How to evaluate double integrals over a region? Evaluate the double integral $\iint_D(1/x)dA$, where D is the region bounded by the circles $x^2+y^2=1$ and $x^2+y^2=2x$
Alright so first I converted to polar coordinates:
$$ x^2 + y^2 = 1 \ \Rightarrow \ r = 1 \ \ , \ \
x^2 + y^2 = 2x \ \Rightarrow \ r^2 = 2r \cos θ \ \Rightarrow \ r = 2 \cos θ \ .  $$
Points of intersection:
$ 2 \cos θ = 1 \ \Rightarrow \ θ = ±π/3 \ , $ 
$ 2 \cos θ > 1 $ for θ in (-π/3, π/3).
So, 
$$ \int \int_D \ (1/x) \ \ dA
 \ \ = \ \ \int_{-π/3}^{π/3} \ \int_1^{2 \cos θ} \ \frac{1}{r \cos θ} \ \ r dr \  dθ  $$
$$ = \ \ \int_{-π/3}^{π/3} \ \int_1^{2 \cos θ} \ \sec θ \ \ dr \  dθ \ \ 
= \ \ \int_{-π/3}^{π/3} \ (2 \cos θ - 1) \sec θ \ \ dθ $$
$$ = \ \ 2 \ \int_0^{π/3} \ (2 - \sec θ) \ \ dθ \ \ , $$
(since the integrand is even)
$$ = \ \ 2 \ (2 θ \ - \ \ln |\sec θ + \tan θ| \ ) \vert_0^{π/3} \ \ = \ \ \frac{4π}{3} \ - \ 2 \ln(2 + √3) \ \ .  $$
I'm not sure this is right. Could someone look over it?

Of course, you can do the problem by using the polar coordinates. If it's understood correctly, you would want to find the right limits for double integrals. I made a plot of the region as follows:

The red colored part is our $D$. So:
$$r|_1^{2\cos\theta},\theta|_{-\pi/3}^{\pi/3}$$
