeing things when I looked up the formula.

    And so now I get a an equation of [tex]\int^{\pi}_0\int^{acos\theta}_0\int^{\sqrt{a^2-r^2}}_{-\sqrt{a^2-r^2}}dzrdrd\theta[/tex]
    which (unless I did my math wrong) gives me a somewhat nice value of [tex]V=(\frac{2a^3}{3})(\frac{4}{3}-\pi)[/tex]
    Is this right now?
     
  8. Nov 18, 2008 #7

    tiny-tim

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    yes, that looks right …

    except doesn't theta go from -π/2 to π/2, as in your original post? :smile:
     
  9. Nov 18, 2008 #8
    That's what I thought at first, but HallsofIvy pointed out that it's actually from 0 to pi. Was I right at first then? Because it does make more sense to me if it is from -pi/2 to pi/2, since I see the limits as lying in the xz-plane.
     
  10. Nov 19, 2008 #9

    tiny-tim

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    Yes! :biggrin:

    The circle has centre (a/2,0) …

    for example, the point (0,1) obviously doesn't lie on it. :wink:
     
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