Surface area

Definition

Let $D \subseteq \mathbb{R}^2$ be a bounded Jordan measurable closed set, and let $r \colon D \to \mathbb{R}^3$ be a parametrised surface that is injective on the interior of $D$. The surface area of $S = r(D)$ is $$\operatorname{area}(S) = \iint_D \| r_u(u, v) \times r_v(u, v) \| \, du \, dv.$$

Remarks

The integrand is the length of the normal vector. It is the area of the parallelogram spanned by $r_u$ and $r_v$, which approximates the image of a small parameter rectangle. For the graph $z = f(x, y)$ over $D$, the area is $\iint_D \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy$. For the unit sphere, $\| r_\varphi \times r_\theta \| = \sin \varphi$, and the area is $\int_0^{2\pi} \int_0^\pi \sin \varphi \, d\varphi \, d\theta = 4\pi$. The value does not depend on the parametrisation. A piecewise smooth surface gets the sum of the areas of its pieces. Inscribed polyhedra do not give a good definition: the Schwarz lantern is a counterexample.

Explanations

L1 human

To find the area of a curved surface, cut the parameter domain into small rectangles. Each one maps to a small curved patch, which is nearly a parallelogram, and you add up the areas of these parallelograms.