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.

-
If you commited no mistake in writing down $\;D\;$ , the integration region is precisely what is inside the rightmost circle outside the coloured zone! Just behold that it must be $\;x^2+y^2\ge 1\;$ , meaning outside (or on the circumference) of the canonical unit circle... – DonAntonio Jan 22 '14 at 21:04
Oh god :( . Well, I guess I'll proceed to kill myself shortly It has been fun guys :( – Cure Jan 22 '14 at 21:08
Welcome to the club, @Dante...and try it again. :) – DonAntonio Jan 22 '14 at 21:09
@Dante You don't need to throw everything away and start from scratch. In fact, now that you know the value of the integral in the colored part, you can take it away from the integral over the semi-circle. The integral over the semi-circle can be done in a single step using polar coordinates centered at $(1,0)$. – Braindead Jan 22 '14 at 21:17

The integral's basically:

$$\int\limits_{1/2}^1\int\limits_{\sqrt{1-x^2}}^{\sqrt{1-(x-1)^2}}\,f(x,y)dydx+\int\limits_1^2\int\limits_0^{\sqrt{1-(x-1)^2}}f(x,y)\,dydx$$

-
I am little confused, isn't the second integral essentially outside the unit circle and inside the right circle? Why is the first integral part of that answer? What region does it describe? – imranfat Jan 22 '14 at 21:29
@imranfat Draw the vertical line $x=1\;$. The first integral is the one over the region to the left of the line, and the second integral is on the right of it....and now I see I meesed up the limits of the inner integral in the first expression! Thanks. – DonAntonio Jan 22 '14 at 21:35
Oh, the line $x=1$ yes that makes sense, but then why is the first integral starting for $x=0$? Looking at the picture, shouldn't it start a $x=0.5$? Of the region the most left point has $x=0.5$ at the top corner where the circles intersect? – imranfat Jan 22 '14 at 21:39
Yes @imranfat . I confused the axis. Thanks. – DonAntonio Jan 22 '14 at 21:40
Man...from a moment I thought I just lost it here.... – imranfat Jan 22 '14 at 21:41

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: $$\int_0^{\pi/3} \int_1^{\cos \theta} r^2 \cdot r\,dr\,d\theta$$

-