This deck covers parametric surfaces and surface area, scalar surface integrals, and oriented surfaces and flux, then reaches the two capstone theorems: the Divergence Theorem of Gauss, and Stokes's Theorem. It targets the classic errors: applying the Divergence Theorem to an open surface, orienting the normal the wrong way, and confusing when to reach for Stokes and when for Divergence.
Subject: Calculus III · 127 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
This is where the whole course comes together. By the end you can:
1. Describe a surface parametrically and set up its surface area.
2. Evaluate a surface integral of a scalar function over a curved surface.
3. Orient a surface and compute the flux of a vector field across it.
4. Use the Divergence Theorem to turn flux out of a closed surface into a triple integral.
5. Use Stokes's Theorem to trade a boundary line integral for a curl surface integral, and see how it generalizes Green's Theorem.
6. Decide, from the wording of a problem, which theorem actually applies.
Section
Warm-up
Discussion prompt
Without looking back: what did a line integral of a vector field measure, and what were the two things you had to produce before you could compute one?
Hint: One was a way to describe the curve. The other turned that description into a length.
Answer:
It measured circulation — how much the field pushed along the curve. You needed a parametrization of the curve, and the speed factor that turned the parameter step into arc length.
Concept
A space curve needed one parameter: as t runs, a point traces a 1-D path.
A surface is 2-D, so it needs two parameters. As they vary, a point sweeps out a sheet in space.
\[ \text{curve: } \mathbf{r}(t) \qquad\qquad \text{surface: } \mathbf{r}(u,v) \]
Picture it
Animation
Shows: A rippled surface being swept out by two parameters, rotating slowly.
The second parameter is the whole story.
Takeaway: A curve had one dial and traced a line through space. A surface has two, and sweeps out a sheet. Everything that follows is bookkeeping for that extra dial.
Prediction
Predict first
A curve needed one parameter. How many does a surface need, and why exactly that many?
Correct: Two — one for each independent direction you can travel while staying on the surface.
Why: Dimension is about freedom of movement, not about the space something sits in. On a curve you can only go forwards or backwards, so one parameter. On a surface you can move in two independent directions, so two parameters, even though the surface lives in three-dimensional space.
Concept
A parametric surface is a vector function of two variables, each component a function of both parameters.
\[ \mathbf{r}(u,v) = \langle\, x(u,v),\; y(u,v),\; z(u,v)\,\rangle \]
parameter domain — The region in the u-v plane over which the parameters are allowed to range. The surface is the image of this region under r.
Intuition
Picture two knobs, u and v. Turning u alone traces one family of curves across the surface; turning v alone traces the crossing family.
Together they lay down a grid on the sheet, the same way latitude and longitude lay a grid on a globe.
Figure (svg): A curved sheet with two families of grid lines crossing it, one labeled u and one labeled v
Step zero
Discussion prompt
Before any formula: you are standing at a corner of a flat sheet. Describe, in plain English, how you would reach every other point on it using exactly two dials.
Hint: What two directions would you walk in, and what would each dial control?
Answer:
Pick a starting point and two directions that lie in the sheet and are not parallel. One dial says how far to walk along the first direction, the other how far along the second. Every point is reachable, and each point exactly once.
Concept
The easiest surface: a flat piece. Start at a point and add multiples of two independent direction vectors.
\[ \mathbf{r}(u,v) = \mathbf{r}_0 + u\,\mathbf{a} + v\,\mathbf{b} \]
This is exactly the vector form of a plane, now read as a surface swept by the two parameters.
Picture it
Animation
Shows: A flat tilted plane patch swept out by two parameters, rotating slowly.
Two dials, one flat sheet.
Takeaway: A plane is flat everywhere, so its stretch factor is the same at every point — which is why its area formula collapses to a single constant times the domain's area.
Concept
Use the spherical angles: one from the top axis, one going around.
\[ \mathbf{r}(\phi,\theta) = \langle a\sin\phi\cos\theta,\; a\sin\phi\sin\theta,\; a\cos\phi\rangle \]
\[ 0 \le \phi \le \pi, \qquad 0 \le \theta \le 2\pi \]
Here the radius is fixed at the constant value; only the two angles move.
Picture it
Animation
Shows: A sphere being swept out as one angle runs pole to pole and the other sweeps around.
Watch where the sweep slows to nothing — those are the poles.
Takeaway: One angle walks from pole to pole, the other spins around the axis. Every point is reached, and the poles are where the parametrization degenerates.
Concept
Any surface written as height over the plane has a free parametrization: use the two coordinates themselves as parameters.
\[ z = g(x,y) \;\Longrightarrow\; \mathbf{r}(x,y) = \langle x,\; y,\; g(x,y)\rangle \]
This special case gives the shortcut formulas we lean on constantly.
Picture it
Animation
Shows: A bowl-shaped graph being swept out with x and y as the two parameters.
The parameters were sitting there all along.
Takeaway: Take x and y themselves as the two parameters and the height comes along for free — that is why a graph never needs a separate parametrization.
Prediction
Predict first
Hold the first parameter fixed and vary only the second. What traces out on the surface, and what does its derivative give you?
Correct: A curve lying in the surface, and its derivative is a vector tangent to the surface.
Why: Freezing one parameter leaves a single free parameter, which is exactly the recipe for a curve. Because that curve never leaves the surface, its velocity vector must lie in the surface, which is what makes it a tangent vector.
Concept
Freeze one parameter and differentiate with respect to the other. Each gives a vector tangent to a grid curve on the surface.
\[ \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u}, \qquad \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} \]
These two tangent vectors span the plane that just kisses the surface at that point.
Picture it
Animation
Shows: The two partial derivatives defined, and the normal built from their cross product.
Two derivatives in, one normal out.
Takeaway: Freeze one dial and differentiate along the other: that is a tangent vector. Do it for each dial and cross the results to leave the surface perpendicularly.
Prediction
Predict first
You now have two vectors that both lie flat in the surface. What single operation produces a vector perpendicular to both?
Correct: The cross product.
