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Area of the symmetric difference of two disks

This post goes back to high school: the area \delta_d of the symmetric difference of two d-dimensional disks when one center is shifted a little bit. Let’s start with d=1. So we have two intervals [-r,r] and [x-r,x+r]. It is easy to see the symmetric difference is of length \delta_1(x)=2x.

Then we move to d=2: two disks L and R of radius r and center distance x=2a<r. So the angle \theta(x) satisfies \cos\theta=\frac{a}{r}.


The symmetric difference is the union of R\backslash L and L\backslash R, which have the same area: \displaystyle (\pi-\theta)r^2+2x\sqrt{r^2-x^2}-\theta r^2=2(\frac{\pi}{2}-\arccos\frac{x}{r})r^2+2x\sqrt{r^2-x^2}. Note that the limit
\displaystyle \lim_{x\to0}\frac{\text{area}(\triangle)}{2a}=\lim_{a\to0}2\left(\frac{r^2}{\sqrt{1-\frac{a^2}{r^2}}}\cdot\frac{1}{r}+\sqrt{r^2-a^2}\right)=4r.
So \delta_2(x)\sim 4rx.

I didn’t try for d\ge3. Looks like it will start with a linear term 2d r^{d-1}x.


Now let {\bf r}(t)=(a\cos t,\sin t) be an ellipse with a>1, and {\bf r}'(t)=(-a\sin t,\cos t) be the tangent vector at {\bf r}(t). Let \omega be the angle from {\bf j}=(0,1) to {\bf r}'(t).
Let s(t)=\int_0^t |{\bf r}'(u)|du be the arc-length parameter and K(s)=|{\bf l}''(s)| be the curvature at {\bf l}(s)={\bf r}(t(s)). Alternatively we have \displaystyle K(t)=\frac{a}{|{\bf r}'(t)|^{3}}.


The following explains the geometric meaning of curvature:

\displaystyle K(s)=\frac{d\omega}{ds}, or equivalently, K(s)\cdot ds=d\omega. (\star).

Proof. Viewed as functions of t, it is easy to see that (\star) is equivalent to K(t)\cdot \frac{ds}{dt}=\frac{d\omega}{dt}.

Note that \displaystyle \cos\omega=\frac{{\bf r}'(t)\cdot {\bf j}}{|{\bf r}'(t)|}=\frac{\cos t}{|{\bf r}'(t)|}. Taking derivatives with respect to t, we get
\displaystyle -\sin\omega\cdot\frac{d\omega}{dt}=-\frac{a^2\sin t}{|{\bf r}'(t)|^3}. Then (\star) is equivalent to

\displaystyle \frac{a}{|{\bf r}'(t)|^{3}}\cdot |{\bf r}'(t)|=\frac{a^2\sin t}{\sin\omega\cdot |{\bf r}'(t)|^3}, or
\displaystyle \sin\omega\cdot |{\bf r}'(t)|=a\sin t. Note that \displaystyle \sin^2\omega=1-\cos^2\omega=1-\frac{\cos^2 t}{|{\bf r}'(t)|^2}. Therefore \displaystyle \sin^2\omega\cdot |{\bf r}'(t)|^2= |{\bf r}'(t)|^2-\cos^2 t=a^2\sin^2 t, which completes the verification.

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