This deck covers vector fields in the plane and in space, gradient fields, divergence and curl, and line integrals of both scalar fields and vector fields, the latter giving work. It targets the classic errors: dropping the speed factor in a scalar line integral, mishandling orientation, and mis-parameterizing a curve.
Subject: Calculus III · 118 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Read and sketch a vector field in the plane and in space, and recognize a gradient field.
2. Compute the divergence and curl of a field and say what each one measures.
3. Parameterize a curve, give it an orientation, and set up the arc-length element correctly.
4. Evaluate a scalar line integral of a function over a curve.
5. Evaluate a vector line integral as the work done by a field along a path, in both dot-product and differential forms.
Warm-up
Discussion prompt
Before we open Week 13 - Vector Fields & Line Integrals: without looking back, what was the main idea of Week 12 - Cylindrical & Spherical Integrals, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers triple integrals in cylindrical and spherical coordinates: what each coordinate means, how to convert between systems, and the all-important volume factors r and rho-squared sin-phi. It targets the classic traps of dropping the volume factor, swapping phi and theta, and choosing the harder coordinate system.
Concept
A plain function sends each point to a number. A vector field sends each point to a whole arrow.
In the plane, a vector field assigns to each point a vector with two components:
\[ \mathbf{F}(x,y) = \langle P(x,y),\, Q(x,y)\rangle \]
vector field — A rule F that assigns a vector to every point of a region. P and Q (and R in space) are ordinary scalar functions called the component functions.
Counterexample
Discussion prompt
A plain function sends each point to a number. A vector field sends each point to a whole arrow.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
In the plane, a vector field assigns to each point a vector with two components:
Intuition
Every vector field is a picture of something flowing or pushing.
At each point of a weather map there is a wind arrow: a direction and a speed. That map is a vector field.
Other everyday fields: the current in a river, the gravitational pull near a planet, the force on a charge in an electric field. In each case, stand at a point and you feel one specific push.
Analogy
Discussion prompt
Explain Think wind, water, and force by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Every vector field is a picture of something flowing or pushing.
Concept
A vector field in space works the same way, with one more component:
\[ \mathbf{F}(x,y,z) = \langle P,\, Q,\, R\rangle = P\,\mathbf{i} + Q\,\mathbf{j} + R\,\mathbf{k} \]
Here P, Q, and R are each functions of x, y, and z. The two notations, angle brackets and the i, j, k basis, mean exactly the same thing.
Explain it
Discussion prompt
Explain Fields in space have three components to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A vector field in space works the same way, with one more component:
Concept
To draw a field, pick a grid of points. At each point, draw the field's vector starting from that point.
The length of each arrow shows the field's magnitude there; the direction of the arrow shows where it points. Long arrows mean a strong field.
Picture it
Figure (svg): Arrows arranged in a counterclockwise circulation pattern around the origin, tangent to circles centered at the origin.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Look at this field, whose arrows circulate around the origin:
Concept
Look at this field, whose arrows circulate around the origin:
\[ \mathbf{F}(x,y) = \langle -y,\, x\rangle \]
Figure (svg): Arrows arranged in a counterclockwise circulation pattern around the origin, tangent to circles centered at the origin.
At the point one unit right of the origin the field points straight up; one unit up it points left. The whole field spins counterclockwise. Keep this picture in mind for curl.
Picture it
Figure (svg): Arrows pointing radially outward from the origin, growing longer with distance from the center.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Now a field whose arrows point straight out from the origin:
Concept
Now a field whose arrows point straight out from the origin:
\[ \mathbf{F}(x,y) = \langle x,\, y\rangle \]
Figure (svg): Arrows pointing radially outward from the origin, growing longer with distance from the center.
The farther out you go, the longer the arrows: the field grows with distance. This outward-spreading picture is what divergence measures.
Concept
You already met one vector field in the differentiation unit: the gradient.
Given a scalar function f, its gradient assigns a vector to each point:
\[ \nabla f = \left\langle f_x,\, f_y,\, f_z \right\rangle \]
gradient field — A vector field F that equals the gradient of some scalar function f, so F = grad f. The function f is called a potential function for F.
Definition probe
Sort into buckets
Every line below is part of the definition of vector field or of gradient field — one or the other, never both. Put each where it belongs.
Intuition
Picture f as the height of a hill above each point of the ground.
The gradient field at each point is the arrow pointing in the direction of steepest climb, and it is always perpendicular to the level curves (the contour lines) of f.
So a gradient field is the arrow-map of the fastest way uphill everywhere. Not every field is a gradient field, but this is an important special family.
Ranking
Put in order
Put the moves of Worked example: build a gradient field into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The first component of the gradient is the partial derivative in x.
