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Consider the integral $$ \int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}}dydx.$$ I need some help understanding how to find the new limits of integration if I'm evaluating the integral in polar coordinates.

I've drawn part of the picture, a circle centered at the origin with a radius of 1, and I understand why the limits for dr are from 0 to 1, but I don't know what to do with the limits for d(theta).

Thank you for your help!

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    $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Tacet Dec 7 '14 at 20:40
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    $\begingroup$ Why don't you want to draw a picture? Carefully setting up the inequalities to use is a way of "almost" doing a picture. $\endgroup$ – GEdgar Dec 7 '14 at 20:45
  • $\begingroup$ So I tried graphing part of it, but I'm not sure what the limits 1 and -1 tell me. :( The graph I have is of a circle with radius 1 $\endgroup$ – Katie Dec 7 '14 at 21:04
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Converting to polar coordinates is not required until after integrating with respect to y. $\displaystyle \int_{-1}^1\int_{\sqrt{1-x^2}}^0\mathbb{d}y\,\mathbb{d}x=\int_{-1}^1-\sqrt{1-x^2}\,\mathbb{d}x=2\int_0^1\sqrt{1-x^2}\,\mathbb{d}x$
(note that the negative sign may be removed since $f(x)=-\sqrt{1-x^2}$ is half of the unit circle $r=1$, and therefore the area above $f(x)$ is equivalent to the area below $\sqrt{1-x^2}$)
Hint: Try making the substitution $x=\sin(\theta)$ such that $\mathbb{d}x=\cos(\theta)\,\mathbb{d}\theta$. No graphing is required, merely an understanding of the key values of $\sin(\theta)$.

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  • $\begingroup$ How would I convert 1 and -1 for integrating with polar coordinates though?? $\endgroup$ – Katie Dec 7 '14 at 21:18
  • $\begingroup$ At what value of $\theta$ on the interval $[0,\,2\pi)$ is $\sin(\theta)=1$? What about $\sin(\theta)=0$? $\endgroup$ – teadawg1337 Dec 7 '14 at 21:24
  • $\begingroup$ Do you mean cos(theta)? Since those are the limits of integration for dx? $\endgroup$ – Katie Dec 7 '14 at 21:26
  • $\begingroup$ Both would work, though making the substitution $x=\sin(\theta)$ would be easier to evaluate. $\endgroup$ – teadawg1337 Dec 7 '14 at 21:38

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