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.
Best Answer
This question is a bit tricky because $R$ is not a circle quadrant, it's an ellipse quadrant. Therefore, the underlying ellipse needs to be transformed into a circle by a substitution – $u=\frac23x$ works: $$\iint_Rxy\,dA=\frac94\iint _Suy\,dA$$ $S$ is now a quarter-disc of radius $4$ around the origin, so polar coordinates can now be used: $$=\frac94\int_0^4\int_0^{\pi/2}r^3\cos\theta\sin\theta\,d\theta\,dr$$ $$=\frac94\int_0^4\frac12r^3\,dr$$ $$=\frac98×\frac14×4^4=72$$