Worked example
Find the gradient field of the potential function:
\[ f(x,y) = x^2 y \]
Differentiate with respect to x, holding y constant
Why: The first component of the gradient is the partial derivative in x.
\[ f_x = 2xy \]
Differentiate with respect to y, holding x constant
Why: The second component is the partial derivative in y; here x squared is a constant multiplier.
\[ f_y = x^2 \]
Assemble the gradient vector field
Why: Stack the partials into a vector at each point.
\[ \nabla f = \langle 2xy,\, x^2\rangle \]
Verify at the point where x = 1 and y = 2
Why: Recompute each partial at the point and confirm the vector is consistent with the formula.
\[ \nabla f(1,2) = \langle 2(1)(2),\, 1^2\rangle = \langle 4,\, 1\rangle \]
Picture it
Animation
Shows: Each line of the worked example "build a gradient field", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Recompute each partial at the point and confirm the vector is consistent with the formula.
Concept
Divergence turns a vector field back into a number at each point: how much the field is spreading out there.
In space, for F with components P, Q, R:
\[ \operatorname{div}\mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \]
A handy shorthand uses the del operator dotted with the field:
\[ \operatorname{div}\mathbf{F} = \nabla \cdot \mathbf{F} \]
Intuition
Imagine the field as a fluid flow and put a tiny box around a point.
If more fluid leaves the box than enters, the point acts like a source and the divergence is positive. If more enters than leaves, it is a sink and divergence is negative.
The outward field of arrows we drew earlier has positive divergence everywhere: fluid is being pumped outward at every point.
Step zero
Discussion prompt
Worked example: compute a divergence — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Take the x-partial of the first component
Answer:
Worked example
Find the divergence of:
\[ \mathbf{F} = \langle x^2,\, y^2,\, z^2\rangle \]
Take the x-partial of the first component
Why: Divergence uses the derivative of P with respect to x only.
\[ \frac{\partial}{\partial x}(x^2) = 2x \]
Take the y-partial of the second component
Why: Match each component to its own variable.
\[ \frac{\partial}{\partial y}(y^2) = 2y \]
Take the z-partial of the third component
Why: The third derivative in the divergence sum.
\[ \frac{\partial}{\partial z}(z^2) = 2z \]
Add the three partials
Why: Divergence is the sum of these matched partial derivatives.
\[ \operatorname{div}\mathbf{F} = 2x + 2y + 2z \]
Verify at the origin
Why: At the origin all terms vanish, matching a field whose arrows shrink to zero there; the result is a scalar, as divergence should be.
\[ \operatorname{div}\mathbf{F}(0,0,0) = 0 \]
Picture it
Animation
Shows: Each line of the worked example "compute a divergence", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the origin all terms vanish, matching a field whose arrows shrink to zero there; the result is a scalar, as divergence should be.
Concept
Curl turns a vector field into another vector field, measuring the axis and strength of rotation at each point.
It is the del operator crossed with the field:
\[ \operatorname{curl}\mathbf{F} = \nabla \times \mathbf{F} \]
Written out in components:
\[ \nabla \times \mathbf{F} = \left\langle R_y - Q_z,\ P_z - R_x,\ Q_x - P_y \right\rangle \]
Intuition
Drop a tiny paddlewheel into the flow at a point.
If the field makes it spin, the field has curl there. The curl vector points along the wheel's axle, and its length tells how fast the wheel turns.
The circulating field we drew earlier spins a paddlewheel counterclockwise, so it has nonzero curl pointing out of the page.
Fill the middle
Fill in the blanks
From Worked example: compute a curl — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle xy,\, yz,\, zx\rangle = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Match names so the curl formula is easy to apply.
Worked example
Find the curl of:
\[ \mathbf{F} = \langle xy,\, yz,\, zx\rangle \]
Identify the components
Why: Match names so the curl formula is easy to apply.
\[ P = xy,\quad Q = yz,\quad R = zx \]
First component: R_y minus Q_z
Why: R has no y, so R_y is 0; Q_z is the z-partial of yz, which is y.
\[ R_y - Q_z = 0 - y = -y \]
Second component: P_z minus R_x
Why: P has no z, so P_z is 0; R_x is the x-partial of zx, which is z.
\[ P_z - R_x = 0 - z = -z \]
Third component: Q_x minus P_y
Why: Q has no x, so Q_x is 0; P_y is the y-partial of xy, which is x.
\[ Q_x - P_y = 0 - x = -x \]
Assemble the curl vector
Why: Combine the three components into one vector field.
\[ \operatorname{curl}\mathbf{F} = \langle -y,\, -z,\, -x\rangle \]
Verify with a spot check at (1,1,1)
Why: Plug the point into both the original partials and the final formula; both give the same vector, confirming the algebra.
\[ \operatorname{curl}\mathbf{F}(1,1,1) = \langle -1,\, -1,\, -1\rangle \]
Picture it
Animation
Shows: Each line of the worked example "compute a curl", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Plug the point into both the original partials and the final formula; both give the same vector, confirming the algebra.
Concept
Divergence and curl connect back to the gradient in two clean identities.
