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Setting up a triple integral in cylindrical coordinates?

  1. Apr 12, 2005 #1
    The problem says to find the volume of material cut from the solid sphere,

    [tex]r^2 + z^2 \le 9[/tex]

    by the cylinder,

    [tex]r = 3\sin\theta[/tex]

    I don't know how to graph the first equation, but I can do the second in polar coordinates. How do I go about converting to use cylindrical coordinates?
     
  2. jcsd
  3. Apr 12, 2005 #2

    SpaceTiger

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    It's a sphere of radius 3 centered at the origin:

    [tex]r^2+z^2=x^2+y^2+z^2=9[/tex]


    It's already in cylindrical coordinates. The equation is implying that [tex]r=3sin(\theta)[/tex] for all z.
     
  4. Apr 12, 2005 #3
    I did a crude sketch and came up with this integral

    Is this correct limits for the problem?

    [tex]\int_{0}^{2\pi}\int_{0}^{3\sin\theta}\int_{-\sqrt{9 - r^2}}^{\sqrt{9 - r^2}}\;dz\;r\;dr\;d\theta[/tex]
     
  5. Apr 12, 2005 #4

    HallsofIvy

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    "Cylindrical coordinates" is simply polar coordinates with "z" added.
    [tex] r= 3 sin \theta[/tex], a circle with center at (0, 3/2), radius 3/2, in polar coordinates is a cylinder running parallel to the z axis in cylindrical coordinates.
    Since r2= x2+ y2, the sphere, x2+ y2+ z2= 9 is r2+ z2= 9 in cylindrical coordinates.
     
  6. Apr 12, 2005 #5

    SpaceTiger

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    Looks good to me.
     
  7. Apr 12, 2005 #6
    Many Thanks

    Thank you both. When I have more time on my hands, I will be sure and return the favor to someone here someday :smile:
     
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