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Integrating over a curved surface: two parameters, the area element $|\mathbf{r}_u \times \mathbf{r}_v|\,du\,dv$, the choice of normal that orients it, and the flux of a field through it.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to parametrise a surface, build its area element from a cross product, compute the surface integral of a scalar function, choose and state an orientation, and compute the flux of a field through a plane rectangle and through the graph of a function.
Line integrals, where a curve was described by one parameter and the speed $|\mathbf{r}'(t)|$ converted parameter steps into length. A surface needs two parameters, and the same job is done by $|\mathbf{r}_u \times \mathbf{r}_v|$, which converts a small parameter rectangle into an area. You also have the change-of-variables formula, of which this is the curved cousin.
A parametrised surface is $\mathbf{r}(u,v)$ for $(u,v)$ in a region of the plane. Its tangent plane at a point is spanned by $\mathbf{r}_u$ and $\mathbf{r}_v$; a normal is perpendicular to it. A surface is orientable when a continuous choice of unit normal $\mathbf{n}$ can be made over the whole of it, and oriented once one is chosen. The flux of a field through an oriented surface is $\iint_S \mathbf{F} \cdot \mathbf{n}\,dS$, also written $\iint_S \mathbf{F} \cdot d\mathbf{S}$.
The area element. A small rectangle $du \times dv$ in the parameter plane maps to a small parallelogram on the surface with edges $\mathbf{r}_u\,du$ and $\mathbf{r}_v\,dv$. Its area is the length of their cross product, so
$$dS = |\mathbf{r}_u \times \mathbf{r}_v|\,du\,dv.$$
The scalar surface integral. For a function $f$ on the surface,
$$\iint_S f\,dS = \iint_D f(\mathbf{r}(u,v))\,|\mathbf{r}_u \times \mathbf{r}_v|\,du\,dv,$$
with $D$ the parameter region. With $f = 1$ this is the surface area; with $f$ a density it is the mass of a sheet.
Orientation. The cross product $\mathbf{r}_u \times \mathbf{r}_v$ is normal to the surface, and the unit normal is $\mathbf{n} = \pm(\mathbf{r}_u \times \mathbf{r}_v)/|\mathbf{r}_u \times \mathbf{r}_v|$. Which sign is a decision, and it must be made and stated: outward for a closed surface, upward for a graph, and whatever the problem says otherwise.
Flux. For an oriented surface,
$$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv.$$
The normalising length cancels between $\mathbf{n}$ and $dS$, so the cross product goes in whole. Flux measures how much of the field crosses the surface per unit time; a field running along the surface contributes nothing.
The graph shortcut. For $z = g(x,y)$ over a region $D$, taking $x$ and $y$ as the parameters gives
$$\mathbf{r}_x \times \mathbf{r}_y = \langle -g_x, -g_y, 1 \rangle, \qquad dS = \sqrt{1 + g_x^2 + g_y^2}\,dA.$$
Another way: picture
Hold a hoop in a stream. Face it square-on to the flow and the water pouring through is maximal; turn it edge-on and nothing goes through at all, even though the stream is as fast as ever. Flux is that measurement, and the dot product with the normal is exactly the facing square-on factor. Turning the hoop over reverses the sign without changing how much water is moving.
Another way: steps
To compute a flux:
The two partial derivatives $\mathbf{r}_u$ and $\mathbf{r}_v$ are the velocities with which the point moves as each parameter is increased. Over a small step they trace a parallelogram whose edges are $\mathbf{r}_u\,du$ and $\mathbf{r}_v\,dv$, and the area of a parallelogram is the length of the cross product of its edges — which is the lesson on the cross product, used for exactly the purpose it was introduced for.
This is also why the change-of-variables formula from the last unit and this one are the same idea. There, a map from one plane region to another stretched area by $|\det J|$; here, a map from a plane region to a curved surface stretches it by $|\mathbf{r}_u \times \mathbf{r}_v|$. For a flat surface in the $xy$ plane the two agree exactly.
For flux, the length cancels: $\mathbf{n}\,dS$ is $(\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv$ with nothing left over. That cancellation is why flux integrals are usually easier than surface area integrals — the square root that makes arc length and surface area hard to compute in closed form never appears.
Leaving out the area factor. $\iint_D f\,du\,dv$ is not a surface integral; the cross product's length is what turns parameter area into real area, and omitting it silently computes the wrong thing.
Normalising and then multiplying by $dS$ separately. For flux, use the cross product whole. Dividing by its length and then multiplying by it again is extra work with an extra chance of error.
