All saddles in the unit ball have area $<2\pi$? 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.

 A: 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.
