# Arc of a circle problem in double integral.

So i have a relatively complicated double integral to solve

$$\iint \frac{y(x\sqrt{3}-y)}{x^2+y^2}dxdy$$ over the following area

$$x^2+y^2=2x\sqrt{3}+2y-3$$

and this actually the circle

$$(x-\sqrt{3})^2+(y-1)^2=1$$

Here's the picture:

Obviously, this calls for polar coordinates, which means that now i have to find bounds of integration, for angle, it is quite easy lower one is zero, upper one is $\frac{\pi}{3}$, but the problem is that i have to set a bounds for radius that are somewhat non-standard, they would not be constants, let me explain how i got the bounds for angle

note that i only added the tangent to the circle which is $y=\sqrt{3}x$, that is how i determined angle bounds, now, i have to find the equation of two arcs of the circle where one is from $0$ to the point where tangent "touches" the circle using short path (R1, the one that is closer to the origin) and the one that is further away from the origin than the first one (R2), i have to find them because they are the bounds for the radius.

I actually tried something, and it was that i simply introduced polar coordinates into the equation of a given circle, but i couldn't do much more because, up to now, whenever i got to this point, i could easily express radius in terms of angle, but now, i should solve quadratic equation to do so, i know how to solve it but i don't know if i would get proper bounds that way. I actually don't understand how those two solutions represent exactly the two arcs of a circle that i am looking for?

There actually are two circles playing here: the unit circle centered at the origin and the unit circle centered at $(\sqrt{3},1)$. By setting $x=\frac{1}{2}\left(\sqrt{3}X+Y\right), y=\frac{1}{2}\left(X-\sqrt{3}Y\right)$ the original integral turns into $$\frac{1}{2}\iint\frac{X^2-3Y^2}{X^2+Y^2}\,dX\,dY$$ over the region $\|(X,Y)-(2,0)\|=1$, which equals $$\frac{1}{2}\int_{0}^{2\pi}\frac{1+4\cos\theta+4\cos^2\theta}{5+4\cos\theta}d\theta\stackrel{\text{symmetry}}{=}\int_{0}^{\pi/2}\frac{10+8\cos^2\theta}{25-16\cos^2\theta}\,d\theta=\frac{\pi}{2}$$ by the substitution $\theta=\arctan u$. If you are interested in the integral over $\|(X,Y)-(2,0)\|\leq 1$ you may easily adapt this argument; the key idea still is that a suitable rotation simplifies a lot the needed computations.
• @cdummie: I mean that the denominator of the original integrand function depends on the distance from the origin, so it is a good idea to align the origin and the point $(\sqrt{3},1)$ along the $x$-axis, for instance. – Jack D'Aurizio Dec 11 '17 at 15:52
Change variables first as $x-\sqrt{3}=u$ and $y-1=v$. Then you get $$\iint \frac{(v+1)((u+\sqrt{3})\sqrt{3}-(v+1))}{(u+\sqrt{3})^2+(v+1)^2}dudv$$ over the standard radius-$1$ circle centered in zero. Now you change to polar coordinates $u=r\cos\theta$ and $v=r\sin\theta$ to get $$\int_0^1 dr\ r\int_0^{2\pi}d\theta \frac{(r \sin (\theta )+1) \left(-r \sin (\theta )+\sqrt{3} r \cos (\theta )+2\right)}{r^2+2 r \left(\sin (\theta )+\sqrt{3} \cos (\theta )\right)+4}\ ,$$ which is not pretty but doable (the answer seems to be $3\pi/8$).