Why: The cross product of two vectors is perpendicular to both of them by construction. Since both tangent vectors lie in the surface, anything perpendicular to both is perpendicular to the surface itself, which is exactly what a normal vector means.
Concept
A vector perpendicular to both tangents is perpendicular to the surface. The cross product delivers exactly that.
\[ \mathbf{N} = \mathbf{r}_u \times \mathbf{r}_v \]
normal vector — A vector perpendicular to the tangent plane of the surface. Its length also measures how much area a small parameter rectangle covers on the surface.
Picture it
Animation
Shows: A vector field flowing left to right across a closed loop, with inflow marked on one side and outflow on the other.
Keep this picture — every flux computation later is this idea with coordinates attached.
Takeaway: Flux is bookkeeping: count what enters, count what leaves, and keep only the difference.
Intuition
A tiny step in each parameter carves out a little parallelogram on the surface, with sides given by the two tangent vectors scaled by the steps.
The area of a parallelogram is the magnitude of the cross product of its sides. So the length of the normal is the local area-stretch factor.
\[ dS = |\mathbf{r}_u \times \mathbf{r}_v|\, du\, dv \]
Concept
We need the normal to actually point somewhere at each point, so the tangent plane is well defined.
smooth surface — A parametric surface whose normal vector is nonzero everywhere on the interior of the parameter domain, so it has a well-defined tangent plane at every point.
Estimation
Predict first
Two tangent vectors meet at a right angle, one of length 3 and one of length 2. Before computing: how much surface area does the little patch they span cover?
Correct: 6 — the area of the parallelogram they span.
Why: The magnitude of a cross product is the area of the parallelogram spanned by the two vectors. When they are perpendicular that is simply the product of their lengths, so three times two gives six.
Concept
Add up all those little parallelograms over the parameter domain.
\[ \text{Area}(S) = \iint_D |\mathbf{r}_u \times \mathbf{r}_v|\, dA \]
Everything else in this deck is a variation on this one idea: integrate something against the surface element.
Picture it
Animation
Shows: A parameter rectangle mapping to a parallelogram, giving the area element.
The formula is derived, not decreed.
Takeaway: A tiny rectangle in parameter space lands as a parallelogram on the surface, and the cross-product magnitude is exactly that parallelogram's area. The formula is not a definition — it is a measurement.
Concept
For a surface given as height over the plane, the cross product magnitude simplifies beautifully.
\[ z = g(x,y): \quad dS = \sqrt{1 + g_x^2 + g_y^2}\; dA \]
The steeper the surface, the larger this factor, so a patch of ground covers more slanted area above it.
Step zero
Discussion prompt
You are about to find the surface area of a patch of a slanted plane. Write your game plan in plain English — what do you produce first, second, third?
Hint: Three products, then one integral.
Answer:
First a parametrization with two parameters and a stated domain. Second the two tangent vectors, one derivative per parameter. Third the cross product and its magnitude, which is the area scaling factor. Then integrate that factor over the domain.
Worked example
Find the area of the part of a tilted plane lying above a rectangle.
\[ z = 1 + 2x + 3y, \qquad 0 \le x \le 2,\; 0 \le y \le 1 \]
Take the partial derivatives of the height
Why: The graph shortcut needs the slopes in each direction.
\[ g_x = 2, \qquad g_y = 3 \]
Build the surface element
Why: Plug the slopes into the square-root factor.
\[ dS = \sqrt{1 + 2^2 + 3^2}\, dA = \sqrt{14}\; dA \]
Integrate over the rectangle
Why: The factor is constant, so the integral is just the factor times the base area.
\[ \text{Area} = \iint_D \sqrt{14}\, dA = \sqrt{14}\,(2 \cdot 1) = 2\sqrt{14} \]
Verify the answer is bigger than the flat base
Why: A tilted patch must have more area than its shadow. The base area is 2, and the factor is greater than one, so the slanted area exceeds 2. It does.
\[ 2\sqrt{14} \approx 7.48 \; > \; 2 \]
Picture it
Animation
Shows: The algebra of the slanted plane area computation, revealed one line at a time.
The same four moves work for every surface — only the last line simplifies here.
Takeaway: Parametrize, take one derivative per parameter, cross them, take the magnitude. For a plane that magnitude is constant, so the integral is just a number times the domain's area.
Prediction
Predict first
For a cone parametrized by radius and angle, do you expect the area scaling factor to depend on the angle?
Correct: No. The cone is rotationally symmetric, so the factor cannot depend on the angle.
Why: If a shape looks identical from every angular direction, no quantity computed from its local geometry may depend on the angle. Noticing that symmetry ahead of time gives you a free check on your algebra when the factor comes out.
Worked example
Find the area of the cone lying above the unit disk.
\[ z = \sqrt{x^2 + y^2}, \qquad x^2 + y^2 \le 1 \]
Differentiate the height
Why: Use the graph shortcut again.
\[ g_x = \frac{x}{\sqrt{x^2+y^2}}, \qquad g_y = \frac{y}{\sqrt{x^2+y^2}} \]
Add the squared slopes
Why: They collapse to one, because the two squared fractions share the denominator that is the numerator sum.
\[ g_x^2 + g_y^2 = \frac{x^2 + y^2}{x^2 + y^2} = 1 \]
Form the surface element
Why: The factor is a constant, so the cone stretches area by the same amount everywhere.
\[ dS = \sqrt{1 + 1}\; dA = \sqrt{2}\; dA \]
Integrate over the unit disk
Why: The constant factor times the disk area.
\[ \text{Area} = \sqrt{2}\iint_D dA = \sqrt{2}\,(\pi \cdot 1^2) = \sqrt{2}\,\pi \]
Check against the lateral-area formula
Why: A cone of base radius one and height one has slant height equal to the square root of two. The formula pi times radius times slant gives the same value.
\[ \pi r \ell = \pi (1)(\sqrt{2}) = \sqrt{2}\,\pi \;\checkmark \]
Picture it
Animation
Shows: A cone swept out by a radius parameter and an angle parameter, rotating.
If your area factor mentions the angle, something went wrong.
