# Double integral bounds of integration polar change of coordinate

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.

• As a few answers hint, $r=1$ is not the correct boundary. With two dimensional integration, it's pretty much a requirement to sketch the region of interest. Ideally, that would have helped you realize that you don't want $r=1$, but something else. May 29, 2019 at 19:22

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: $$=\frac94\int_0^4\int_0^{\pi/2}r^3\cos\theta\sin\theta\,d\theta\,dr$$ $$=\frac94\int_0^4\frac12r^3\,dr$$ $$=\frac98×\frac14×4^4=72$$

Let $$x = 6r\cos\theta$$ and $$y =4r\sin\theta$$

So, $$\frac{x^2}{6^2}+\frac{y^2}{4^2}=r^2$$

Now, $$r^2\le1$$

$$x\ge0$$ and $$y\ge0\implies0\le r\le1$$ and $$0\le\theta\le\pi/2$$

$$dxdy = 24rdrd\theta$$

$$I = \int^{\frac{\pi}{2}}_0\int^{1}_{0}r^2(24\sin\theta\cos\theta)(24\ r)dr\ d\theta = 576 \int^{\frac{\pi}{2}}_0\int^{1}_{0}r^2(\sin\theta\cos\theta)(\ r)dr\ d\theta = 576\frac{1}{8} = 72$$

(You've already found $$\frac{1}{8}$$, so I just multiplied it by $$576$$)

$$x = 6r\cos\theta$$ $$y =4r\sin\theta$$

Partial derivatives:

$$x_r = 6\cos\theta$$, $$y_r = 4\sin\theta$$

$$x_{\theta} = - 6r\sin\theta$$ , $$y_{\theta} = 4r\cos\theta$$

$$J = \begin{vmatrix}6\cos\theta & 4\sin\theta \\ - 6r\sin\theta & 4r\cos\theta\end{vmatrix} = 24r\cos^2\theta+24r\sin^2\theta = 24r$$

So, $$dxdy = |J|drd\theta = 24r\ dr\ d\theta$$

• ., Could you please explain how you determined $dxdy=24rdrd\theta$ I see that $6rcos(\theta)(4rcos(\theta))=r^2(24sin(\theta)cos(\theta$)) but I don't know how to get the other $24r$. Thank you for your help.
– jack
May 29, 2019 at 4:21
• I've added it in the answer :) May 29, 2019 at 4:26

The entire problem is that the domain is 1 quarter of an ellipse, and not a circle. Thus, we must parameterize the domain as we would parameterize an ellipse.

As $$\frac{x^2}{36}+\frac{y^2}{16}=1$$ can be parameterized as $$x=6\cos\theta, y = 4\sin\theta$$, the parameterization of the ellipse is $$x=6r\cos\theta, y = 4r\sin\theta$$ where $$r\le 1$$, $$\theta \in [0, \frac{\pi} {2}]$$

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.