Surface integral

Definition

Let $r \colon D \to \mathbb{R}^3$ be a parametrised surface as in the definition of surface area, let $S = r(D)$, and let $f \colon S \to \mathbb{R}$ be continuous. The surface integral of $f$ over $S$ is $$\iint_S f \, dS = \iint_D f\big( r(u, v) \big) \, \| r_u(u, v) \times r_v(u, v) \| \, du \, dv.$$

Remarks

The expression $dS = \| r_u \times r_v \| \, du \, dv$ is the surface element. With $f = 1$, the integral is the area of $S$. The value does not depend on the parametrisation of $S$. For a surface made of several pieces, add the integrals over the pieces. For the graph $z = g(x, y)$ over $D$, $\iint_S f \, dS = \iint_D f(x, y, g(x, y)) \sqrt{1 + g_x^2 + g_y^2} \, dx \, dy$. Example: on the unit sphere $S$, symmetry gives $\iint_S x^2 \, dS = \iint_S y^2 \, dS = \iint_S z^2 \, dS$. These add up to the area $4\pi$, so $\iint_S z^2 \, dS = 4\pi / 3$.

Explanations

L1 human

A surface integral adds up the values of a function over a curved surface, each weighted by a small piece of area. If $f$ is a mass density per unit area, the surface integral is the total mass of the surface.