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!
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 $\sin(\theta)$.