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?

First make a sketch/plot of the area that you need.

Now you see that you need the x bounds to be variable and the y bounds to be fixed. Find the intersections with the y-axis, these will be your limits for y (in this case $(\frac{1}{2},-\sqrt{\frac{3}{2}}), (\frac{1}{2},\sqrt{\frac{3}{2}})$. Your limits for x will be $x=\sqrt{1-y^2}$ (the 'right' part of the left circle in the plot) and $x=1-\sqrt{1-y^2}$ (the 'left' part of the right circle in the plot). Hence, the integral that you need to evaluate is $$\int\limits_{-\sqrt{\frac{3}{2}}}^{\sqrt{\frac{3}{2}}}\quad\int\limits_{1-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x} dy dx$$ (you can interchange the role of $x$ and $y$ using Fubini's theorem if desired in general, but since we have the integral that has 'functions' as its limits as the inner integral here, this form is preferable). 
