*Math Online's 1000th Page - 7:04 PM CST on April 4th, 2015*

# Surface Integrals of Vector Fields

We have already computed surface integrals of scalar functions of a surface $\delta$. We will now take a look at surface integrals of vector fields over a surface $\delta$.

We will explain surface integrals of vector fields with the term known as flux. To visualize what flux really is, imagine a smooth orientable surface $\delta$ (that is also permeable) in $\mathbb{R}^3$ whose normal field is $\hat{N}$, and suppose that this surface is submerged in a fluid with an associated velocity field, $\mathbf{V}$. The rate for which fluid passes through the surface $\delta$ in accordance to the velocity field is the flux of that field over the surface $\delta$.

Let $P$ be a point on $\delta$ and let $dS$ be the area element around $P$. The fluid that passes through this area element $dS$ from time $t$ to time $t + dt$ forms a cylinder. The base of this cylinder is the area element $dS$, while the height of this cylinder is $\| \mathbf{V}(P) \| \: dt \cos \theta$ where $\theta$ is the angle between $\hat{N}(P)$ and $\mathbf{V}(P)$.

The total rate at which fluid passes through $dS$ can be obtained with the following surface integral:

(1)Definition: Let $\delta$ be an orientable surface with unit normal $\hat{N}$ and let $\mathbf{F}$ be a continuous vector field. Then the Flux of $\mathbf{F}$ across $\delta$ is $\iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS$. |

*The notation $\iint_{\delta} \mathbf{F} \cdot d\vec{S}$ is also used to denote the flux of a vector field, where $d \vec{S}$ is the Vector Area Element can be obtained from the regular area element $dS$ with the formula $d\vec{S} = \hat{N} \cdot dS$.*

Let's now try to find a formula for computing surface integrals over vector fields.

Suppose that the surface $\delta$ is parameterized as $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$. Recall from the Surface Integrals page that $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$ is normal to the surface $\delta$, and that:

(2)The corresponding area element for the surface $\delta$ was then $dS = \biggr \| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \biggr \| \: du \: dv$. We also have that:

(3)Thus the *vector* area element for our surface integral will be (choosing the sign to reflect the appropriate orientation of $\delta$):

Thus if $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a continuous vector field on $\mathbb{R}^3$ then the flux inward/outward of $\mathbf{F}$ over the surface $\delta$ with unit normal $\hat{N}$ and parameterized by $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ is:

(5)