The curl of any gradient field is the zero vector: gradient fields never spin.
\[ \operatorname{curl}(\nabla f) = \mathbf{0} \]
The divergence of any curl is zero: a pure rotation has no source.
\[ \operatorname{div}(\operatorname{curl}\mathbf{F}) = 0 \]
Concept
So far a field lives everywhere. Now we walk a specific path through it and add up what happens along the way.
That sum is a line integral. There are two flavors: integrating a scalar function over a curve, and integrating a vector field over a curve. To do either, we first need a good way to describe the curve.
Concept
To parameterize a curve is to describe it as the tip of a moving position vector, traced out as a single parameter runs over an interval.
\[ \mathbf{r}(t) = \langle x(t),\, y(t)\rangle,\qquad a \le t \le b \]
parameterization — A vector function r(t) whose tip sweeps out the curve exactly once as t runs from a to b. The starting and ending values of t must be chosen so the tip lands exactly on the endpoints.
Estimation
Predict first
Parameterize the straight segment from the origin to the point (1, 2, 3).
Commit before you compute: what does Worked example: parameterize a segment come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify both endpoints
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Plug in the ends of the interval; they must land exactly on the two given points.
Worked example
Parameterize the straight segment from the origin to the point (1, 2, 3).
Start at the first point and add t times the displacement
Why: A segment from point A to point B is A plus t times the vector from A to B; at t equals 0 you sit at A.
\[ \mathbf{r}(t) = \langle 0,0,0\rangle + t\langle 1,2,3\rangle = \langle t,\, 2t,\, 3t\rangle \]
Choose the parameter range
Why: We need t equal to 1 to reach the far endpoint, so t runs from 0 to 1.
\[ 0 \le t \le 1 \]
Verify both endpoints
Why: Plug in the ends of the interval; they must land exactly on the two given points.
\[ \mathbf{r}(0) = \langle 0,0,0\rangle,\quad \mathbf{r}(1) = \langle 1,2,3\rangle \]
Picture it
Animation
Shows: Each line of the worked example "parameterize a segment", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Plug in the ends of the interval; they must land exactly on the two given points.
Missing information
Discussion prompt
Parameterize the circle of radius 2 centered at the origin, traced counterclockwise.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Points on a circle of radius 2 have coordinates (2 cos t, 2 sin t).
Worked example
Parameterize the circle of radius 2 centered at the origin, traced counterclockwise.
Use cosine and sine scaled by the radius
Why: Points on a circle of radius 2 have coordinates (2 cos t, 2 sin t).
\[ \mathbf{r}(t) = \langle 2\cos t,\, 2\sin t\rangle \]
Let t run once around
Why: A full counterclockwise loop needs t to sweep a full turn.
\[ 0 \le t \le 2\pi \]
Verify the radius stays 2
Why: The magnitude of r(t) should equal the radius for every t; the Pythagorean identity confirms it.
\[ \sqrt{(2\cos t)^2 + (2\sin t)^2} = \sqrt{4} = 2 \]
Picture it
Animation
Shows: Each line of the worked example "parameterize a circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The magnitude of r(t) should equal the radius for every t; the Pythagorean identity confirms it.
Concept
A curve is not just a set of points; a line integral also needs a direction of travel along it.
The parameterization sets the orientation: as t increases, the tip moves in the positive direction. Reversing the curve is written with a minus sign.
\[ -C : \text{ the same curve traced in the opposite direction} \]
Whether orientation matters to the answer depends on which kind of line integral you are computing. Hold that thought; it is a classic trap.
Concept
First flavor: integrate a scalar function f along a curve C.
\[ \int_C f(x,y)\, ds \]
The symbol ds stands for an infinitesimal piece of arc length along the curve, not a piece of the parameter t. Getting ds right is the whole game.
Intuition
Two pictures make the scalar line integral concrete.
Picture one: erect a curtain whose base is the curve C and whose height at each point is f. The integral is the area of that curtain.
Picture two: let f be the density of a bent wire shaped like C. The integral is the wire's total mass. Either way, you are summing f times little lengths of curve.
Concept
To turn ds into something you can integrate, convert it to the parameter t.
A small step in t moves the tip by the velocity vector times dt; the length of that little move is the speed times dt:
\[ ds = |\mathbf{r}'(t)|\, dt \]
So the scalar line integral becomes an ordinary single integral in t:
\[ \int_C f\, ds = \int_a^b f(\mathbf{r}(t))\, |\mathbf{r}'(t)|\, dt \]
The factor giving the magnitude of the derivative is the speed. Miss it and you are summing over t, not over length.
