Integral of $\frac1{|x|^3}$ on a circular segment

Question

How can I evaluate $$\int \int _C |(x,y)|^{-3} dx dy$$ Where $C$ is the part between the chord $AB$ and the arc $AB$, and $|(x,y)| = \sqrt{x^2 + y^2}$? The radius of the circle is $R$.

I tried using polar coordinates, but I can’t find a way to represent $C$ with polar coordinates.

Answer 1

The tricky part of doing this in polar coordinates is the chord $AB$. Triangle $OAB$ is isosceles, so its angles $\angle OAB$ and $\angle OBA$ are both $75^\circ = \frac{5 \pi}{12}$. If $P$ is a point on segment $AB$ then from triangle $OBP$ with $OB = R$ and $\angle OBP = \angle OBA$, we have

$$ \angle POB + \angle OBP + \angle BPO = \pi $$

$$ \angle BPO = \frac{7 \pi}{12} - \angle POB $$

And from the law of sines,

$$ \frac{OB}{\sin \angle BPO} = \frac{OP}{\sin \angle OBP} $$

$$ OP = R\, \frac{\sin\left(\frac{7 \pi}{12} - \angle POB\right)}{\sin \frac{7 \pi}{12}} $$

If $\theta$ is the usual counter-clockwise angle from the $+x$-axis, then $\angle POB = \theta - \frac{\pi}{6}$ and $OP$ is the needed lower bound of $r$, so the original integral becomes

$$ \int_{\pi/6}^{\pi/3} \int_{R \sin(3 \pi/4 - \theta) / \sin(7 \pi/12)}^R r^{-3} r\, dr\, d\theta $$

Answer 2

The function over which you are integrating is rotationally symmetric around the origin $(0,0)$, thus we can rotate the coordinate system in such a ways that $A$, $B$ are at $\pm15^\circ$ from an axis and, say, $x=\mathrm{const}$ for points on $\gamma$. One such rotation is such that

$$\begin{align} A &= (\cos 15^\circ, \sin 15^\circ) = (\cos(\pi/12), \sin(\pi/12)) \\ &= \tfrac14 (\sqrt6+\sqrt2, \sqrt6-\sqrt2) =:(c,s)\\ B &= (\cos 15^\circ, -\sin 15^\circ) = (c,-s) \end{align}$$ and all points in the chord have $x=\cos15^\circ = c$. Then $$\begin{align} \iint_C |(x,y)|^{-3}dxdy &= \int_{+\sin(\pi/12)}^{-\sin(\pi/12)} \frac{dy}{(y^2+\cos^2(\pi/12))^{3/2}} \\ &= \int_{+s}^{-s} \frac{dy}{(y^2+c^2)^{3/2}} \\ &= \frac{y}{c^2(y^2+c^2)^{1/2}} \Big|_{y=s}^{y=-s}\\ &= \frac{-2s}{c^2(s^2+c^2)^{1/2}} = -2\frac{s}{c^2} \\ &= -2 \frac{\sqrt6-\sqrt2}{2+\sqrt3} \end{align}$$ where in the second-last line we use $s^2+c^2=1$.