A parametric surface is a continuous map $\mathbf{r} : D \to \mathbb{R}^3$ from a region $D \subseteq \mathbb{R}^2$ to $\mathbb{R}^3$: $\mathbf{r}(u, v) = (x(u,v), y(u,v), z(u,v))$ The vectors $\mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u}$ and $\mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v}$ are tangent to the surface, and the surface element is $d\mathbf{S} = (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv, \quad dS = \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv$

The scalar surface integral of $f$ over the surface $S$ parametrized by $\mathbf{r}(u,v)$ is $\iint_S f\,dS = \iint_D f(\mathbf{r}(u,v)) \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv$ For the graph $z = g(x,y)$, this becomes $\iint_D f(x, y, g(x,y)) \sqrt{1 + g_x^2 + g_y^2}\,dA$.

The flux of a vector field $\mathbf{F}$ through an oriented surface $S$ is $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv$ The sign depends on the choice of orientation (which side of $S$ is "outward").