Parametrising a surface
Describe the surface by a vector function of two parameters,
\mathbf{r}(u, v) = \big(x(u,v),\ y(u,v),\ z(u,v)\big), \qquad (u, v) \in D,
so a flat parameter region D in the
uv-plane is wrapped onto S. Holding
v fixed and varying u traces a curve on
the surface with tangent \mathbf{r}_u = \partial \mathbf{r}/\partial u;
likewise \mathbf{r}_v is the tangent along
v. These two tangents span the tangent plane, and a tiny
parameter rectangle du\,dv maps to a tiny parallelogram on
S with area
dS = \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert \, du\, dv.
The cross product \mathbf{r}_u \times \mathbf{r}_v is perpendicular
to the surface; its length is the area-magnification of the parametrisation, and its
direction (once normalised) is the unit normal
\mathbf{n}.
Scalar surface integral
\iint_S f \, dS = \iint_D f\big(\mathbf{r}(u,v)\big)\, \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert \, du\, dv.
Flux of a vector field
The flux of \mathbf{F} through
S sums the component of \mathbf{F} that
pierces the surface, \mathbf{F} \cdot \mathbf{n}, over the whole
sheet. Because \mathbf{n}\, dS = (\mathbf{r}_u \times \mathbf{r}_v)\, du\, dv,
the normalising length cancels and the messy \lvert\cdot\rvert
disappears:
\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{n}\, dS = \iint_D \mathbf{F}\big(\mathbf{r}(u,v)\big) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\, du\, dv.
Worked example: flux through a tilted plane patch
Find the upward flux of \mathbf{F} = (0,\ 0,\ z) through the part
of the plane z = 1 - x - y lying above the unit square
0 \le x \le 1,\ 0 \le y \le 1.
Step 1 — parametrise by x and
y. A graph z = g(x,y) is its
own parametrisation with u = x, v = y:
\mathbf{r}(x, y) = \big(x,\ y,\ 1 - x - y\big), \qquad (x, y) \in [0,1] \times [0,1].
Step 2 — tangent vectors. Differentiate \mathbf{r}
in each parameter:
\mathbf{r}_x = (1,\ 0,\ -1), \qquad \mathbf{r}_y = (0,\ 1,\ -1).
Step 3 — the normal via the cross product.
\mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{vmatrix} = \big(0\cdot(-1) - (-1)\cdot 1,\ \ (-1)\cdot 0 - 1\cdot(-1),\ \ 1\cdot 1 - 0\big) = (1,\ 1,\ 1).
Its z-component is positive, so it already points
upward — the orientation we want. (For a graph
z = g this is always
(-g_x,\ -g_y,\ 1); here
g_x = g_y = -1, confirming (1, 1, 1).)
Step 4 — dot the field into the normal. On the surface
z = 1 - x - y, so
\mathbf{F} = (0,\ 0,\ 1 - x - y), and
\mathbf{F} \cdot (\mathbf{r}_x \times \mathbf{r}_y) = (0,\ 0,\ 1 - x - y) \cdot (1,\ 1,\ 1) = 1 - x - y.
Step 5 — reduce to a double integral over the unit square:
\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^1 \!\! \int_0^1 (1 - x - y)\, dx\, dy.
Step 6 — integrate in x.
\int_0^1 (1 - x - y)\, dx = \left[\, x - \tfrac{x^2}{2} - yx \,\right]_0^1 = 1 - \tfrac{1}{2} - y = \tfrac{1}{2} - y.
Step 7 — integrate in y.
\int_0^1 \left(\tfrac{1}{2} - y\right) dy = \left[\, \tfrac{y}{2} - \tfrac{y^2}{2} \,\right]_0^1 = \tfrac{1}{2} - \tfrac{1}{2} = 0.
The net upward flux is 0: as much of the field crosses upward on
one half of the patch as crosses (relatively) less on the other, and they cancel. Change the
field to a constant \mathbf{F} = (0,0,1) and the same machinery
gives flux \iint_D 1\, dx\, dy = 1 — the area of the square's
shadow, exactly as a uniform upward flow should read.
Let S be a smooth surface parametrised by
\mathbf{r}(u,v) over D, with
\mathbf{n} = (\mathbf{r}_u \times \mathbf{r}_v)/\lvert \mathbf{r}_u \times \mathbf{r}_v \rvert.
-
The area element is
dS = \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert\, du\, dv.
-
Scalar integral:
\iint_S f\, dS = \iint_D f(\mathbf{r})\,\lvert \mathbf{r}_u \times \mathbf{r}_v \rvert\, du\, dv.
-
Flux:
\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\, du\, dv
— the normalising length cancels.
-
Flux is signed: reversing the orientation (choosing
-\mathbf{n}) flips its sign.
A flux integral is meaningless until you choose a side. At each point a
surface has two unit normals, \pm\mathbf{n}, and the flux
through it flips sign with the choice — flow "out" through one face is flow "in" through the
other. For a closed surface (a sphere, a box) the convention is the
outward normal, the one pointing away from the enclosed solid; for an open
patch you simply declare which way counts as positive (here, upward).
A surface admitting a consistent global choice is orientable. Most are —
but the Möbius strip is the famous exception: slide its normal once around the loop and it
returns reversed, so "flux through a Möbius strip" is undefined. We quietly assume every
surface is orientable, the standing hypothesis of
Stokes' theorem.
Read \mathbf{F} as the velocity of a fluid (volume per unit area
per unit time). Then \mathbf{F} \cdot \mathbf{n}\, dS is the
volume crossing the little patch dS each second: only the
component along the normal gets through — fluid sliding parallel to the
surface (\mathbf{F} \perp \mathbf{n}) contributes nothing. Summed
over S, the flux is the total flow rate through the membrane: how
much water passes through a net, how much field crosses a Gaussian surface, how much heat
leaves a wall. This single picture is what
the divergence theorem
ties back to the
divergence
inside.