Let $M$ be the *saddle surface* in $\mathbb R^3$ defined by $x^2-y^2+z=0$. For any $r\geq 0$ and $(x\_0,y\_0,z\_0)\in\mathbb R^3$, let $rM+(x\_0,y\_0,z\_0)$ denotes the surface obtained by scaling $M$ by $r$ and then *translating* by $(x\_0,y\_0,z\_0)$. (Note that by $rM$, with $r=0$, we mean $\lim\_{r\to 0} rM$, which is two perpendicularly intersecting plane.) And let $B$ be the unit ball.


Is it true that


***$$\mathrm{area}\left[(rM+(x\_0,y\_0,z\_0))\cap B\right]\leq 2\pi,$$***
with equality holds ***if and only if*** $r=0$ and $(x\_0,y\_0,z\_0)=(0,0,0)$ (which gives two intersecting equatorial disks)?


**Edit:** Using the first variation formula, I can show some partial results:


* For any fixed $z\_0$, the area of $(r(M+(0,0,z\_0)))\cap B$ decreases as $r$ increases, for all $r\in(0,+\infty)$.
* $(r,x\_0,y\_0,z\_0)=(0,0,0,0)$ is a strict local maximum.



All saddles in the unit ball have area $<2\pi$?
It is actually next to trivial if you choose the right parameterization (and rather puzzling if you don't, so it can make a decent take-home exam problem in multivariate calculus).


I'll use the line cover $x(s,t)=(s+t,s-t,4st)$. The area element is then
$2\sqrt{1+8s^2+8t^2}\,ds\,dt< 2(\sqrt{1+8s^2}+\sqrt{1+8t^2})\,ds\,dt$.


Now for a ball $B$ of radius $R$ centered at $(u,v,w)$ and for fixed $s$, we have the line in $t$ whose moving speed is $\sqrt{2+16s^2}=\sqrt 2\sqrt{1+8s^2}$ and whose square distance from the center of the ball is at least
$$
\min\_t[(s+t-u)^2+(s-t-v)^2]=2(s-\tfrac{u+v}2)^2=2(s-s\_0)^2.
$$
Thus, integrating in $t$ first, we have
$$
\int\_{s,t:x(s,t)\in B}2\sqrt{1+8s^2}\,dt\,ds\le \int\_{s\in\mathbb R}2\sqrt 2\sqrt{[R^2-2(s-s\_0)^2]\_+}\,ds=\pi R^2
$$
(time = line length/speed; length = $2\sqrt{R^2-\text{(line distance to the center)}^2}$).


The other integral is done in exactly the same way, only you need to integrate with respect to $s$ first. Hence, the area is $<2\pi R^2$, which is equivalent to the requested bound after scaling.


