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?

To do multiple integrals over a region, you should use Fubini's theorem:
$$\iint_Df(x,y)dx\space{dy}=\int_a^b\int_{c(y)}^{d(y)}f(x,y)dx\space{dy}$$
or
$$\iint_Df(x,y)dx\space{dy}=\int_a^b\int_{c(x)}^{d(x)}f(x,y)dy\space{dx}$$
Define your region as having four bounds. This can be done one of two ways: a type I region has constant vertical bounds, and the horizontal bounds are a function of $y$. A type II region has constant horizontal bounds, and its vertical bounds are a function of $x$. Determine which type of region it is, and use Fubini's theorem accordingly.