Takeaway: The cone is rotationally symmetric, so nothing computed from its local geometry may depend on the angle — a free check on the algebra you just did.
Section
Real world
Discussion prompt
A curved metal sheet is thicker in some places than others, so its density varies from point to point. You want its total mass. Turn that into a mathematical recipe — what are you adding up, and over what?
Hint: Chop it into tiny patches. What is the mass of one patch?
Answer:
Each tiny patch has mass equal to its density times its area. Summing over all patches and taking the limit gives the integral of the density function over the surface — a scalar surface integral.
Concept
Now let a function have a value at every point of the surface, and add it up weighted by area.
\[ \iint_S f\, dS = \iint_D f(\mathbf{r}(u,v))\,|\mathbf{r}_u \times \mathbf{r}_v|\, dA \]
If the function is the constant one, this reduces to surface area. Everything is one machine.
Intuition
Think of the surface as a thin metal shell whose density changes from spot to spot.
Multiply density by the tiny area of each patch and sum: that total is the mass. The surface integral of a scalar is that sum.
Picture it
Animation
Shows: The scalar surface integral built up from tiny patch masses.
One integral, many jobs, depending on the weight.
Takeaway: Each tiny patch contributes density times area. Sum them and you have the scalar surface integral — with surface area itself as the case where the density is one.
Analogy
Match the pairs
Each integral you have met so far has the same shape: something to add up, times a factor that converts parameter steps into real geometric size. Match each integral to its conversion factor.
Why: Every integral in this course is a sum of contributions times a conversion factor that turns abstract parameter steps into real geometric size. The factor gets richer as the dimension rises, but the structure never changes.
Concept
For a surface given as height over the plane, drop in the same square-root factor.
\[ \iint_S f\, dS = \iint_D f(x,y,g(x,y))\,\sqrt{1 + g_x^2 + g_y^2}\; dA \]
The one habit to build: never write only dA where a surface wants dS. The stretch factor is the whole point.
Step zero
Discussion prompt
Before touching algebra: what is the one extra thing a scalar surface integral needs that a surface-area computation did not?
Answer:
The function being integrated, rewritten in terms of the two parameters. Surface area is the special case where that function is the constant one.
Worked example
Evaluate the integral of the height function over a tilted plane above the unit square.
\[ \iint_S z\, dS, \quad S:\; z = x + y, \;\; 0 \le x \le 1,\; 0 \le y \le 1 \]
Find the surface element
Why: Both slopes are one, so the factor is the square root of three.
\[ g_x = 1,\; g_y = 1 \;\Rightarrow\; dS = \sqrt{1+1+1}\, dA = \sqrt{3}\, dA \]
Replace z by its formula on the surface
Why: On the surface the height equals x plus y, so the integrand becomes that sum.
\[ \iint_S z\, dS = \sqrt{3}\iint_D (x + y)\, dA \]
Integrate the inside over x first
Why: Hold y constant and integrate the sum across the unit interval.
\[ \int_0^1 (x+y)\, dx = \left[\tfrac{x^2}{2} + yx\right]_0^1 = \tfrac{1}{2} + y \]
Integrate the result over y
Why: Now sweep y across the unit interval.
\[ \int_0^1 \left(\tfrac{1}{2} + y\right) dy = \tfrac{1}{2} + \tfrac{1}{2} = 1 \]
Multiply by the constant factor
Why: The surface element factor comes back out front.
\[ \iint_S z\, dS = \sqrt{3}\cdot 1 = \sqrt{3} \]
Verify by an average-value sanity check
Why: The height ranges from zero to two with average one over the square, and the surface area is the square root of three; average height times area gives the same product.
\[ \bar{z}\cdot \text{Area} = 1 \cdot \sqrt{3} = \sqrt{3} \;\checkmark \]
Picture it
Animation
Shows: The scalar surface integral formula broken into its three practical steps.
Three steps, and you have already done all three.
Takeaway: Rewrite the integrand in the parameters, multiply by the stretch factor, integrate over the domain. Surface area is the case where the integrand is the constant one.
Missing information
Discussion prompt
Someone hands you: a cylinder of radius 2, and a density that grows with height. They ask for the total mass. What critical piece of information have they not given you?
Hint: You can describe the surface, but can you say where it stops?
Answer:
The height range — where the cylinder starts and ends. Without a domain for the parameters there is nothing to integrate over, and the answer would be infinite. A parametrization is not complete until its domain is stated.
Worked example
Integrate the height over the side of a unit cylinder of height one.
\[ \iint_S z\, dS, \quad S:\; x^2 + y^2 = 1,\; 0 \le z \le 1 \]
Parametrize the side
Why: Go around with an angle and up with the height.
\[ \mathbf{r}(\theta, z) = \langle \cos\theta,\; \sin\theta,\; z\rangle \]
Compute the tangent vectors and their cross product
Why: One tangent goes around the circle, the other straight up.
\[ \mathbf{r}_\theta \times \mathbf{r}_z = \langle \cos\theta,\; \sin\theta,\; 0\rangle, \qquad |\mathbf{r}_\theta \times \mathbf{r}_z| = 1 \]
Set up and evaluate
Why: The surface element is simply the angle times height, so the integral separates.
\[ \int_0^{2\pi}\!\!\int_0^1 z\, dz\, d\theta = 2\pi \cdot \tfrac{1}{2} = \pi \]
Verify with the average height
Why: The average height on the cylinder is one half and the side area is two pi; their product matches.
\[ \tfrac{1}{2}\cdot 2\pi = \pi \;\checkmark \]
Picture it
Animation
Shows: A cylinder swept out by an angle wrapping around and a height running up.
A parametrization is not finished until its domain is stated.
Takeaway: The angle wraps all the way around and the height runs between stated limits — without those limits there is no region to integrate over and the answer would be infinite.
Section
Prediction
Predict first
Take a long strip of paper, give one end a half twist, and glue the ends together. Start painting one side without lifting the brush. What happens?
Correct: You paint the entire strip. It has only one side.
Why: This is the Mobius strip, and it is the reason orientation has to be assumed rather than taken for granted. Flux is defined in terms of a consistent choice of normal direction, so a surface with no such consistent choice has no flux at all.
