Finding surface area S using area of projection of S??

Question

I was going over my calculus textbook and came across a question about surface area.

and question is as follows.

Let S be a parallelogram not parallel to any of the coordinate planes.

Let $S_1,S_2,S_3$ denote the areas of the projections of S on the three coordinates planes.

Find the area of S in terms of $S_1, S_2, S_3$ ?

In general, I was trying to show this is true by using double integration over the projection of S but couldn't do it.

Can anyone help me please?

Thanks in advance

Answer 1

If $P$ is a parallelogram in $\mathbf{R}^{3}$ with edges $v_{1}$ and $v_{2}$, the (unsigned) area of $P$ is $\|v_{1} \times v_{2}\|$, the magnitude of the cross product. If that doesn't finish the job, here are some additional hints:

To get a general formula for the area of $S$ in terms of the areas of $S_{1}$, $S_{2}$, and $S_{3}$, write $v_{i} = (a_{i}, b_{i}, c_{i})$ for $i = 1$, $2$ (use any three convenient letters for the components). The projection of $S$ to the $(x, y)$-plane has edges $(a_{i}, b_{i}, 0)$, and analogously for $S_{2}$ and $S_{3}$. Calculate $(a_{1}, b_{1}, 0) \times (a_{2}, b_{2}, 0)$ and its magnitude, and similarly for the other two coordinate planes. Then examine the cross product $v_{1} \times v_{2}$, and identify its magnitude in terms of the three magnitudes you found for the projected parallelograms.