Step zero
Discussion prompt
Worked example: a scalar line integral — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Compute the velocity vector
Answer:
Worked example
Evaluate the integral of f along the segment C described below.
\[ f(x,y) = xy,\qquad \mathbf{r}(t) = \langle 3t,\, 4t\rangle,\ \ 0 \le t \le 1 \]
Compute the velocity vector
Why: Differentiate each component of r(t) with respect to t.
\[ \mathbf{r}'(t) = \langle 3,\, 4\rangle \]
Compute the speed
Why: The speed is the magnitude of the velocity; this is the factor multiplying dt.
\[ |\mathbf{r}'(t)| = \sqrt{3^2 + 4^2} = 5 \]
Write f along the curve
Why: Substitute x = 3t and y = 4t into f.
\[ f(\mathbf{r}(t)) = (3t)(4t) = 12t^2 \]
Assemble and integrate
Why: Multiply f along the curve by the speed, then integrate over the t-interval.
\[ \int_0^1 12t^2 \cdot 5\, dt = 60\int_0^1 t^2\, dt = 60 \cdot \tfrac{1}{3} \]
Verify the arithmetic
Why: Evaluate the antiderivative at the limits; the speed factor of 5 is what turns 4 into the correct 20.
\[ 60 \cdot \tfrac{1}{3} = 20 \]
Picture it
Animation
Shows: Each line of the worked example "a scalar line integral", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Evaluate the antiderivative at the limits; the speed factor of 5 is what turns 4 into the correct 20.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting shortcut: treat ds as if it were just dt and forget the speed factor.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Integrating over t instead of over arc length measures the wrong thing and gives the wrong number.
Always convert ds using the speed of the parameterization.
Why: Integrating over t instead of over arc length measures the wrong thing and gives the wrong number.
Trap
The tempting shortcut: treat ds as if it were just dt and forget the speed factor.
\[ \int_C xy\, ds \stackrel{?}{=} \int_0^1 12t^2\, dt \]
This drops the magnitude of the derivative
Why: Integrating over t instead of over arc length measures the wrong thing and gives the wrong number.
\[ \int_0^1 12t^2\, dt = 4 \quad (\text{wrong}) \]
Always convert ds using the speed of the parameterization.
\[ ds = |\mathbf{r}'(t)|\, dt = 5\, dt \]
Keep the speed factor of 5
Why: The segment covers 5 units of length per unit of t, so each slice of the integral is 5 times larger than the dt version.
\[ \int_0^1 12t^2 \cdot 5\, dt = 20 \quad (\text{right}) \]
Notation
Annotate
From Trap: integrating dt instead of ds — read this one piece at a time. What is each part doing?
On: \( \int_C xy\, ds \stackrel{?}{=} \int_0^1 12t^2\, dt \)
Missing information
Discussion prompt
Integrate f over the quarter of the unit circle in the first quadrant, going counterclockwise.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Differentiate the components; on the unit circle the speed is exactly 1.
Worked example
Integrate f over the quarter of the unit circle in the first quadrant, going counterclockwise.
\[ f(x,y) = xy,\qquad \mathbf{r}(t) = \langle \cos t,\, \sin t\rangle,\ \ 0 \le t \le \tfrac{\pi}{2} \]
Find the velocity and speed
Why: Differentiate the components; on the unit circle the speed is exactly 1.
\[ \mathbf{r}'(t) = \langle -\sin t,\, \cos t\rangle,\qquad |\mathbf{r}'(t)| = 1 \]
Write f along the curve
Why: Substitute x = cos t and y = sin t.
\[ f(\mathbf{r}(t)) = \cos t \sin t \]
Set up the integral
Why: Even though the speed is 1 here, we still write it in so the method is airtight.
\[ \int_0^{\pi/2} \cos t \sin t \cdot 1\, dt \]
Substitute u for sin t
Why: Let u equal sin t so du equals cos t dt, turning the integral into a simple power.
\[ \int_0^{1} u\, du = \tfrac{1}{2}u^2 \Big|_0^1 \]
Verify the value
Why: Evaluate the antiderivative at the new limits u equal 0 and u equal 1.
\[ \tfrac{1}{2}(1)^2 - \tfrac{1}{2}(0)^2 = \tfrac{1}{2} \]
Picture it
Animation
Shows: Each line of the worked example "scalar integral over a quarter circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Evaluate the antiderivative at the new limits u equal 0 and u equal 1.
Concept
Second flavor: integrate a vector field F along a curve C. Its meaning is work.
If F is a force field and you move a particle along C, the total work done is the line integral of F over C. Only the part of the force pointing along the path counts.
\[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \]
Intuition
Push a cart down a hallway. A sideways shove into the wall does no useful work; only the forward part of your push moves the cart.
That is exactly the dot product: at each point we keep only the component of F pointing along the direction of travel, then add those up over the whole path.
Concept
Writing the unit tangent vector as T, the along-path component of F is F dotted with T, and we integrate that over arc length:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (\mathbf{F} \cdot \mathbf{T})\, ds \]
In terms of the parameter, dr is the velocity times dt, so the practical formula is:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt \]
Notice: here the velocity vector itself appears, not its magnitude. The dot product already selects the along-path piece, so there is no separate speed factor to remember.
