TITLE: Double integral bounds of integration polar change of coordinate
QUESTION [8 upvotes]: I hope that someone can help me determine the the bounds of integration for this problem.
Evaluate $$\iint\limits_{R}xydA$$ where, $$R={(x,y): \frac{x^2}{36}+\frac{y^2}{16}\leq1, x\geq0,y\geq0}$$
my attempt, r=1, since $${(x,y): \frac{x^2}{36}+\frac{y^2}{16}\leq1}$$The region of integration is in the first quadrant since$${x\geq0,y\geq0}$$I change from cartesion to polar $$f(x,y)=xy $$$$f(r,\theta)=rcos(\theta)rsin(\theta)$$$$dA=rdrd\theta$$ So from everything above I thought the bounds would be r=0 to r=1 and $\theta=0$ to $\theta=\frac{\pi}{2}$ When I put everything together I get $$\int\limits_{0}^\frac{\pi}{2}\int\limits_{0}^{1}rcos(\theta)rsin(\theta)rdrd\theta$$
If I compute the integral above I get $\frac{1}{8}$, I know that $\frac{1}{8}$ is not the correct answer since the question is multiple choice and the options are (a) 5 (b) 25 (c) 55 (d) 72 (e) 73. I don't know where to go from here I assume that my error has to do with $(x,y): \frac{x^2}{36}+\frac{y^2}{16}\leq1$ since I don't use $0\leq{y}\leq4$ and $0\leq{x}\leq6$ I'm just not sure how to incorporate the fractions are they part of r?
Any help would be greatly appreciated, thank you.

REPLY [3 votes]: If you had $x^2+y^2=1$, then you would have a circle of radius $1$. If instead of $x^2+y^2=1$ you had $\left(\frac x r\right)^2+\left(\frac y r\right)^2=1$, then the radius would be $r$. But you for a circle, you can't have two different radii, so that leads to the conclusion that this can't be a circle in x-y coordinates; instead, it's an ellipse. To have a radius-1 circle, you need to do a change of coordinates: $u = \frac x 6$ $v = \frac y 4 $. Then you can do another change of coordinates to $r$ and $\theta$ in terms of $u$ and $v$. Keep in mind that you need to take into account the Jacobians for both transformations. Or you can do it all in one transformation as in Ak19's answer.