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Convert the integral from rectangular to cylindrical coordinates and solve

I think I know how to do this, but I just want to double check my method. So assuming I have the below problem:

$$\int^2_0\int^\sqrt{2x-x^2}_0xy dy dx$$

Since:

$$x=rcos\theta$$

$$y=rsin\theta$$

Is it then true that the integral becomes:

$$\int^2_0\int^\sqrt{2rcos\theta-r^2cos^2\theta}_0rcos\theta\; rsin\theta\; r\;dr\; d\theta$$

$$\int^2_0\int^\sqrt{2rcos\theta-r^2cos^2\theta}_0r^3cos\theta\; sin\theta\; dr\; d\theta$$

$$\int^2_0 \frac{r^4}{4}cos\theta\; sin\theta\; |^{\sqrt{2rcos\theta-r^2cos^2\theta}}_0\;d\theta$$

$$\frac{1}{4}\int^{2}_{0}(r\;cos\theta\; (2-r\;cos\theta))^4cos\theta\; sin\theta\; d\theta$$

Is this correct thus far?

EDIT

Attempting to convert to cylindrical coordinates again:

$$y=\sqrt{2x-x^2}$$

$$r\;sin\theta\;=\sqrt{2r\;cos\theta-r^2cos^2\theta}$$

$$r^2\;sin^2\theta=2r\;cos\theta-r^2cos^2\theta$$

$$r^2sin^2\theta-2r\;cos\theta+r^2cos^2\theta=0$$

$$r^2(sin^2\theta+cos^2\theta)=2r\;cos\theta$$

$$r=2cos\theta$$

$$\int^2_0\int^{2cos\theta}_0r^3\;cos\theta\;sin\theta\;dr\;d\theta$$

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    $\begingroup$ $(1)$ Just look at your formula after the polar transformation. That couldn't possibly be right, could it? Afterall, after you take the integral wrt $r$ you'll still have $r$'s left. $(2)$ Does the question ask you to use cylindrical coordinates? This works out pretty simply in Cartesian coordinates. $\endgroup$ – got it--thanks Nov 27 '15 at 23:30
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    $\begingroup$ Either way, when doing a coordinate transformation you don't just blindly plug in expressions in the bounds of integration. You draw the region and parametrize it in the new coordinates. $\endgroup$ – got it--thanks Nov 27 '15 at 23:41
  • $\begingroup$ @gotit--thanks thanks for the comments! I wouldn't say I was doing it 'blindly' but these are the conversions I was given for cartesian coordinates. $\endgroup$ – hax0r_n_code Nov 27 '15 at 23:46
  • $\begingroup$ @gotit--thanks hello, I tried following your suggestions and reworking this problem. Can you tell me if I now have the integral setup properly? $\endgroup$ – hax0r_n_code Nov 28 '15 at 5:39
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I think the first limits are correct, but not the second, the angle would be only between 0 and pi/2 based on the area if you draw on a graph.

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  • $\begingroup$ How were you able to determine the angle being between $0$ and $\frac{\pi}{2}$? $\endgroup$ – hax0r_n_code Nov 28 '15 at 14:07
  • $\begingroup$ @inquisitor You just have to plot/ tabulate some values of $r=2\cos(\theta)$ at a few values of $\theta$ and determine where it should start and end based on the plot/ table you make. $\endgroup$ – got it--thanks Nov 28 '15 at 15:00

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