Double integral bounds of integration polar change of coordinate

Question

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.

Answer 1

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

Answer 2

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 $576$)

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

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

Answer 3

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 $r\le 1$, $\theta \in [0, \frac{\pi} {2}]$

Answer 4

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.