Computing Radon transform of indicator function

Question

Let $x\in\mathbb{R}^{2}$ and $a\in\mathbb{R}^{2}$. I want to compute $$\mathcal{R}1_{B_{1}(a)}(\theta,s):=\int_{x\cdot\theta=s}1_{B_{1}(a)}(x)dx=\int_{\theta^{\perp}}1_{B_{1}(a)}(s\theta+y)dy.$$ I'm not so familiar with working over hyperplanes and such so I was wondering how I would "absorb" my indicator function into the integral?

Answer 1

$\theta \in S^1$ (it might help to write $\vec \theta$ instead) is a unit vector on the unit circle, and hence defines a direction (a line, if you like). $\theta^\perp$ is the line orthogonal to the direction $\theta$. The translation of $\theta^\perp$ by $s\theta$, where $s \in \mathbb R$ is a magnitude, gives the "line of interest"; the radon transform you desire gives the length of [the intersection of the "line of interest" and the ball $B_1(a)$]. This is the red segment's length.

So $\int_{\theta^\bot} 1_{B_1(a)} (s\theta + y)\, dy$ is the length of the red segment. The integral is equal to $\int_{y_0}^{y_1} 1 \, dy = y_1 - y_0$, where $y_0$ is the distance from $s\theta$ to the left endpoint of the red segment, and $y_1$ is the distance from $s\theta$ to the right endpoint of the red segment.