# Change of Variables

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• Nov 18th 2010, 09:38 PM
ragnar
Change of Variables
So I'm asked to find $\displaystyle \displaystyle \int \int_{R} xy \, dA$ where $\displaystyle R$ is the region in the first quadrant bounded by $\displaystyle y = x, y = 3x, xy = 1,$ and $\displaystyle xy = 3$ using the change $\displaystyle x = u/v$ and $\displaystyle y = v$.

I find the equations $\displaystyle \sqrt{u} = v, \sqrt{3u} = v, 1 = u,$ and $\displaystyle 3 = u$. The Jacobian is $\displaystyle 1/v \cdot 0 - (u \cdot - \frac{1}{v^{2}} \cdot 1) = \frac{u}{v^{2}}$. So I set up the integral $\displaystyle \displaystyle \int_{1}^{3} \int_{\sqrt{u}}^{\sqrt{3u}} u \cdot \frac{u}{v^{2}} \, dv du = \int_{1}^{3} u^{2} (\frac{1}{\sqrt{u}} - \frac{1}{\sqrt{3u}}) du = (1-\frac{1}{\sqrt{3}})\int^{3}_{1} u^{\frac{3}{2}} du = (1-\frac{1}{\sqrt{3}}) \cdot \frac{2}{5} \cdot (3^{\frac{5}{2}} - 1) = (1-\frac{1}{\sqrt{3}}) \cdot \frac{2}{5} \cdot (9\sqrt{3} -1)$. Now this looks very far from the answer in the book. But I don't see the mistake.
• Nov 18th 2010, 10:12 PM
matheagle
stewart 16.9 #15

the jacobian is 1/v, you have the 0 and 1 backwards in your matrix of partial derivatives
• Nov 18th 2010, 10:26 PM
ragnar
Gah, stupid mistake, thanks for catching it!
• Nov 18th 2010, 10:31 PM
matheagle
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.