Surface Integrals of Vector Fields
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.7: “Surface Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-7-surface-integrals |
| Textbook used in class | Stewart, Calculus, Section 16.7: “Surface Integrals” (Examples 7, 8) |
Quick Reference
Flux integral of $\mathbf{F}$ over oriented surface $S$: $$\iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v))\cdot(\mathbf{r}_u\times\mathbf{r}_v)\,dA.$$
For a graph $z = g(x,y)$ with upward orientation: $$\iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D \mathbf{F}(x,y,g(x,y))\cdot\langle -g_x,-g_y,1\rangle\,dA.$$
Reversing orientation negates the integral: $\iint_{-S}\mathbf{F}\cdot d\mathbf{S} = -\iint_S \mathbf{F}\cdot d\mathbf{S}$.
Motivation
Imagine a fluid with velocity field $\mathbf{F}(x,y,z)$ and a permeable membrane shaped like a surface $S$. The flux integral $\iint_S \mathbf{F}\cdot d\mathbf{S}$ measures the net volume of fluid crossing the membrane per unit time -- positive when the fluid moves in the direction of the chosen normal, negative when it flows the other way.
The scalar surface integral $\iint_S f\,dS$ does not care which side of $S$ you call “up.” The vector version does: you must choose an orientation (an outward or upward normal) before the integral has a definite sign.
Key Concept
The vector surface element is $d\mathbf{S} = (\mathbf{r}_u\times\mathbf{r}_v)\,dA$. The cross product carries both the magnitude (area of the surface patch) and the direction (the normal vector). Taking the dot product with $\mathbf{F}$ picks out the component of the field that points through the surface, exactly as the dot product $\mathbf{F}\cdot\hat{\mathbf{n}}$ picks out the normal component of a vector.
For a graph $z = g(x,y)$ parametrized by $(x,y)$, one finds $\mathbf{r}_x\times\mathbf{r}_y = \langle -g_x, -g_y, 1\rangle$, which points upward (positive $z$-component). If the problem asks for the downward orientation, use $\langle g_x, g_y, -1\rangle$ instead.
Worked Example
Compute the flux $\iint_S \mathbf{F}\cdot d\mathbf{S}$ where $\mathbf{F} = \langle 0, 0, z\rangle$ and $S$ is the paraboloid $z = 1-x^2-y^2$, $z\geq 0$, with upward orientation.
The graph is $z = g(x,y) = 1-x^2-y^2$ over the disk $D: x^2+y^2\leq 1$.
Partial derivatives: $g_x = -2x$, $g_y = -2y$.
Upward vector element: $\langle -g_x,-g_y,1\rangle = \langle 2x, 2y, 1\rangle$.
Evaluate $\mathbf{F}$ on the surface: $z = 1-x^2-y^2$, so $\mathbf{F} = \langle 0, 0, 1-x^2-y^2\rangle$.
Dot product: $$\mathbf{F}\cdot\langle 2x,2y,1\rangle = 0\cdot 2x + 0\cdot 2y + (1-x^2-y^2)\cdot 1 = 1-x^2-y^2.$$
Convert to polar ($x = r\cos\theta$, $y = r\sin\theta$, $dA = r\,dr\,d\theta$): $$\iint_S\mathbf{F}\cdot d\mathbf{S} = \int_0^{2\pi}\int_0^1 (1-r^2)\,r\,dr\,d\theta = 2\pi\int_0^1(r-r^3)\,dr = 2\pi\left[\frac{r^2}{2}-\frac{r^4}{4}\right]_0^1 = 2\pi\cdot\frac{1}{4} = \frac{\pi}{2}.$$
The net upward flux through the paraboloid is $\pi/2$.
the order of $\mathbf{r}_u\times\mathbf{r}_v$ does not matter. The order matters exactly for orientation. $\mathbf{r}_u\times\mathbf{r}_v$ and $\mathbf{r}_v\times\mathbf{r}_u$ point in opposite directions. If you compute $\mathbf{r}_v\times\mathbf{r}_u$ when the problem asks for the outward flux, you will get the inward flux -- a sign error with no arithmetic mistake anywhere. Always check: does the cross product you computed point in the correct direction for the chosen orientation?
Common Misconceptions
the flux integral $\iint_S \mathbf{F}\cdot d\mathbf{S}$ is always positive when $\mathbf{F}$ has positive components.
This is the concept-image-conflicts-definition error. The flux integral measures the net flow through the surface in the direction of the chosen normal. If $\mathbf{F}$ points generally against the normal direction, the dot product $\mathbf{F}\cdot(\mathbf{r}_u \times \mathbf{r}_v)$ is negative, giving a negative integral even when all components of $\mathbf{F}$ are positive. For instance, the field $\mathbf{F} = \langle 0, 0, 1\rangle$ gives positive flux through a surface with upward normal but negative flux through the same surface with downward normal.
Leveled Practice
Problem 1. Let $\mathbf{F} = \langle x, y, z\rangle$ and $S$ the portion of the plane $z = 4$ over $0\leq x\leq 1$, $0\leq y\leq 1$, with upward orientation.
Show answer
Graph $z = 4$, so $g_x = g_y = 0$. Upward element $\langle 0, 0, 1\rangle\,dA$.
$\mathbf{F}$ on $S$: $\langle x,y,4\rangle$.
$\mathbf{F}\cdot\langle 0,0,1\rangle = 4$.
$\iint_S\mathbf{F}\cdot d\mathbf{S} = \int_0^1\int_0^1 4\,dx\,dy = 4$.
Problem 2. Repeat with the same surface but downward orientation. What changes?
Show answer
The element becomes $\langle 0,0,-1\rangle\,dA$, so the integral becomes $-4$. Only the sign changes.