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Math Help - area of annulus

  1. #1
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    area of annulus

    find area between 2 circles: x^2+y^2=1 and x^2+y^2=4

    using (x+y) da

    I got the answer of (3*pi)/2 is this a correct answer? To check I just took the difference between the two circles and then got the half of that. Does this seem right?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by caliguy View Post
    find area between 2 circles: x^2+y^2=1 and x^2+y^2=4

    using (x+y) da

    I got the answer of (3*pi)/2 is this a correct answer? To check I just took the difference between the two circles and then got the half of that. Does this seem right?
    \int\int_A(x+y)\,dA

    Use polar coordinates.

    x=r\cos\theta
    y=r\sin\theta
    dA=r\,dr\,d\theta
    r=1..2
    \theta=0..2\pi

    \int_0^{2\pi}\int_1^2 r^2(\cos\theta+\sin\theta)\,dr\,d\theta

    When I integrate this, I got zero. It's possible I made a sign goof somewhere though.
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  3. #3
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    The integral over the region is zero because

     \int\int_{\omega} x ~dA = ( x-center of gravity ) ( area of the region )


     \int\int_{\omega} y ~dA = ( y-center of gravity ) ( area of the region )

    the center of gravity is  (0,0)
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