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Math Help - Polar Coordinates problem area of region

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    Polar Coordinates problem area of region

    Find the area of the region inside: r = 9 sinθ but outside: r = 1
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    Quote Originally Posted by muddyjch View Post
    Find the area of the region inside: r = 9 sinθ but outside: r = 1
    r= 9 sin(\theta) is a circle with center at (0, 4/2) and radius 4/2 while r= 1 is a circle with center at (0, 0) and radius 1.

    The two curves intersect where [/tex]r= 1= 9 sin(\theta)[/tex] or sin(\theta)= 1/9. For 0\le\theta\le\pi, that is satified by \theta= sin^{-1}(1/9) and \theta= \pi- sin^{-1}(1/9).

    For \theta between those two limits, r ranges from 1 up to 9 sin(\theta). The "differential of area" in polar coordinates is r drd\theta so the area is given by \int_{\theta= sin^{-1}(1/9)}^{\pi- sin^{-1}(1/9)}\int_{r= 1}^{9 sin(\theta)} r dr d\theta.
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