How would one approximate the surface area of a curved shape as on these ETFE pillows?

Question

I need to get the surface area of these ETFE pillows:

Answer 1

There are surely different types of modelization, some of them based on physics. Here is one based on a visually satisfying similarity. Indeed (white) cross sections appear to be well modelized by parabolas ($z=k(W^2-y^2)$); the way these coefficients $k$ vary as a function of $x$ is rather arbitrary; here we have taken:

$$z=f(x,y)=\underbrace{\frac{h}{W^2} \sqrt{\cos(\pi \frac{x}{2L})}}_{k}(W^2-y^2) \ \ \text{for} \ \ -L<x<L, \ -W<y<W $$

where $2L$ is the length and $2W$ the width of the mattress and $h$ the measured maximal height in the middle of the mattress.

It remains to use the following formula for the surface area (see here):

$$A=\int_{x=-L}^L \int_{y=-W}^W \sqrt{1+\left(\partial f/\partial x\right)^2 + \left(\partial f/\partial y\right)^2}dx dy$$

For given values of $L$ and $W$, using a numerical software, one gets a numerical value for $A$.

Remarks: One could instead have considered NURBS surfaces.

Here is the family of parabolas with common roots $\pm a$ (here with $a=1$) with common equation $y=k(a^2-x^2)=k(a-x)(a+x)$. Here $k$ is linearly increasing. In the case of the mattress, $k$ must stay a long time almost constant then decrease rather sharply.