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Math Help - Intergral in Polar Coordinate

  1. #1
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    Intergral in Polar Coordinate

    I need help solving this problem.

    Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles x^2+y^2=144 and  x^2-12x+y^2=0
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  2. #2
    MHF Contributor matheagle's Avatar
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    Its a good idea to know what these circles look like.
    The first is easy, its x^2+y^2=144 or r=12, which is a circle of radius 12.
    if you add 36 to both sides of the second you get (x-6)^2+y^2=36.
    Now that's a circle that has diameter 12. It traces from (0,0) to (12,0).
    NOW that we know what it looks like we can integrate it from zero to \pi/2.
    The outer radius is 12 and the inner one is that messy circle which we need to solve for in terms of r.
    I would do the cal 3 integral of rdrd \theta.

    Actually, forget about it.
    It's one circle inside another, just use geometry.
    Unless I drew it incorrectly, it's one quarter of the area of the bigger circle of radius 12,
    minus half the area of the smaller circle of radius 6...
    \pi\biggl({144\over 4}-{36\over 2}\biggr).
    Last edited by matheagle; March 24th 2009 at 06:45 PM.
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