Asymptotic behaviour of the area of a 2-dimensional flat subset of $\mathbb{R}^3$

Question

I am interested by the area of the $2$-dimensional flat subset of $\mathbb{R}^3$ defined by the following equations with one parameter $t>1$:

1. $x,y,z>0$ (positive octant)
2. $x+y+z=t$ (hyperplane equation)
3. $xyz<1$ (points below the cubic surface $xyz=1$)

Sorry for not being able to give you a nice picture. But I can comment the shape of this object. The two first conditions define a triangle, and the third condition dig a hole in this triangle. What interest me is the asymptotic behaviour of the area when $t\to+\infty$.

An answer to this first question would be already really nice. An answer to the following more general question would be even better. In $\mathbb{R}^d$, what is the asymptotic behaviour when $t\to+\infty$ of the $(d-1)$-volume of the set defined by

1. $x_1,\ldots,x_d>0$
2. $x_1+\cdots+x_d=t$ (hyperplane equation)
3. $x_1\cdots x_d<1$ (points below the hypersurface $x_1\cdots x_d=1$)

The previous geometric description holds. The first conditions define a $(d-1)$-simplex. The last condition dig a hole in this simplex.

If you don't have a full answer, ideas of how attacking these computations are welcome.

Answer 1

This is a report of a failed attempt.

Using the area-preserving transformation

$$ \left(\begin{array}{c} u \\ v \\ w \end{array}\right) = \left(\begin{array}{ccc} \tfrac{1}{\sqrt{2}} & 0 & -\tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{6}} & \sqrt{\tfrac{2}{3}} & -\tfrac{1}{\sqrt{6}} \\ \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} \end{array}\right) \left(\begin{array}{c} x \\ y \\ z \end{array}\right) + \left(\begin{array}{c} 0 \\ \tfrac{t}{\sqrt{6}} \\ -\tfrac{t}{\sqrt{3}} \end{array}\right), $$

your triangle, defined by $x+y+z=t$ and $x,y,z>0$, becomes the one in the $u,v$-plane with vertices

$$ \left(\begin{array}{c} u \\ v \\ w \end{array}\right) = \left(\begin{array}{c} \pm \tfrac{t}{\sqrt{2}} \\ 0 \\ 0 \end{array}\right),\ \left(\begin{array}{c} 0 \\ t \sqrt{\tfrac{3}{2}} \\ 0 \end{array}\right). $$

I had hoped that the condition $xyz < 1$ would come out as something simple, but is is

$$ \frac{t^2 v}{2 \sqrt{6}}-\frac{u^2 v}{\sqrt{6}}-\frac{t v^2}{3}+\frac{v^3}{3 \sqrt{6}} < 1. $$

One could, for example, solve this for $v$ to obtain a formula for the bottom of the rounded hole in the triangle, but the expression is pretty nasty. Integrating it---even approximately---doesn't seem like it would be very fun.