Constructing a function with constant line integral in $\mathbb{R}^n$? Suppose, $\Omega \subset \mathbb{R}^n$ is a bounded convex set. If, there is an integrable function $f:\Omega \to \mathbb{R}$ s.t. $$\int_{\Omega \cap \ell} f = 1$$ for every line $\ell \subset \mathbb{R}^n$ with $\mathcal{H}^1(\ell \cap \Omega) > 0$. Then, can we claim $\Omega$ is a ball and $f$ is a radial function? 
My main question is: How'd we go about constructing such a function if there exists one?
Any help/hint is appreciated! Thanks.
 A: For the unit ball in $\mathbb{R}^n$ you can consider the function
$$f(x)=\frac{1}{\sqrt{1-|x|^2}}.$$
Then, if you take a line at distance $h$ from the origin, using the arclength parameter $t$ on the segment the integral becomes
$$\int_{-\sqrt{1-h^2}}^{\sqrt{1-h^2}} \frac{1}{\sqrt{1-(h^2+t^2)}}dt=
\frac{1}{\sqrt{1-h^2}}\int_{-\sqrt{1-h^2}}^{\sqrt{1-h^2}} \frac{1}{\sqrt{1-\frac{t^2}{1-h^2}}}dt=\int_{-1}^1\frac{1}{\sqrt{1-y^2}}dy=\pi$$
by the change of variables $y=\frac{t}{\sqrt{1-h^2}}$.
How to come up with the function? Well, first of all we consider for simplicity a radial function $f(x)=g(|x|)$. Now in order for $f$ to have the same integral on all segments, a sufficient condition is that on each interval $f$ looks the same: if a segment is shorter than another, the function gets multiplied by the right factor to preserve the integral.
Following this idea, consider a segment at distance $h$ from the origin. Then the segment's length is $2\sqrt{1-h^2}$, and therefore the value $g(h)$ (middle point of the segment) must be $\frac{1}{\sqrt{1-h^2}}$ times higher than $g(0)$ (middle point of the segment through the origin) and thus
$$g(h)=\frac{g(0)}{\sqrt{1-h^2}}.$$
Luckily this works, as shown above.
