<s>[INST] Double integral -- tricky? 
If $f(x,y) = x^2+y^2$ and $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\geq1, x^2+y^2-2x\leq0 \text{ and } y\geq0\}$, find $\displaystyle\int\displaystyle\int_D f$.

$D$ looks like the intersection between two circumferences, as I draw below:

Polar coordinates seems a obvious choice, but using them I'd have unwanted bits, this is, if I split the region in two sections (by $x=\frac{1}{2}$) after computing the red and blue region I would have twice $R1$ and $R2$, a problem that could be solved computing each one of them and then substract them of the final answer -this is, after adding the integral of the red and blue sections-.
I believe the answer would be given by
$$\displaystyle\int \displaystyle\int_D f = \displaystyle\int\displaystyle\int _{\text{RED}} f + \displaystyle\int\displaystyle\int _{\text{BLUE}} f - \displaystyle\int\displaystyle\int _{\text{R1}} f - \displaystyle\int\displaystyle\int _{\text{R2}} f$$
$R1$ and $R2$ could be described as follows:
$$R1 = \{(x,y):0\leq y\leq \sqrt{3}x,\;0\leq x\leq 1/2\}$$
$$R2 = \{(x,y):0\leq y\leq \sqrt{3}-\sqrt{3}x,\;1/2 \leq x\leq 1\}$$
Which gives
$$\displaystyle\int\displaystyle\int _{\text{R1}} f =\displaystyle\int_0^{1/2}\displaystyle\int_0^{\sqrt{3}x} (x^2+y^2)\; dy dx = 2\sqrt{3}\displaystyle\int_0^{1/2}x^3 = \displaystyle\frac{1}{52\sqrt{3}}.$$
$$\displaystyle\int\displaystyle\int_{R2} f = \displaystyle\int_{1/2}^1\displaystyle\int_0^{\sqrt{3}-\sqrt{3}x}(x^2+y^2)\; dydx \\ = \displaystyle\int_{1/2}^1(x^2(\sqrt{3}-\sqrt{3}x)dx + \frac{1}{3}\int_{\frac{1}{2}}^1 (\sqrt{3}-\sqrt{3}x)^3 dx \\ = \left[\sqrt{3}\displaystyle\int_{1/2}^1 x^2 -\sqrt{3}\displaystyle\int_{1/2}^1 x^3\right]  + \frac{1}{3}\int_{\frac{1}{2}}^1 (\sqrt{3}-\sqrt{3}x)^3 dx \\ = \sqrt{3}\left[\left(\displaystyle\frac{1}{3}-\displaystyle\frac{1}{24}\right)+\left(\displaystyle\frac{1}{4}-\displaystyle\frac{1}{64}\right) \right] + \displaystyle\frac{1}{3}\left(\displaystyle\frac{1}{2}\right) \\= \frac{\sqrt{3}}{16}+\frac{1}{6} = \frac{1}{48}(3\sqrt{3}+8).$$
Now changing to polar coordinates $x=r\cos \theta$, $y= r\sin \theta$ to compute the red and blue:
$$\displaystyle\int\displaystyle\int _{\text{RED}} f = \displaystyle\int_0^{\pi/3}\displaystyle\int_0^1 r^3\;drd\theta = \frac{1}{4}\displaystyle\int_0^{\pi/3}d\theta = \frac{1}{12}\pi$$
Setting auxiliary axis $u = x-1, v = y$ I have $\displaystyle\frac{\partial (x,y)}{\partial (u,v)}=1$ and after with $u=r\cos \theta$, $v=r\sin \theta$
$$\displaystyle\int\displaystyle\int _{\text{BLUE}} f = \displaystyle\int_{2\pi/3}^{\pi}\displaystyle\int_0^1 r^3\;drd\theta = \frac{1}{12}\pi.$$
Finally 
$$\displaystyle\int \displaystyle\int_D f = \frac{1}{6}\pi -\displaystyle\frac{1}{52\sqrt{3}} -\frac{1}{48}(3\sqrt{3}+8) .$$
Now I'd like to know if the solution is right, but fundamentally, can this integral be computed without splitting the $D$?
P.S: Legitimate question, but couldn't resist the pun.
 [/INST] There is no need to split this integral into pieces.  The inequality $x^2 + y^2 \geq 1$ is the same as $r \geq 1$, and the inequality $x^2 + y^2 - 2x \leq 0$ is the same as $r^2 - 2 r \cos \theta \leq 0$ or just $r \leq 2\cos \theta$.  The circles $r = 1$ and $r = 2 \cos \theta$ intersect at $\theta = \frac{\pi}{3}$, so the integral is:
$$\int_0^{\pi/3} \int_1^{\cos \theta} r^2 \cdot r\,dr\,d\theta$$
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