Surface Area

Surface area is simply how much material it takes to cover a curved surface, like the fabric for a tent roof or the paint for a car hood. The formula $S = \iint_D\sqrt{1 + f_x^2 + f_y^2}\,dA$ looks like a lot, but it's one idea: $dA$ is a tiny patch of the shadow on the floor, and $\sqrt{1 + f_x^2 + f_y^2}$ (the tilt factor) is how many times bigger the piece of surface above it is. Flat ground has $f_x = f_y = 0$ and a factor of $1$; steep ground has a big factor. Add up shadow patches times tilt factors and you have the area.

Picture a sheet of plywood lying flat, then propped up on one side. Its shadow shrinks, but the plywood doesn't, so the area of a tilted piece is always more than the area of its shadow. The question is how much more.

Where the formula comes from. Over a tiny rectangle $dx$ by $dy$, the surface is nearly its tangent plane. Walk $dx$ in the $x$-direction along that plane and you also climb $f_x\,dx$; walk $dy$ in the $y$-direction and you climb $f_y\,dy$. So the patch is a parallelogram with sides $\langle dx, 0, f_x dx\rangle$ and $\langle 0, dy, f_y dy\rangle$. Its area is the length of their cross product, $\langle -f_x, -f_y, 1\rangle dx\,dy$, which is $\sqrt{1 + f_x^2 + f_y^2}\,dx\,dy$.

It's the two-variable version of arc length's $\sqrt{1 + f'(x)^2}$, from Calculus 2. There, a tiny run $dx$ with rise $f'\,dx$ had length $\sqrt{1 + f'^2}\,dx$; here a patch has two slopes, one in each direction, and both go under the square root. Polar coordinates help whenever $D$ is round.

A common mistake. It's tempting to write $\sqrt{1 + f_x + f_y}$, or to skip the square root because the slopes are small. The slopes get squared, because they come from a length (the cross product's), and for a steep surface the square root makes a big difference: a cone at $45°$ has factor $\sqrt2$, about $41\%$ more area than its shadow.

How much paint covers a curved car panel, how much heat a radiator surface sheds, how much material a dome needs: surface area integrals.