Surface Area

The surface area of z = f(x, y) over a region D, via a double integral.

Different flavor from the rest of this section — geometry, not mass. For a surface z = f(x, y) sitting over a region D, its area is:

The square root is a stretch factor: a tilted patch of surface covers more area than its flat shadow on D, and the steeper the tilt, the more stretch.

A(S) = \iint_D \sqrt{1 + \left(f_x\right)^2 + \left(f_y\right)^2}\, dA

Surface area of z = f(x,y) over D

$f_x, f_y$ Partial derivatives of f — the surface's slopes; feed the stretch factor.
$dA$ An infinitesimal patch of area in the shadow region D.
$A(S)$ The actual area of the surface S, not just the flat shadow D.

Where the square root comes from: f_x and f_y are the surface's slopes in the two directions. If the surface is flat (f_x = f_y = 0), the root is 1 and you just get the area of D, as you'd hope.

The trap here: it's f_x² and f_y² under the root, plus one, all square-rooted. Not √(1 + f_x + f_y), not 1 + √(f_x² + f_y²). Miss the structure and every surface-area problem is wrong.

How to solve one of these

1.
Compute the partial derivatives first. Find $f_x$ and $f_y$ , square them, add 1, take the root — then integrate the result over $D$ .
2.
Set up the region as in Type I, Type II, or polar form. Applications don't change how you integrate over the region $D$ , only what you integrate.

A(S) = \iint_D \sqrt{1+f_x^2+f_y^2}\, dA

Stretch the shadow region's area by how tilted the surface is at each point, then add it all up.