Recipe

Find the surface area of, and flux through, a parametric surface

Use when: you have a parametric surface \mathbf{r}(u,v) over a parameter domain, and need either its surface area or the flux of a field \mathbf{F} through it

Goal: either the surface's true area, or the numerical flux \iint_S\mathbf{F}\cdot d\mathbf{S}, correctly oriented

Taught in Flux integrals

1. 1

   Differentiate the parametrization with respect to each parameter in turn.

   Example

   For the cylinder \mathbf{r}(u,v) = (\cos v,\ \sin v,\ u), 0\le u\le2, 0\le v\le2\pi: \mathbf{r}_u = (0,0,1) and \mathbf{r}_v = (-\sin v,\cos v,0).

   Why: these two tangent vectors are what both the area-stretch factor and the surface's normal are built from.

2. 2

   Take the cross product of the two tangent vectors.

   Example

   \mathbf{r}_u\times\mathbf{r}_v = (-\cos v,\ -\sin v,\ 0), which has length 1 and points toward the z-axis (inward).

   • only the surface area is needed — continue to Step 3.
   • a flux integral through a field \mathbf{F} is needed — skip to Step 4.

   Why: this single computation supplies both the area-stretch factor (its length) and the surface's normal direction (its direction), whichever one the problem actually needs.

3. 3

   Find the magnitude of the cross product, then integrate that magnitude over the u,v domain. This is the final answer for the surface-area branch.

   Example

   |\mathbf{r}_u\times\mathbf{r}_v| = 1 for every u,v, so S = \displaystyle\int_0^{2\pi}\!\!\int_0^2 1\,du\,dv = 4\pi.

   Why: this is exactly the surface-area construction from earlier in this chapter, weighting each patch of the uv-domain by how much it stretches.

4. 4

   Compare the direction of \mathbf{r}_u\times\mathbf{r}_v against the orientation the problem asks for (outward, inward, upward, and so on). Negate the cross product if it points the wrong way.

   Example

   For an outward-oriented flux through this cylinder, \mathbf{r}_u\times\mathbf{r}_v=(-\cos v,-\sin v,0) points inward, so use -(\mathbf{r}_u\times\mathbf{r}_v) = (\cos v,\sin v,0) instead.

   Why: reversing orientation flips the sign of the whole flux integral, so this has to be settled before any dot product is taken.

5. 5

   Evaluate \mathbf{F} at \mathbf{r}(u,v), then dot it against the (unnormalized, correctly oriented) cross product from Step 4.

   Example

   Let \mathbf{F}=(2x,2y,0). On the surface, \mathbf{F}(\mathbf{r}(u,v)) = (2\cos v,\,2\sin v,\,0), and \mathbf{F}\cdot(\cos v,\sin v,0) = 2\cos^2v+2\sin^2v = 2.

   Why: only the field's component along the normal contributes to flux, and dotting against the unnormalized cross product already carries the correct area-stretch weight, with no separate |\cdot| factor needed.

6. 6

   Integrate the Step 5 result over the u,v domain. This is the final answer for the flux branch.

   Example

   \displaystyle\iint_S\mathbf{F}\cdot d\mathbf{S} = \int_0^{2\pi}\!\!\int_0^2 2\,du\,dv = 8\pi.

   Why: this is the definition of the flux integral itself, once the correctly oriented normal direction is already in hand.