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?
 A: 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:
$$r|_1^{2\cos\theta},\theta|_{-\pi/3}^{\pi/3}$$
A: 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). 
A: 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.
A: The "lunette" region shown in the other posts has a little "trap" if one chooses to treat it in Cartesian coordinates.

The intersection points given by Dreamer are correct, but setting the limits of integration for $ \ y \ $ there does not fully cover your region.$^*$
  For $ \ | y | \ > \ \frac{\sqrt{3}}{2} \ , $ there is still a portion of the "right-hand" circle which extends to $ \ |y| \ = \ 1 \ . $  In order to deal with it, we need a second integral where that circle defines both bounds of integration in $ \ x \ . $ Inverting the equation for the circle yields $ \ x \ = \ 1 \ \pm \ \sqrt{1 - y^2} \ . $ 
$^*$ Reading that post again, I see that Dreamer's integration is covering the lenticular region enclosed by both circles, which is what I'd initially interepreted the problem statement to mean.  Your own solution is for the region outside the circle centered on the origin (also the one B.S. shows in red).  So there remains the question of the region to which the problem refers.
So we need to construct two integrals to complete the job (again using the symmetry about the $ \ x-$ axis):
$$ 2 \ \left[ \ \int\limits_0^{\sqrt{\frac{3}{2}}} \ \int\limits^{1+\sqrt{1-y^2}}_{\sqrt{1-y^2}} \ \frac{1}{x} \ \ dx \ dy \ \ + \ \ \int\limits^1_{\sqrt{\frac{3}{2}}} \ \ \int\limits^{1+\sqrt{1-y^2}}_{1-\sqrt{1-y^2}} \ \frac{1}{x} \ \ dx \ dy \ \right] $$ 
This is just sufficiently horrid, and the polar form can be integrated so simply, that the polar integration would seem to be the method intended by the poser of this problem.