Concept
To measure flow across a surface, you must first decide which way is out. A surface has two sides, and the normal can point either way.
orientation — A consistent choice of one of the two unit normal directions across the whole surface. Choosing it turns the surface into an oriented surface.
Picture it
Animation
Shows: A saddle surface with normal vectors drawn along it, all pointing consistently to one side.
Every arrow agrees. That agreement is what orientable means.
Takeaway: Choosing a side is choosing a normal direction. Flip every arrow and the flux flips sign — the geometry is untouched, only the bookkeeping reverses.
Concept
Normalize the cross-product normal to get a unit vector pointing to the chosen side.
\[ \mathbf{n} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{|\mathbf{r}_u \times \mathbf{r}_v|} \]
Swapping the parameter order, or negating, flips the normal to the other side. That sign choice is the orientation.
Notation
Annotate
You will meet this expression constantly. Take it apart before you use it.
On: \( \iint_S \mathbf{F}\cdot \mathbf{n}\, dS \)
Concept
A surface that fully encloses a solid region, with no edges, is called closed. Its standard positive orientation is the outward normal.
closed surface — A surface with no boundary curve that completely encloses a region of space, like a sphere or the six faces of a box. Positive orientation means the normal points away from the enclosed solid.
Concept
Flux measures how much of a vector field passes through the surface. Take the part of the field along the normal and integrate over the surface.
\[ \text{Flux} = \iint_S \mathbf{F}\cdot\mathbf{n}\, dS \]
Only the component crossing the surface counts; flow that slides along the surface contributes nothing.
Picture it
Animation
Shows: A vector field crossing a closed loop, with the loop highlighted.
The dot product is doing the discarding.
Takeaway: Only the component pointing straight through the boundary is counted. Field sliding along the boundary contributes exactly nothing.
Edge cases
Discussion prompt
The field at some point is exactly parallel to the surface — it slides along it and never pokes through. What does the flux integrand do at that point, and why should that feel right?
Hint: Think about what the dot product does to perpendicular vectors.
Answer:
The integrand is zero there. A field parallel to the surface is perpendicular to the normal, and the dot product of perpendicular vectors vanishes. That matches the physical picture exactly: nothing is crossing, so nothing should be counted.
Intuition
Imagine wind, or a moving fluid, and hold up a net. Flux is the net rate at which stuff passes through the mesh.
Wind blowing straight through counts fully; wind skimming sideways along the net passes through nothing. The dot product with the normal captures exactly that.
Concept
In practice, do not normalize and re-multiply. The unit vector's denominator cancels the surface element, leaving the raw cross product.
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\, dS = \iint_D \mathbf{F}\cdot(\mathbf{r}_u \times \mathbf{r}_v)\, dA \]
This is the working formula: dot the field into the cross product and integrate over the parameter domain.
Picture it
Animation
Shows: A bowl-shaped surface with all its normal vectors pointing the opposite way.
Compare against the earlier normals clip.
Takeaway: This is the same surface as before with every normal reversed. Nothing geometric changed — but every flux computed across it now carries the opposite sign.
Concept
For an upward-oriented surface given as height over the plane, the cross product takes a memorable form.
\[ \mathbf{r}_x \times \mathbf{r}_y = \langle -g_x,\; -g_y,\; 1\rangle \]
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\, dS = \iint_D \big(-P\,g_x - Q\,g_y + R\big)\, dA \]
Step zero
Discussion prompt
Plan the flux computation before starting it. What are the four things you must produce, in order?
Answer:
One: a parametrization of the triangle with its domain. Two: the two tangent vectors. Three: their cross product, then a decision about whether its direction matches the orientation you were asked for, flipping the sign if not. Four: dot the field with that vector and integrate over the domain.
Worked example
Find the upward flux of the position field across the first-octant piece of a plane.
\[ \mathbf{F} = \langle x, y, z\rangle, \quad S:\; x + y + z = 1 \text{ in the first octant} \]
Write the surface as a graph
Why: Solve the plane for the height so the graph formula applies.
\[ z = g(x,y) = 1 - x - y, \qquad g_x = -1,\; g_y = -1 \]
Identify the projected region
Why: The shadow in the plane is the triangle where both coordinates are nonnegative and their sum is at most one.
\[ D:\; x \ge 0,\; y \ge 0,\; x + y \le 1 \]
Assemble the integrand
Why: Use the graph flux formula with the field components P, Q, R equal to x, y, and the height.
\[ -P g_x - Q g_y + R = -x(-1) - y(-1) + (1-x-y) \]
Simplify
Why: The x and y terms cancel, leaving a constant integrand.
\[ x + y + 1 - x - y = 1 \]
Integrate the constant over the triangle
Why: The integral of one is just the area of the triangle, which has legs of length one.
\[ \iint_D 1\, dA = \tfrac{1}{2}(1)(1) = \tfrac{1}{2} \]
Verify the sign is positive
Why: The position field points outward from the origin and the chosen normal points up and away from the origin, so a positive flux is exactly what we expect.
\[ \text{Flux} = \tfrac{1}{2} > 0 \;\checkmark \]
Picture it
Animation
Shows: The flux-across-a-triangle computation with the orientation check highlighted.
Highlighted: the step that most answers lose a sign on.
Takeaway: The cross product gives a normal, but not necessarily the one you were asked for. Checking its direction against the stated orientation is the step that decides the sign.
Anomaly
Predict first
Two students compute the flux of the same field across the same surface. One reports 12, the other reports negative 12. Neither made an arithmetic error. What happened?
Correct: They chose opposite normal directions.
Why: Flux is only defined relative to a choice of orientation, and the two choices differ by a sign. This is why a flux problem is not fully stated until it says which way the normal points, and why an answer that differs from the book by only a minus sign usually means an orientation slip rather than bad algebra.
