Is the radial projection map area increasing? Let $S$ be a hypersurface enclosed inside the unit sphere in $R^n$. Assume that every ray $\{t x: t \geq 0 \}$ intersects $S$ at most once.
Is it always true that ${\rm Area}(S) \leq {\rm Area}(P(S))$?
Here $P$ is the radial projection map onto $S^{n-1}$, i.e. $P(x) = x/\|x\|$.
(I am mostly interested in the 2-dimensional case.)
Thanks.
 A: The answer is negative: the area of $P(S)$ is at most the area of the unit sphere, while the area of $S$ can be made arbitrarily high.
An $S$ contained in the unit sphere and star-shaped at $0$ can be parametrized by the radius in polar coordinates: $S=\{\phi(u)u : \lVert u\rVert=1\}$ where $\phi$ is any smooth function from the unit sphere to $(0,1)$. Now, the area of $S$ is something like 
$$\int \phi^{n-1}\sqrt{\lVert \nabla \phi\rVert^2+1}$$
(a bit late here, so I might have gotten the formula wrong but in any case the integrand goes to infinity with $\nabla\phi$).
Taking $\phi$ with value in say $[\frac13,\frac23]$ and with a lot of variation (e.g. making fingers or wrinkles) we can easily make the area of $S$ arbitrarily high.
A: Here is another counterexample. This is in $\mathbb{R}^3$ for simplicity, but the same argument works in any dimension. Let
$$
S_\epsilon=\{(x,y,z):\, x^2+y^2\geq 0.01,\ z=\epsilon,\ z^2+y^2+z^2\leq 1\}.
$$
This is a disc parallel to the equator plane, $\epsilon$ above the equator, and with a small disc of radius $0.1$ removed. $P(S_\epsilon)$ is a small strip above the equator so $\operatorname{Area}(P(S_\epsilon))\to 0$ as $\epsilon\to 0^+$. Therefore
$$
\lim_{\epsilon\to 0^+}\frac{\operatorname{Area(P(S_\epsilon))}}{\operatorname{Area}(S_\epsilon)}=0.
$$ 