Not deciding the orientation. The cross product's direction depends on which parameter was called $u$; swapping them negates it. Check where it points and say so. An orientation, an order of integration and an order of factors are part of the answer, not part of the handwriting. Reversing a curve flips the sign of the work along it, swapping the factors of a cross product flips the vector, and exchanging the two limits of an inner integral changes what region was integrated over. Say which one you chose, every time.
Assuming every surface can be oriented. A Möbius band cannot: following a continuous normal all the way round returns it upside down, and flux through it is undefined rather than merely hard.
The habit from elementary work is that a quantity spread over a region is its density times the area, and flux looks like it should obey the same rule. It does not, for two reasons that both matter.
First, only the component along the normal counts. A gale blowing parallel to a window puts no air through it. The dot product is not a technical decoration; it is the whole measurement.
Second, flux carries a sign, and the sign depends on a choice somebody made. The same field through the same surface gives $+12$ or $-12$ according to which side was called positive. For a closed surface the convention is outward, so a positive total flux means net escape and a negative one net entry — and the next lesson's divergence theorem is precisely the statement that net escape equals total divergence inside.
Find the upward flux of $\mathbf{F} = \langle 0, 0, 5 \rangle$ through the part of $z = 2x + 3y$ above the unit square $0 \le x, y \le 1$.
A graph, so take $x$ and $y$ as the parameters.
The normal is $\langle -2, -3, 1 \rangle$, whose third entry is positive, so it already points upward and needs no sign change.
The graph formula gives the upward normal directly.
Dotting: $\mathbf{F} \cdot \langle -2,-3,1 \rangle = 5$, constant, so the flux is $5$ times the area of the unit square, namely $5$. The tilt of the plane made no difference: a steeper surface has more area but presents it at a shallower angle, and the two effects cancel exactly for a vertical field.
A constant integrand over a unit square.
Take $z = \sqrt{x^2 + y^2}$ over the disc of radius $1$. Then $g_x = x/\sqrt{x^2+y^2}$ and $g_y = y/\sqrt{x^2+y^2}$.
The graph formula again, this time for area.
So $g_x^2 + g_y^2 = 1$ and $dS = \sqrt{2}\,dA$: the cone stretches area by the same factor everywhere.
A constant stretch is the happy case.
The area is therefore $\sqrt{2}$ times the area of the disc, which is $\pi\sqrt{2}$. Notice the shape of the work: the square root that usually blocks a surface area integral collapsed because the cone has the same slope at every point, and recognising that in advance saved the whole computation.
Surface area is a scalar integral with $f = 1$.
Find the upward flux of $\mathbf{F} = \langle x, y, z \rangle$ through the square $0 \le x \le 2$, $0 \le y \le 2$ at height $z = 3$.
A flat horizontal surface: the normal is $\mathbf{k}$.
On that surface $\mathbf{F} \cdot \mathbf{k} = z = 3$, constant, because every point of the square is at the same height.
So the flux is $3 \times 4 = 12$. The $x$ and $y$ components of the field are large near the corners and contribute nothing at all — they run parallel to the square. Flux is never about how big a field is, only about how much of it crosses.
Write $\mathbf{r}_u \times \mathbf{r}_v$ as a row, for the surface $\mathbf{r}(u,v) = \langle u,\; v,\; -u + 3v \rangle$.
This task has no paper form; do it on a device.
Put the stages of computing the flux of a field through a surface in order, for a surface of area $4$.
Number the steps in order (write the number in the box):
Find the upward flux of $\mathbf{F} = \langle 6, 5, 1 \rangle$ through the rectangle $0 \le x \le 6$, $0 \le y \le 3$ in the plane $z = 0$.
Answer:
Evaluate $\iint_S (x + y)\,dS$ over the rectangle $0 \le x \le 2$, $0 \le y \le 6$ in the plane $z = 0$.
Answer:
Write $\mathbf{r}_u \times \mathbf{r}_v$ as a row, for the surface $\mathbf{r}(u,v) = \langle u,\; v,\; -2u - 4v \rangle$.
This task has no paper form; do it on a device.
Two engineers describe the same patch of surface with different parametrisations, both oriented upward, and each computes the flux of the same field. One uses a parameter square of side $5$. What must agree?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the stages of computing the flux of a field through a surface in order, for a surface of area $6$.
Number the steps in order (write the number in the box):
You can turn a surface into a parameter region, produce its area element and its normal, and compute both kinds of surface integral with the orientation stated. Next: the two theorems that convert these integrals into others, and finish the subject.
10. Your turn: flux through the top of a box, step 3