Trap
Computing the outward flux of the position field across the unit sphere, but taking the inward normal.
\[ \mathbf{F} = \langle x, y, z\rangle, \qquad \mathbf{n}_{\text{wrong}} = -\frac{\langle x,y,z\rangle}{1} \]
The field points straight out, so dotting it with an inward normal makes every contribution negative and the answer comes out wrong.
\[ \iint_S \mathbf{F}\cdot\mathbf{n}_{\text{wrong}}\, dS = -4\pi \]
Outward flux uses the outward normal. On the unit sphere that is the position direction itself.
\[ \mathbf{n} = \frac{\langle x,y,z\rangle}{1}, \qquad \mathbf{F}\cdot\mathbf{n} = x^2+y^2+z^2 = 1 \]
Now the integrand is a clean positive one, and integrating over the sphere gives the correct outward flux.
\[ \iint_S 1\, dS = 4\pi(1)^2 = 4\pi \]
Comparison
Comparison matrix
Fill in what changes between the three surface integrals you now know.
| What you integrate | Does orientation matter? | |
|---|---|---|
| Surface area | the constant 1 | no |
| Scalar surface integral | a scalar function | no |
| Flux | the normal component of a vector field | yes |
Section
Warm-up
Discussion prompt
Without looking: divergence takes a vector field and returns what kind of object, and what does its sign tell you at a point?
Answer:
It returns a scalar — one number at each point. Positive means the field is spreading out from that point, so it acts as a source. Negative means it is collapsing inward, a sink. Zero means whatever flows in also flows out.
Concept
Divergence turns a vector field into a scalar: the sum of the rates at which each component grows in its own direction.
\[ \nabla\cdot\mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \]
Picture it
Animation
Shows: A vector field spreading outward from the origin, then replaced by a rotating field, with a loop drawn in both cases.
Watch the loop, not the arrows.
Takeaway: Divergence asks only one question: does more leave the little loop than enters it? A purely rotating field has divergence zero even though nothing is standing still.
Intuition
At a point, positive divergence means the field is a source: more flows out of a tiny box than into it. Negative divergence means a sink.
So divergence is the outflow per unit volume, measured right at that point.
Hypothesis
Predict first
A solid region is full of little sources and sinks. Guess the relationship between the total flux out through its skin and the sum of all the sources inside.
Correct: They are equal. That statement is the Divergence Theorem.
Why: Everything produced inside must eventually leave through the boundary — there is nowhere else for it to go. Adding up the sources throughout the solid and measuring what crosses the skin are two ways of counting the same thing, which is precisely why the theorem holds.
Concept
Add up the local outflow throughout a solid, and it must equal the total flow escaping across the boundary surface.
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\, dS = \iiint_E \nabla\cdot\mathbf{F}\; dV \]
The surface must be the complete closed boundary of the solid, oriented outward.
Intuition
Cut the solid into tiny cells. Flow crossing an interior wall leaves one cell and enters its neighbor, so it cancels.
Only flow across the outer skin survives. Summing every cell's outflow, which is its divergence times its volume, leaves exactly the flux through the outer surface.
Picture it
Animation
Shows: A field spreading outward from the interior, crossing a loop drawn around it.
There is nowhere else for it to go.
Takeaway: Nothing can accumulate inside. Whatever the interior produces has to cross the boundary, which is why summing sources and measuring outflow give the same number.
Definition probe
Sort into buckets
The Divergence Theorem needs a closed surface bounding a solid, with a field that is smooth everywhere inside. Sort each situation by which requirement it violates.
Concept
Three conditions, all required: the surface is closed, it bounds a solid region, and the normal points outward.
It is the tool of choice when you want a flux across a closed surface and the divergence is simpler than the surface itself.
Estimation
Predict first
A field has constant divergence 3 everywhere. The solid is a cube of side 2. Estimate the total flux out of it before doing any integral.
Correct: 24 — the divergence times the volume, so 3 times 8.
Why: When divergence is constant the volume integral collapses to a multiplication: the constant times the volume of the region. The cube of side two has volume eight, so the flux is three times eight. Having this number in hand before you integrate turns the computation into a check rather than a leap of faith.
Worked example
Find the outward flux across the surface of the unit cube.
\[ \mathbf{F} = \langle x^2, y^2, z^2\rangle, \quad E:\; [0,1]\times[0,1]\times[0,1] \]
Recognize the shortcut
Why: Direct flux would need all six faces. The divergence is a simple polynomial, so the triple integral is far easier.
Compute the divergence
Why: Differentiate each component in its own variable and add.
\[ \nabla\cdot\mathbf{F} = 2x + 2y + 2z \]
Integrate over the cube
Why: By symmetry each of the three terms integrates the same way over the unit cube.
\[ \iiint_E (2x+2y+2z)\, dV = 3\int_0^1\!\!\int_0^1\!\!\int_0^1 2x\, dx\, dy\, dz \]
Evaluate one term and triple it
Why: The integral of two x over the unit cube is one, and there are three matching terms.
\[ 3\left(2\cdot\tfrac{1}{2}\cdot 1 \cdot 1\right) = 3(1) = 3 \]
Verify a single face is consistent
Why: On the face where x equals one, the outward normal is the x direction and the field's x component is one, contributing area one; the zero-coordinate faces contribute nothing, and the pattern across all faces sums to three.
\[ \text{Flux} = 3 \;\checkmark \]
Picture it
Animation
Shows: The cube flux computation collapsing to divergence times volume.
This is the whole reason to convert.
Takeaway: When the divergence is constant, the volume integral is just that constant times the volume — so six separate face integrals become a single multiplication.
Discrimination
Sort into buckets
For each problem, decide whether it is easier to attack the surface directly or to convert it into a volume integral. Do not solve anything.
Worked example
Find the outward flux of the position field across the sphere of radius a.
\[ \mathbf{F} = \langle x, y, z\rangle, \quad S:\; x^2 + y^2 + z^2 = a^2 \]
Compute the divergence
Why: Each component contributes a one.
\[ \nabla\cdot\mathbf{F} = 1 + 1 + 1 = 3 \]
Integrate the constant over the ball
Why: A constant times the volume of the enclosed ball.
\[ \iiint_E 3\, dV = 3\cdot \tfrac{4}{3}\pi a^3 = 4\pi a^3 \]
Verify against the direct surface calculation
Why: On the sphere the outward normal is the position over its length, so the field dotted with the normal equals the radius a; multiplying by the sphere's area gives the same result.
\[ a \cdot 4\pi a^2 = 4\pi a^3 \;\checkmark \]
Picture it
Animation
Shows: The Divergence Theorem written out, with the plain-language reading highlighted underneath.