Concept
Expanding the dot product component by component gives the differential form:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C P\, dx + Q\, dy + R\, dz \]
To evaluate it, replace each differential using the parameterization:
\[ dx = x'(t)\, dt,\quad dy = y'(t)\, dt,\quad dz = z'(t)\, dt \]
This is the same integral as the dot-product form, just written differently. Use whichever matches how the problem is stated.
Worked example
Find the work done by the field along the curve below.
\[ \mathbf{F} = \langle y,\, x\rangle,\qquad \mathbf{r}(t) = \langle t,\, t^2\rangle,\ \ 0 \le t \le 1 \]
Compute the velocity vector
Why: Differentiate each component of r(t); this is the dr piece.
\[ \mathbf{r}'(t) = \langle 1,\, 2t\rangle \]
Write F along the curve
Why: Substitute x = t and y = t squared into the field.
\[ \mathbf{F}(\mathbf{r}(t)) = \langle t^2,\, t\rangle \]
Take the dot product
Why: Dot the field with the velocity to keep only the along-path component.
\[ \langle t^2,\, t\rangle \cdot \langle 1,\, 2t\rangle = t^2 + 2t^2 = 3t^2 \]
Integrate over the interval
Why: Integrate the scalar 3 t squared from 0 to 1.
\[ \int_0^1 3t^2\, dt = t^3 \Big|_0^1 = 1 \]
Verify a different way
Why: This field is the gradient of xy, so the work should equal xy at the end minus xy at the start; that matches, confirming the answer.
\[ xy\big|_{(1,1)} - xy\big|_{(0,0)} = 1 - 0 = 1 \]
Picture it
Animation
Shows: Each line of the worked example "work along a parabola", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: This field is the gradient of xy, so the work should equal xy at the end minus xy at the start; that matches, confirming the answer.
Ranking
Put in order
Put the moves of Worked example: the P dx + Q dy form into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. From x = t and y = t squared, dx is dt and dy is 2t dt.
Worked example
Evaluate the line integral written in differential form along the parabola below.
\[ \int_C (x+y)\, dx + (x-y)\, dy,\qquad \mathbf{r}(t) = \langle t,\, t^2\rangle,\ \ 0 \le t \le 1 \]
Convert the differentials
Why: From x = t and y = t squared, dx is dt and dy is 2t dt.
\[ dx = dt,\qquad dy = 2t\, dt \]
Substitute for x and y in the coefficients
Why: Replace x with t and y with t squared everywhere.
\[ x+y = t + t^2,\qquad x-y = t - t^2 \]
Combine into a single t-integral
Why: Multiply each coefficient by its differential and add.
\[ \int_0^1 \big[(t+t^2) + (t-t^2)(2t)\big]\, dt \]
Expand the integrand
Why: Distribute the 2t and collect like terms before integrating.
\[ \int_0^1 \big(t + 3t^2 - 2t^3\big)\, dt \]
Verify by evaluating the antiderivative
Why: Integrate term by term and plug in the limits 0 and 1.
\[ \left[\tfrac{t^2}{2} + t^3 - \tfrac{t^4}{2}\right]_0^1 = \tfrac{1}{2} + 1 - \tfrac{1}{2} = 1 \]
Picture it
Animation
Shows: Each line of the worked example "the P dx + Q dy form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Integrate term by term and plug in the limits 0 and 1.
Hypothesis
Predict first
Worked example: work in space is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Compute the velocity vector
Why: Differentiate each component of the segment's parameterization.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Find the work done by the radial field moving a particle along the segment from the origin to (1, 2, 3).
\[ \mathbf{F} = \langle x,\, y,\, z\rangle,\qquad \mathbf{r}(t) = \langle t,\, 2t,\, 3t\rangle,\ \ 0 \le t \le 1 \]
Compute the velocity vector
Why: Differentiate each component of the segment's parameterization.
\[ \mathbf{r}'(t) = \langle 1,\, 2,\, 3\rangle \]
Write F along the curve
Why: Substitute x = t, y = 2t, z = 3t into the field.
\[ \mathbf{F}(\mathbf{r}(t)) = \langle t,\, 2t,\, 3t\rangle \]
Dot and simplify
Why: Multiply matching components and add.
\[ \langle t,2t,3t\rangle \cdot \langle 1,2,3\rangle = t + 4t + 9t = 14t \]
Integrate over the interval
Why: Integrate 14t from 0 to 1.
\[ \int_0^1 14t\, dt = 7t^2 \Big|_0^1 = 7 \]
Verify with the potential
Why: The field is the gradient of half of x squared plus y squared plus z squared; evaluating that potential at the endpoints gives the same 7.
\[ \tfrac{1}{2}(1^2+2^2+3^2) - 0 = \tfrac{14}{2} = 7 \]
Picture it
Animation
Shows: Each line of the worked example "work in space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The field is the gradient of half of x squared plus y squared plus z squared; evaluating that potential at the endpoints gives the same 7.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Assuming that reversing the direction of travel flips the sign of every line integral, scalar ones included.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Arc length ds is always positive no matter which way you walk, so the scalar integral does not change sign.
