My Math Forum Change of variables/ Transformations part 2

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April 9th, 2018, 08:27 AM   #1
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Change of variables/ Transformations part 2

I am not sure how I should set my u and v expressions into the u-v plane for this question.
How should I look at the expression to set u and v expressions?
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April 12th, 2018, 12:56 PM   #2
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Quote:
 Originally Posted by Alexis87 I am not sure how I should set my u and v expressions into the u-v plane for this question. How should I look at the expression to set u and v expressions?
Let $x= 2u \cos(v)$ and $y= 3u \sin(v)$ so that $\left(\frac{x}{2}\right)^2+ \left(\frac{x}{3}\right)^2= 1$ becomes $u^2= 1$. That is a circle with center (0, 0) and radius 1. (If you had not insisted on "u" and "v" I would have called those variables "r" and "$\theta$" as in polar coordinates.)

To integrate over that disk, take u going from 0 to 2 and v from 0 to $2\pi$. $9x^2+ 4y^2$ becomes $36u^2 \cos^2(v)+ 36u^2 \sin^2(v)= 36u^2$. Finally the Jacobian is $\left|\begin{array}{cc} 2\cos(v) & -2u \sin(v) \\ 3\sin(v) & 3u \cos(v)\end{array}\right|= 6u \cos^2(v)+ 6u \sin^2(v)= 6u$ so the integral becomes $\int_0^1\int_0^{2\pi} e^{36u^2}6u dudv= 12\pi\int_0^1 ue^{36u^2}du$. To do that last integral let $y= 36u^2$.

Last edited by skipjack; April 13th, 2018 at 04:07 AM.

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