Read the second line out loud — that is the whole idea.
Takeaway: A boundary integral equals a derivative integrated over the inside. Every big theorem in this course is that same sentence at a different dimension.
Trap
Wanting the outward flux across just the paraboloid cap (open at the bottom), a student jumps straight to a triple integral of the divergence.
\[ \mathbf{F} = \langle x, y, 1\rangle, \quad S:\; z = 1 - x^2 - y^2,\; z \ge 0 \]
The cap alone is not a closed surface, so the theorem does not apply to it. The triple integral secretly answers a different question.
\[ \iiint_E \nabla\cdot\mathbf{F}\, dV = \iiint_E 2\, dV = 2\cdot\tfrac{\pi}{2} = \pi \]
Close the region first: add the flat disk at the bottom, apply the theorem to the closed solid, then subtract the disk's flux.
\[ \Phi_{\text{cap}} + \Phi_{\text{disk}} = \iiint_E \nabla\cdot\mathbf{F}\, dV = \pi \]
The disk sits at height zero with outward normal pointing down, so its flux is the negative of its area. Subtracting a negative adds it back.
\[ \Phi_{\text{disk}} = -\pi \;\Rightarrow\; \Phi_{\text{cap}} = \pi - (-\pi) = 2\pi \]
Constraint
Discussion prompt
You want the flux across an open bowl, but the Divergence Theorem needs a closed surface. You are not allowed to parametrize the bowl. What else could you do?
Hint: You are allowed to add a surface of your own choosing.
Answer:
Cap it. Add a flat lid so the bowl plus lid is closed, apply the theorem to that whole closed surface, then subtract the flux through the lid — which is easy, because a flat disk with a constant normal is the simplest surface integral there is.
Concept
That fix is a reusable move. To get flux across an open surface with the Divergence Theorem, seal it with a simple lid to make a closed solid.
Apply the theorem to the whole closed boundary, compute the easy lid's flux directly, and subtract it out.
Picture it
Animation
Shows: An open hemispherical bowl rotating, with its missing flat lid implied.
The lid is the cheapest surface integral there is.
Takeaway: An open surface bounds no solid, so the theorem has nothing to convert to. Add a flat lid, apply the theorem to the closed result, then subtract the lid's flux — which is easy, because a flat disk has a constant normal.
Counterexample
Discussion prompt
Claim: if a field has divergence zero everywhere inside a region, then no field lines pass through the region at all. Construct a counterexample.
Hint: A uniform wind has divergence zero.
Answer:
A constant field — a steady wind blowing in one direction — has divergence zero everywhere, yet field lines pass straight through any region you like. Divergence zero does not mean nothing flows. It means whatever enters also leaves, so the net is zero.
Section
Warm-up
Predict first
Divergence turned a vector field into a scalar. What does curl turn a vector field into?
Correct: Another vector field.
Why: Curl has to record not just how much something spins but which axis it spins about, and an axis needs a direction. A single number could not carry that, so the output must be a vector, pointing along the axis of rotation with length equal to the rate.
Concept
Curl turns a vector field into another vector field that measures local spinning.
\[ \nabla\times\mathbf{F} = \left\langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \right\rangle \]
Picture it
Animation
Shows: A shear field where arrows are longer higher up, with a small paddle wheel placed in it that begins rotating.
The field does not have to look swirly to have curl.
Takeaway: Curl is local. Every arrow here points the same way and nothing circles anything, yet the wheel still turns — because the field is faster on one side of it than the other.
Intuition
Drop a small paddle wheel into the flow. If it spins, the field has curl there, and the axis it spins about is the curl's direction.
The faster the spin, the larger the curl. Where the flow just carries the wheel along without turning it, the curl is zero.
Prediction
Predict first
The Divergence Theorem traded a boundary integral for an integral over the inside. If Stokes does the same thing one dimension down, what should it relate?
Correct: A line integral around the boundary curve to a surface integral of curl over the surface it bounds.
Why: The pattern is always the same: something integrated over a boundary equals a derivative of it integrated over the inside. Drop the Divergence Theorem by one dimension and the solid becomes a surface, its skin becomes a curve, and divergence becomes curl.
Concept
The circulation of a field around the boundary curve equals the flux of its curl through any surface the curve bounds.
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot\mathbf{n}\, dS \]
It trades a possibly nasty line integral for a surface integral, or the other way around, whichever side is easier.
Error analysis
Annotate
A student applies Stokes and gets the right magnitude but the wrong sign. Their work is below. Find the step that broke.
Concept
The two sides only agree when the curve and the normal are matched consistently.
Point the thumb of your right hand along the chosen normal; your fingers curl in the direction the boundary must be traversed.
Figure (svg): A disk with an upward normal arrow and a counterclockwise arrow around its boundary
Intuition
Tile the surface with tiny loops. Neighboring loops share an edge, and their circulations along that shared edge run opposite and cancel.
Only the outer rim survives. So the total spin inside, added up, equals the circulation around the single boundary curve.
Analogy
Match the pairs
Stokes is Green's Theorem with the flat plane bent into a surface. Match each piece of Green's Theorem to what it becomes.
Why: Green's Theorem is the special case of Stokes where the surface happens to lie flat in the plane and the normal points straight up. When that happens the dot product with the normal picks out exactly one component of curl, which is the scalar curl you already knew.
Concept
Flatten the surface into a region of the plane with an upward normal, and Stokes collapses to Green's Theorem in circulation form.
\[ \oint_C P\, dx + Q\, dy = \iint_D (Q_x - P_y)\, dA \]
Green's Theorem is just the flat, two-dimensional case of Stokes.
Picture it
Animation
Shows: A rotating vector field in the plane with a closed loop, standing in for Green's Theorem.