Reversing orientation flips the sign of a vector line integral only.
Why: Arc length ds is always positive no matter which way you walk, so the scalar integral does not change sign.
Trap
Assuming that reversing the direction of travel flips the sign of every line integral, scalar ones included.
\[ \int_{-C} f\, ds \stackrel{?}{=} -\int_{C} f\, ds \]
This is false for scalar integrals
Why: Arc length ds is always positive no matter which way you walk, so the scalar integral does not change sign.
\[ \int_{-C} f\, ds = \int_{C} f\, ds \quad (\text{unchanged}) \]
Reversing orientation flips the sign of a vector line integral only.
\[ \int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_{C} \mathbf{F} \cdot d\mathbf{r} \]
The tangent direction reverses, but arc length does not
Why: In the vector integral the tangent vector T points the opposite way, so the dot product changes sign; in the scalar integral there is no direction to flip.
\[ \mathbf{T} \to -\mathbf{T},\qquad ds \to ds \]
Translation
\( \mathbf{T} \to -\mathbf{T},\qquad ds \to ds \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Parameterizing the segment from the origin to (2, 4) but letting t run too far.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This parameterization reaches (4, 8), not (2, 4), so the integral is taken over the wrong curve.
Choose the range so the endpoints land exactly right.
Why: This parameterization reaches (4, 8), not (2, 4), so the integral is taken over the wrong curve.
Trap
Parameterizing the segment from the origin to (2, 4) but letting t run too far.
\[ \mathbf{r}(t) = \langle 2t,\, 4t\rangle,\quad 0 \le t \le 2 \ \ (\text{wrong}) \]
At t equal to 2 you overshoot the endpoint
Why: This parameterization reaches (4, 8), not (2, 4), so the integral is taken over the wrong curve.
\[ \mathbf{r}(2) = \langle 4,\, 8\rangle \neq \langle 2,\, 4\rangle \]
Choose the range so the endpoints land exactly right.
\[ \mathbf{r}(t) = \langle 2t,\, 4t\rangle,\quad 0 \le t \le 1 \ \ (\text{right}) \]
Always check the ends before integrating
Why: At t equal to 1 the tip lands on (2, 4), so the parameterization covers exactly the intended segment.
\[ \mathbf{r}(0) = \langle 0,0\rangle,\qquad \mathbf{r}(1) = \langle 2,\, 4\rangle \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
In the plane the formulas shrink. Divergence keeps only the first two partials.
\[ \operatorname{div}\mathbf{F} = P_x + Q_y \]
Curl in the plane returns a single number, the out-of-plane component:
\[ (\operatorname{curl}\mathbf{F})_z = Q_x - P_y \]
A positive scalar curl means the field spins counterclockwise there; a negative one means clockwise.
Estimation
Predict first
Find the divergence of the planar field below.
Commit before you compute: what does Worked example: divergence in the plane come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify at the point (1, 2)
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Substitute the point to confirm a single scalar value, as divergence must produce.
Worked example
Find the divergence of the planar field below.
\[ \mathbf{F} = \langle x^2 y,\, x y^2\rangle \]
Take the x-partial of the first component
Why: Differentiate x squared y with respect to x, treating y as constant.
\[ \frac{\partial}{\partial x}(x^2 y) = 2xy \]
Take the y-partial of the second component
Why: Differentiate x y squared with respect to y, treating x as constant.
\[ \frac{\partial}{\partial y}(x y^2) = 2xy \]
Add the two partials
Why: Divergence in the plane is the sum of these two.
\[ \operatorname{div}\mathbf{F} = 2xy + 2xy = 4xy \]
Verify at the point (1, 2)
Why: Substitute the point to confirm a single scalar value, as divergence must produce.
\[ \operatorname{div}\mathbf{F}(1,2) = 4(1)(2) = 8 \]
Picture it
Animation
Shows: Each line of the worked example "divergence in the plane", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substitute the point to confirm a single scalar value, as divergence must produce.
Step zero
Discussion prompt
Worked example: curl of the rotational field — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Identify P and Q
Answer:
Worked example
Compute the scalar curl of the circulating field we sketched earlier.
\[ \mathbf{F} = \langle -y,\, x\rangle \]
Identify P and Q
Why: Read off the components before applying the plane curl formula.
\[ P = -y,\qquad Q = x \]
Compute Q_x minus P_y
Why: The x-partial of x is 1; the y-partial of negative y is negative 1.
\[ (\operatorname{curl}\mathbf{F})_z = 1 - (-1) = 2 \]
Verify against the picture
Why: The scalar curl is a positive 2, matching the counterclockwise spin we drew; sign and picture agree.
\[ (\operatorname{curl}\mathbf{F})_z = 2 > 0 \]
Picture it
Animation
Shows: Each line of the worked example "curl of the rotational field", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The scalar curl is a positive 2, matching the counterclockwise spin we drew; sign and picture agree.
