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?
 A: Draw a picture. You will note that part of the region is in the second quadrant. 
If you want to use rectangular coordinates, it will be necessary to see where circles meet. That is not at $x=1$. If we solve the system of two equations, pretty quickly we get $y=\frac{1}{2}$, which gives $x=\pm \frac{\sqrt{3}}{2}$. Your integral setup would be almost right if you had integrated from $x=-\frac{\sqrt{3}}{2}$ to $x=\frac{\sqrt{3}}{2}$. The $y$ should range from $1-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$. 
However, it might be preferable to take advantage of the symmetry and integrate from $x=0$ to $x=\frac{\sqrt{3}}{2}$, and double the result.
But as you indicated, polar may be better. The two circles have polar equations $r=1$ and $r=2\sin\theta$. The curves meet where $2\sin\theta=1$, so in the first quadrant at $r=1$, $\theta=\frac{\pi}{6}$ and also at $r=1$, $\theta=\frac{5\pi}{6}$.
Let's find the first quadrant area and double. Up to $\frac{\pi}{6}$, our bounding curve is $r=2\sin\theta$, and then up to $\frac{\pi}{2}$ it is $r=1$. So half our area is 
$$\int_0^{\pi/6} 2r\sin\theta \,dr\,d\theta+\int_{\pi/6}^{\pi/2}r\,dr\,d\theta.$$
(The second integral could be replaced by a simple geometric argument.)
Added: For the first integral, integrate first with respect to $r$. We get $\sin\theta$. Then integrate with respect to $\theta$. We get $1-\frac{\sqrt{3}}{2}$. 
The second integral gives the area of a circular sector with angle $\frac{\pi}{2}-\frac{\pi}{6}$, which is $\frac{\pi}{3}$, one-sixth of a circle, so area $\frac{\pi}{6}$. Thus half our area is $1-\frac{\sqrt{3}}{2}+\frac{\pi}{6}$. 
A: Solve the system: $x^2 + y^2 = 1 = x^2 + (y - 1)^2$ we have: $x =\dfrac{\sqrt{3}}{2}$ and $y = \dfrac{1}{2}$, and $x = -\dfrac{\sqrt{3}}{2}$ and $y = \dfrac{1}{2}$. So the set up for the integral is:$\int_\frac{-\sqrt{3}}{2}^{\frac{\sqrt{3}}{2}} \int_{1 -\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} 1dydx$ which can be easily computed.
A: (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
$$ \frac{2}{3} \pi (1)^2 - 2 \frac{(1)^2 \sqrt{3}}{4} = \frac{2}{3} \pi - \frac{\sqrt{3}}{2}$$
