Finding area of a region between two circles Let $S$ be the region in the plane that is inside the circle $(x-1)^2 + y^2 = 1$ and outside the circle $x^2 + y^2 = 1 $. I want to calculate the area of $S$.
Try:
first, the circles intersect when $x^2 = (x-1)^2 $ that is when $x = 1/2$ and so $y =\pm \frac{ \sqrt{3} }{2} $. So, using washer method, we have
$$Area(S) = \pi \int\limits_{- \sqrt{3}/2}^{ \sqrt{3}/2} [ (1+ \sqrt{1-y^2})^2 - (1-y^2) ] dy $$
is this the correct setting for the area im looking for?
 A: The shape $S$ is extremely unwieldy for integration, and has to be cut up in several parts for that purpose. Use elementary geometry instead:
The area in question is a full unit disk minus one third of such a disk, minus two small circular segments. The latter are a sixth of a disc with  an equilateral triangle removed. It follows that
$${\rm area}(S)=\pi-{\pi\over3}-2\left({\pi\over6}-{\sqrt{3}\over4}\right)={\pi\over3}+{\sqrt{3}\over2}\ .$$
A: It will be  complicated if you tried to solve such problems  with cartesian coordinates  it's much easier to solve by polar coordinates  ,  and in that type of problems try to draw  them if you draw you will find that the area of the upper half is same of the area of lower half so you can calculate 1 of them and multiply it by 2 . You brought  the point of iterse ction bring it in terms of r and theta then integrate 
A: I would recommend using a polar coordinate system, i.e.
$$ x = rcos\theta $$ 
$$ y=rsin\theta $$
which implies: $ \rightarrow x^2 + y^2 = r^2sin^2\theta + r^2cos^2\theta = r^2(sin\theta + cos\theta) = r^2 $.
To use this method, you must find the intersection points (as you have already done) in order to find the angle $ \theta $. If you substitute $ x = \frac{1}{2} $ or $ y = \pm \frac{\sqrt3}{2} $ into either of the above polar equations, you get $ \theta = \pm \frac{\pi}{3} $.
The general formula for finding the area under a curve using polar coordinates is:
$$ A = \int_{a}^{b} \frac{1}{2} r^2 d\theta $$
If you are interested in finding the area between two curves, the formula becomes:
$$ A = \int_{a}^{b} \frac{1}{2} \big( (r_{outer})^2 - (r_{inner})^2 \big)d\theta $$
In these formulas, $ a $ is the lower bound of the angle $ \theta $, and $ b $ is the upper bound for the angle $ \theta $. Finally, $ r $ is the radius equation that defines region that your integral is "sweeping over."
In this case, we have: $ A = \frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \big( (r_{outer})^2 - (r_{inner})^2 \big)d\theta $
To find the outer radius and the inner radius, you must use the fact that $ x^2 + y^2 = r^2 $.
Circle 1 (outer circle $ \rightarrow (x-1)^2 + y^2 = 1 $ ):
$ (x-1)^2 + y^2 = 1 \rightarrow x^2 - 2x + 1 + y^2 = 1 \rightarrow x^2 + y^2 = 2x \rightarrow r^2 = 2rcos\theta \rightarrow r = 2cos\theta $
Circle 2 (inner circle $ \rightarrow x^2 + y^2 = 1 $ ): 
$ x^2 + y^2 = 1 \rightarrow r^2 = 1 $
Putting this all together, we get:
$ A = \frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \big( (r_{outer})^2 - (r_{inner})^2 \big)d\theta = \frac{1}{2} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \big( (2cos\theta)^2 - (1)^2 \big)d\theta = \frac{1}{6} (3\sqrt3 + 2\pi) \approx 1.91 $
