# How do you find the limits of integration without drawing a picture?

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).

• For some basic information about writing math at this site see e.g. here, here, here and here. – Tacet Dec 7 '14 at 20:40
• Why don't you want to draw a picture? Carefully setting up the inequalities to use is a way of "almost" doing a picture. – GEdgar Dec 7 '14 at 20:45
• 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 – Katie Dec 7 '14 at 21:04

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 $\sin(\theta)$.
• At what value of $\theta$ on the interval $[0,\,2\pi)$ is $\sin(\theta)=1$? What about $\sin(\theta)=0$? – teadawg1337 Dec 7 '14 at 21:24
• Both would work, though making the substitution $x=\sin(\theta)$ would be easier to evaluate. – teadawg1337 Dec 7 '14 at 21:38