Same theorem, one dimension down.
Takeaway: Flatten a Stokes surface into the plane and the normal points straight up, so the dot product picks out exactly one component of curl — the scalar curl you already knew.
Break the constraint
Discussion prompt
A flat disk and a tall balloon-like dome both have the same circle as their boundary. Stokes says the surface integral of curl is the same over both. Does that bother you? Argue yourself into or out of it.
Hint: What does Stokes say each of them equals?
Answer:
It should not bother you, because Stokes says each equals the same line integral around the shared circle. Anything equal to the same thing is equal to the other. The practical payoff is large: when a surface is unpleasant, swap it for any easier surface with the same boundary.
Concept
The right-hand side of Stokes depends only on the boundary curve, not on which surface you cap it with.
So you may replace an awkward surface with the simplest one sharing that boundary, usually a flat disk. This freedom is where Stokes saves the most work.
Picture it
Animation
Shows: A rotating vector field with a loop drawn in it.
Everything interior cancels in pairs.
Takeaway: Neighbouring paddle wheels push against each other and cancel. Only the spin at the outer rim has nothing to cancel against — and that is the circulation.
Step zero
Discussion prompt
Before computing anything with Stokes, there is a decision to make that determines how hard the rest will be. What is it?
Answer:
Which side to compute. Stokes gives an equation with a line integral on one side and a surface integral on the other, and you get to pick whichever is easier. If the boundary is one simple circle, do the line integral. If the field's curl is simple but the curve is awkward, do the surface integral — and choose the friendliest surface with that boundary.
Worked example
Compute the circulation of a rotating field around the unit circle in the plane, oriented counterclockwise.
\[ \mathbf{F} = \langle -y, x, 0\rangle, \quad C:\; x^2 + y^2 = 1,\; z = 0 \]
Take the curl
Why: Only the last component of the curl survives here.
\[ \nabla\times\mathbf{F} = \langle 0, 0, 1 - (-1)\rangle = \langle 0, 0, 2\rangle \]
Cap with the flat disk
Why: The disk shares the boundary and its upward normal matches the counterclockwise curve by the right-hand rule.
\[ S:\; x^2 + y^2 \le 1,\; z = 0, \qquad \mathbf{n} = \langle 0,0,1\rangle \]
Integrate the curl flux
Why: The curl dotted with the upward normal is the constant two, so the integral is two times the disk area.
\[ \iint_S 2\, dS = 2\,(\pi \cdot 1^2) = 2\pi \]
Verify by the direct line integral
Why: Parametrize the circle and integrate the field along it; the integrand simplifies to one, giving the same answer and confirming Stokes.
\[ \int_0^{2\pi} (\sin^2 t + \cos^2 t)\, dt = \int_0^{2\pi} 1\, dt = 2\pi \;\checkmark \]
Picture it
Animation
Shows: Stokes's Theorem with the two strategy lines revealed underneath it.
Deciding which side to compute is most of the work.
Takeaway: The theorem is an equation, so you may compute whichever side is cheaper. A simple boundary favours the line integral; a simple curl favours the surface integral.
Worked example
Find the circulation around the triangle where a plane meets the first octant, oriented to match an upward normal. Integrating along three edges directly would be tedious.
\[ \mathbf{F} = \langle y, z, x\rangle, \quad C = \partial S,\; S:\; x + y + z = 1 \text{ in the first octant} \]
Take the curl
Why: Each component of the curl works out to negative one.
\[ \nabla\times\mathbf{F} = \langle 0-1,\; 0-1,\; 0-1\rangle = \langle -1, -1, -1\rangle \]
Get the upward surface element of the plane
Why: Write the plane as height over the triangle; the upward cross product is the vector with the two negated slopes and a one.
\[ z = 1 - x - y, \quad \mathbf{r}_x \times \mathbf{r}_y = \langle 1, 1, 1\rangle \]
Dot the curl into the surface element
Why: The dot product of the curl with that vector is a constant.
\[ (\nabla\times\mathbf{F})\cdot(\mathbf{r}_x\times\mathbf{r}_y) = -1 - 1 - 1 = -3 \]
Integrate over the triangle
Why: The constant times the area of the projected triangle, whose legs are one.
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D (-3)\, dA = -3\cdot\tfrac{1}{2} = -\tfrac{3}{2} \]
Verify the sign makes sense
Why: The curl points opposite to the upward normal, so their dot product is negative, and a negative circulation is exactly what a curl pointing against the normal produces.
\[ (\nabla\times\mathbf{F})\cdot\mathbf{n} < 0 \;\Rightarrow\; \oint_C \mathbf{F}\cdot d\mathbf{r} < 0 \;\checkmark \]
Picture it
Animation
Shows: A dome-shaped surface rotating, sharing its circular boundary with a flat disk.
Swap the surface, keep the boundary.
Takeaway: The dome and the flat disk share a circle, and Stokes says each equals the same line integral around it. So they equal each other, and you may always swap an awkward surface for a friendlier one.
Two truths and a lie
Sort into buckets
Sort each statement by whether it actually holds.
Trap
Using the upward normal on the disk but traversing the boundary clockwise.
\[ \mathbf{F} = \langle -y, x, 0\rangle, \quad \mathbf{n} = \langle 0,0,1\rangle \]
The right-hand rule pairs an upward normal with a counterclockwise loop. Going clockwise flips the sign, so the circulation comes out negative when it should be positive.
\[ \oint_{C_{\text{wrong}}} \mathbf{F}\cdot d\mathbf{r} = -2\pi \]
The magnitude looks right, but the sign is wrong, and a wrong sign is a wrong answer.
Match the orientation by the right-hand rule: thumb up along the normal, fingers curl counterclockwise.
\[ \oint_{C_{\text{right}}} \mathbf{F}\cdot d\mathbf{r} = \iint_S 2\, dS = 2\pi \]
Now the boundary direction and the surface normal agree, and Stokes gives the correct positive value.
