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Math Help - Change of Variables

  1. #1
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    Change of Variables

    So I'm asked to find \displaystyle \int \int_{R} xy \, dA where R is the region in the first quadrant bounded by y = x, y = 3x, xy = 1, and xy = 3 using the change x = u/v and y = v.

    I find the equations \sqrt{u} = v, \sqrt{3u} = v, 1 = u, and 3 = u. The Jacobian is 1/v \cdot 0 - (u \cdot - \frac{1}{v^{2}} \cdot 1) = \frac{u}{v^{2}}. So I set up the integral \displaystyle \int_{1}^{3} \int_{\sqrt{u}}^{\sqrt{3u}} u \cdot \frac{u}{v^{2}} \, dv du = \int_{1}^{3} u^{2} (\frac{1}{\sqrt{u}} - \frac{1}{\sqrt{3u}}) du = (1-\frac{1}{\sqrt{3}})\int^{3}_{1} u^{\frac{3}{2}} du = (1-\frac{1}{\sqrt{3}}) \cdot \frac{2}{5} \cdot (3^{\frac{5}{2}} - 1) = (1-\frac{1}{\sqrt{3}}) \cdot \frac{2}{5} \cdot (9\sqrt{3}  -1). Now this looks very far from the answer in the book. But I don't see the mistake.
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  2. #2
    MHF Contributor matheagle's Avatar
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    stewart 16.9 #15

    the jacobian is 1/v, you have the 0 and 1 backwards in your matrix of partial derivatives
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    Gah, stupid mistake, thanks for catching it!
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  4. #4
    MHF Contributor matheagle's Avatar
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    was easy, given that I recognize it, since I assign it every year
    AND having the solution manual here, so I can easily find errors.
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