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Thread: ∫∫ (x^2)(y)dA

  1. #1
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    ∫∫ (x^2)(y)dA

    find ∫∫ (x^2)(y)dA , where R is the upper half of annulus bounded in between (x^2) +(y^2) = 1 and (x^2) +(y^2) = 4 ..

    Here's my working , i gt my ans = 62/5 instead of the ans given by the author which is 62/15 .... Is my ans wrong ? .
    ∫∫ (x^2)(y)dA-img_20160909_121033.jpg
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  2. #2
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    Re: ∫∫ (x^2)(y)dA

    wrong picture
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    Re: ∫∫ (x^2)(y)dA

    Quote Originally Posted by romsek View Post
    wrong picture
    sorry , here it is∫∫ (x^2)(y)dA-img_20160910_094333.jpg
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    Re: ∫∫ (x^2)(y)dA

    answer is $\dfrac {62}{15}$

    your integral setup is correct

    you forgot a multiplication of 3 in the denominator
    Last edited by romsek; Sep 9th 2016 at 08:00 PM.
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    Re: ∫∫ (x^2)(y)dA

    Quote Originally Posted by romsek View Post
    answer is $\dfrac {62}{15}$

    your integral setup is correct

    you forgot a multiplication of 3 in the denominator
    at where ? why 3 is needed ?
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  6. #6
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    Re: ∫∫ (x^2)(y)dA

    Quote Originally Posted by xl5899 View Post
    at where ? why 3 is needed ?
    review your calcs, it's pretty obvious where
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