Area bounded by two circles $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$ Consider the area enclosed by two circles: $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$
Calculate this area using double integrals:
I think I have determined the region to be $D = \{(x,y)| 0 \leq y \leq 1, \sqrt{1-y^2} \leq x \leq \sqrt{1 - (1-y)^2}\}$
Now I can't seem to integrate this. Is this region wrong? Should the integral just be $\int_0^1 \int_{\sqrt{1-y^2}}^{\sqrt{1- (1-y)^2}} dx dy$?
Do I need to convert this to polar form?

(I recognize you asked for a method using double integrals; I'm leaving this here as "extra")
Using geometry, the area we want is the area of four one-sixth sectors of a circle with $r = 1$ subtracted by the area of two equilateral triangles of side length $s = 1$. This would be
$$ \frac{2}{3} \pi (1)^2 - 2 \frac{(1)^2 \sqrt{3}}{4} = \frac{2}{3} \pi - \frac{\sqrt{3}}{2}$$
