# Area of a surface confined by a sphere

Let $$S$$ be a hypersurface enclosed inside the unit sphere in $$R^n$$. We may assume that every ray $$\{t x: t \geq 0 \}$$ intersects $$S$$ at most once.

Under what extra condition is $${\rm Area}(S) \leq {\rm Area}(S^{n-1})$$ ?

(I am mostly interested in the 2-dimensional case.)

Thanks.

• Should be able to use Radon transforms. Try the paper of Alvarez Paiva and Fernandez. – Ben McKay Feb 15 at 15:58
• I think you mean the unit ball, don't you? – Pietro Majer Feb 15 at 16:01
• If you regard the hypersurface $S$ as a radial graph over a domain $D\subseteq S^{n-1}$, i.e, $$S = \{ r(p)\,p\ |\ p\in D\,\}$$ for some function $r:D\to[0,1]$, then the standard formula for the $(n{-}1)$-content of $S$ is $$A_{n-1}(S) = \int_D r^{n-2}(r^2+|\nabla r|^2)^{1/2} \,\mathrm{d}V_{n-1},$$ so it's a question of what you want to assume about the function $r$. – Robert Bryant Feb 15 at 18:22
• Thanks very much. It seems like a useful tool. I am trying to find a reference for this formula. (For instance, do you mean $D$ is somehow centered at the origin? I also wonder how to interpret the gradient of $r$ exactly, as it is defined on $D$.) BTW, i refined the question in: mathoverflow.net/questions/352836/… – Thomas Feb 16 at 4:21
• $D$ is a domain (i.e., open set) in $S^{n-1}$. (I imagine that you want $D$ to be all of $S^{n-1}$, but you didn't say that.) I'm not sure what you mean by 'centered at the origin'. If $D$ is a hemisphere in $S^{n-1}$, is it 'centered at the origin'? Gradient has its usual meaning for a function defined on a hypersurface: Extend the function to be constant in the normal direction and compute the gradient as usual in $n$-space. I saw the other question (and, really, you should have just modified or added to this one, they are so close). What do you mean by 'elliptic'? – Robert Bryant Feb 16 at 10:05