Fill the middle
Fill in the blanks
From Worked example: another gradient field — finish the line. Write what belongs on the right of the equals sign before you look.
\nabla f = \langle 2x,\, 2y\rangle
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Each partial gives one component of the gradient.
Worked example
Find the gradient field of this potential.
\[ f(x,y) = x^2 + y^2 \]
Differentiate in x and in y
Why: Each partial gives one component of the gradient.
\[ f_x = 2x,\qquad f_y = 2y \]
Assemble the field
Why: This is a radial field: every arrow points straight away from the origin.
\[ \nabla f = \langle 2x,\, 2y\rangle \]
Verify at (1, 1)
Why: The vector there points outward along the diagonal, exactly perpendicular to the circular level curve through the point.
\[ \nabla f(1,1) = \langle 2,\, 2\rangle \]
Picture it
Animation
Shows: Each line of the worked example "another gradient field", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The vector there points outward along the diagonal, exactly perpendicular to the circular level curve through the point.
Intuition
The level curves of the last potential are circles centered at the origin.
Its gradient points radially outward, straight across each circle. That is the geometry to remember: a gradient field is always perpendicular to the level curves and points toward higher values.
Concept
A vector field is only defined on the region where all its component functions make sense.
We call a field continuous on a region when each component function is continuous there. Line integrals are set up over curves that stay inside this region.
Concept
A curve that starts and ends at the same point is closed. The line integral around a closed curve gets a special symbol.
\[ \oint_C \mathbf{F} \cdot d\mathbf{r} \]
circulation — The line integral of a vector field around a closed curve. It measures the net tendency of the field to push a particle all the way around the loop.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of vector field, gradient field, circulation as Week 13 - Vector Fields & Line Integrals uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Imagine floating a leaf around a closed loop in a stream.
If the current tends to carry it forward the whole way around, the circulation is positive. If the pushes cancel out over the loop, the circulation is zero. This is what a closed vector line integral measures.
Concept
The direction of travel at each point is captured by the unit tangent vector.
\[ \mathbf{T} = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \]
It is the velocity vector scaled to length 1. This is the T that appears in the dot-product form of the work integral.
Worked example
Find the unit tangent vector for the segment below.
\[ \mathbf{r}(t) = \langle 3t,\, 4t\rangle \]
Differentiate to get the velocity
Why: The tangent direction comes from the derivative.
\[ \mathbf{r}'(t) = \langle 3,\, 4\rangle \]
Divide by the speed
Why: Scaling by the magnitude, which is 5, makes the vector length 1.
\[ \mathbf{T} = \tfrac{1}{5}\langle 3,\, 4\rangle = \left\langle \tfrac{3}{5},\, \tfrac{4}{5}\right\rangle \]
Verify the length is 1
Why: Check the magnitude of T; it must equal 1 for a genuine unit vector.
\[ \sqrt{\left(\tfrac{3}{5}\right)^2 + \left(\tfrac{4}{5}\right)^2} = \sqrt{\tfrac{25}{25}} = 1 \]
Picture it
Animation
Shows: Each line of the worked example "find the unit tangent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check the magnitude of T; it must equal 1 for a genuine unit vector.
Concept
The value of a scalar line integral depends only on the curve and the function, not on how you happened to parameterize it.
Walk the same curve faster or slower and the speed factor and the dt-interval adjust to compensate, leaving the answer the same. That is why the arc-length element ds, not dt, is the natural thing to integrate.
Explain it
Discussion prompt
Explain The scalar integral ignores the parameterization to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The value of a scalar line integral depends only on the curve and the function, not on how you happened to parameterize it.
Concept
If a path is built from several smooth pieces joined end to end, integrate over each piece and add the results.
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{C_1} \mathbf{F} \cdot d\mathbf{r} + \int_{C_2} \mathbf{F} \cdot d\mathbf{r} + \cdots \]
The same splitting works for scalar line integrals. Parameterize each piece separately and keep each orientation consistent with the overall direction of travel.
Analogy
Discussion prompt
Explain Curves made of pieces by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
If a path is built from several smooth pieces joined end to end, integrate over each piece and add the results.
Estimation
Predict first
A wire lies along the segment below with linear density equal to y. Find its mass.
Commit before you compute: what does Worked example: mass of a wire come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the value
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Evaluate the antiderivative; the speed factor of 5 is again essential to the result.
