<s>[INST] 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?
 [/INST] 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 $
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