Trap
Asked for the work a field does around a closed loop, a student reaches for the Divergence Theorem because the letter F appears.
\[ W = \oint_C \mathbf{F}\cdot d\mathbf{r} \;\overset{?}{=}\; \iiint_E \nabla\cdot\mathbf{F}\, dV \]
This is the wrong tool. The Divergence Theorem needs a closed surface and produces a flux, not a circulation, and there is no solid region in sight, only a curve.
A boundary curve plus a circulation is the signature of Stokes. Use the curl and any capping surface.
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot\mathbf{n}\, dS \]
Read the object: a curve and circulation means Stokes; a closed surface and flux means Divergence.
Section
Connect it up
Draw it
Draw the family. Put the Fundamental Theorem of Calculus, Green's Theorem, Stokes's Theorem and the Divergence Theorem on the page, and connect them by what plays the role of the boundary in each.
Concept
The Fundamental Theorem of Line Integrals, Green's Theorem, Stokes's Theorem, and the Divergence Theorem are one family.
Each says: integrating a derivative over a region equals evaluating the original object on the boundary of that region. The dimensions rise, but the sentence is the same.
Picture it
Animation
Shows: The Fundamental Theorem, Green, Stokes and Divergence written one above the other.
Read down the left column, then down the right.
Takeaway: Boundary on the left, derivative over the interior on the right — four times, at four dimensions. Learn the sentence and the four theorems stop being four things.
Discrimination
Sort into buckets
For each setup, name the tool. Deciding this correctly is most of the exam.
Pattern
Read what is being integrated and over what
Why: The object (curve, surface, or solid) and the quantity (work, flux) pin down the theorem before any computation.
Curve plus work or circulation, in the plane
Why: Use Green's Theorem, or the Fundamental Theorem of Line Integrals if the field is a gradient.
Boundary curve plus circulation, in space
Why: Use Stokes's Theorem: convert to a curl flux over any capping surface.
Closed surface plus outward flux
Why: Use the Divergence Theorem: convert to a triple integral of the divergence.
| Object | Quantity | Theorem |
|---|---|---|
| Closed surface | Outward flux | Divergence |
| Boundary curve | Circulation | Stokes |
| Plane region boundary | Circulation or flux | Green |
| Two endpoints | Work of a gradient | FTLI |
Picture it
Animation
Shows: A three-line decision rule for choosing between the theorems.
Deciding correctly is most of the exam.
Takeaway: Ask what the boundary is. A closed surface means Divergence, a closed curve means Stokes or Green, and neither means parametrize and integrate directly.
Pattern
1. Decide the orientation
Why: Pick the required normal direction; for a closed surface it is outward. The sign of the whole answer rides on this.
2. Check for a shortcut
Why: If the surface is closed, the Divergence Theorem may replace the surface integral with an easier triple integral.
3. Parametrize if computing directly
Why: Find the tangent vectors and their cross product; for a graph use the negated-slopes vector.
4. Dot and integrate
Why: Integrate the field dotted into the cross product over the parameter domain, keeping the orientation's sign.
Explain it
Discussion prompt
Explain what flux measures to a student who has finished Calculus II but has never seen a vector field. No notation at all — only words and everyday images.
Hint: Wind through an open window is a good place to start.
Answer:
A serviceable version: imagine wind blowing through an open window. Flux is how much air passes through the window each second. Wind blowing straight through counts fully; wind sliding sideways past the opening counts for nothing; wind blowing back in counts negatively. Add all of that up across the whole window and that total is the flux. If you can say this without notation, you understand it.
Check
Set up the outward flux of the position field across the surface of the unit cube.
\[ \mathbf{F} = \langle x, y, z\rangle, \quad E:\; [0,1]^3 \]
Check your understanding
What is the outward flux across the whole surface of the cube?
Answer: A
Why: The divergence is one plus one plus one, which is three. The triple integral of the constant three over the unit cube of volume one gives three times one, so the flux is 3.
Check
The curve is the unit circle in the plane, oriented counterclockwise.
\[ \mathbf{F} = \langle -y, x, 3z\rangle, \quad C:\; x^2 + y^2 = 1,\; z = 0 \]
Check your understanding
Using Stokes's Theorem, what is the circulation around C?
Answer: A
Why: The z-component of the curl is the x-partial of x minus the y-partial of negative y, which is one plus one, so two. Integrating the constant two over the unit disk of area pi gives two pi.
Check
You must find the total outward flux of a field across the closed surface of a solid box, and the field's divergence is a simple constant.
Check your understanding
Which theorem is the efficient choice for this setup?
Answer: A
Why: A closed surface plus an outward flux, with a simple divergence, is exactly the Divergence Theorem's setup: it converts the flux into a triple integral of the divergence over the solid.
Check
You are computing a flux directly from a parametrization with tangent vectors in each parameter.
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\, dS = \iint_D \mathbf{F}\cdot(?)\, du\, dv \]
Check your understanding
What replaces the question mark in the flux formula?
Answer: A
Why: Flux needs a vector surface element that carries direction, and that is the cross product of the two tangent vectors, which points along the normal and scales by area.
Exit ticket
Predict first
Rate yourself honestly. Which of these would you least want to see on a test tomorrow?
Correct: Whichever you picked is what tonight's homework should start with.
Why: There is no wrong answer here. The value is in naming the weak spot while the material is still fresh, because a vague sense of unease turns into a specific practice problem only once you say out loud which part it attaches to.
Recap
You built up from the surface element and reached the two capstone theorems of the course.
Surface area, scalar surface integrals, and flux are all one machine: integrate something against the surface element.
The Divergence Theorem turns outward flux across a closed surface into a triple integral of the divergence.
Stokes's Theorem turns circulation around a boundary curve into a curl flux over any capping surface, and Green's Theorem is its flat special case.
The habits that keep you correct: confirm the surface is closed before invoking Divergence, orient the normal outward, match the boundary direction to the normal by the right-hand rule, and read the object and quantity to choose the theorem.
| Setup | Use | It becomes |
|---|---|---|
| Closed surface, outward flux | Divergence Theorem | Triple integral of divergence |
| Boundary curve, circulation | Stokes's Theorem | Curl flux over a capping surface |
| Plane region | Green's Theorem | Double integral |
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