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Math Help - Calculus 3 - Double Integral

  1. #1
    Member VitaX's Avatar
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    Calculus 3 - Double Integral

    As seen in the picture below that's what I have to integrate. You can see I got \frac{1}{2} as my answer yet they say the correct answer is \frac{3}{2}. I don't see how they possibly got that answer. Where did I go wrong?

    =\int_{0}^{3}[\frac{x^2}{2} - \frac{xy}{3}]_{\frac{y}{3}}^{\frac{y}{3}+1} dy

    =\int_{0}^{3}[\frac{1}{2}(\frac{y}{3}+1)^2 -\frac{y}{3}(\frac{y}{3}+1) - (\frac{1}{3}(\frac{y}{3})^2 - \frac{y}{3}(\frac{y}{3}))] dy

    =\int_{0}^{3}[-\frac{y^2}{9} + \frac{1}{2}] dy

    =[-\frac{y^3}{27} + \frac{y}{2}]_{0}^{3} \rightarrow [-\frac{27}{27} + \frac{3}{2}] \rightarrow \frac{1}{2}
    Attached Thumbnails Attached Thumbnails Calculus 3 - Double Integral-calculus-3-double-integral.png  
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  2. #2
    Member VitaX's Avatar
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    Ah I see my error its when I was evaluating the lower limit.
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