Worked example
A wire lies along the segment below with linear density equal to y. Find its mass.
\[ \mathbf{r}(t) = \langle 3t,\, 4t\rangle,\ \ 0 \le t \le 1,\qquad \rho(x,y) = y \]
Recognize mass as a scalar line integral
Why: Total mass is the integral of density over arc length along the wire.
\[ m = \int_C \rho\, ds \]
Find the speed
Why: The velocity is the vector 3 and 4, whose magnitude is 5, the ds-to-dt factor.
\[ |\mathbf{r}'(t)| = 5 \]
Write density along the wire and integrate
Why: Density y equals 4t on the curve; multiply by the speed and integrate.
\[ m = \int_0^1 (4t)(5)\, dt = 20\int_0^1 t\, dt \]
Verify the value
Why: Evaluate the antiderivative; the speed factor of 5 is again essential to the result.
\[ 20 \cdot \tfrac{1}{2} = 10 \]
Picture it
Animation
Shows: Each line of the worked example "mass of a wire", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Evaluate the antiderivative; the speed factor of 5 is again essential to the result.
Step zero
Discussion prompt
Worked example: work in differential form in space — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Convert every differential
Answer:
Worked example
Evaluate the space line integral along the segment where x, y, and z all equal t.
\[ \int_C z\, dx + x\, dy + y\, dz,\qquad \mathbf{r}(t) = \langle t,\, t,\, t\rangle,\ \ 0 \le t \le 1 \]
Convert every differential
Why: Since each coordinate equals t, each differential is dt.
\[ dx = dy = dz = dt \]
Substitute the coordinates
Why: On this curve x, y, and z each equal t, so every coefficient becomes t.
\[ z\, dx + x\, dy + y\, dz = t\, dt + t\, dt + t\, dt = 3t\, dt \]
Integrate over the interval
Why: Integrate 3t from 0 to 1.
\[ \int_0^1 3t\, dt = \tfrac{3}{2}t^2 \Big|_0^1 \]
Verify the value
Why: Evaluate at the limits to finish.
\[ \tfrac{3}{2}(1)^2 - 0 = \tfrac{3}{2} \]
Picture it
Animation
Shows: Each line of the worked example "work in differential form in space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: On this curve x, y, and z each equal t, so every coefficient becomes t.
Concept
Divergence is a single number at each point, and its sign tells a physical story.
| Sign of divergence | Meaning at that point |
|---|---|
| Positive | source: net outflow, fluid created |
| Negative | sink: net inflow, fluid drained |
| Zero | incompressible: as much in as out |
Comparison
Comparison matrix
From Reading the sign of divergence: refill the Meaning at that point column from what you know. The rest of the table is as it appeared.
| Sign of divergence | Meaning at that point |
|---|---|
| Positive | source: net outflow, fluid created |
| Negative | sink: net inflow, fluid drained |
| Zero | incompressible: as much in as out |
Concept
In space the curl is a vector, and its direction is set by the right-hand rule.
Curl the fingers of your right hand in the direction the field circulates; your thumb points along the curl vector. The length of that vector measures how fast the local rotation is.
Counterexample
Discussion prompt
In space the curl is a vector, and its direction is set by the right-hand rule.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Curl the fingers of your right hand in the direction the field circulates; your thumb points along the curl vector. The length of that vector measures how fast the local rotation is.
Ranking
Put in order
Put the moves of Worked example: curl of a gradient is zero into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Use the components of the gradient field.
Worked example
Take the gradient field of the first potential and test its curl.
\[ f(x,y) = x^2 y,\qquad \nabla f = \langle 2xy,\, x^2\rangle \]
Identify P and Q from the gradient
Why: Use the components of the gradient field.
\[ P = 2xy,\qquad Q = x^2 \]
Compute Q_x minus P_y
Why: The x-partial of x squared is 2x; the y-partial of 2xy is 2x.
\[ (\operatorname{curl}\nabla f)_z = 2x - 2x = 0 \]
Verify the identity holds
Why: The scalar curl is zero, confirming the general fact that gradient fields never spin.
\[ \operatorname{curl}(\nabla f) = 0 \]
Picture it
Animation
Shows: Each line of the worked example "curl of a gradient is zero", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The scalar curl is zero, confirming the general fact that gradient fields never spin.
Pattern
1. Parameterize C as r(t) on an interval from a to b
Why: Check the endpoints so the range is exactly right.
2. Compute the velocity and its magnitude, the speed
Why: The speed is the factor that converts ds into dt; never drop it.
3. Substitute the parameterization into f
Why: Write f as a function of t alone before integrating.
4. Integrate f along the curve times the speed over t
Why: This is the whole scalar line integral in one clean single integral.
\[ \int_a^b f(\mathbf{r}(t))\, |\mathbf{r}'(t)|\, dt \]
Pattern
1. Parameterize C as r(t) with the correct orientation
Why: Direction matters here: increasing t must match the way you want to travel.
2. Compute the velocity vector r'(t)
Why: This is dr; you use the whole vector, not its magnitude.
3. Substitute the parameterization into F
Why: Write the field along the curve as a function of t.
4. Dot F with the velocity, then integrate over t
Why: The dot product selects the along-path force; integrate it to get the work.
\[ \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt \]
Fill the middle
Fill in the blanks
From Recipe: divergence and curl — finish the line. Write what belongs on the right of the equals sign before you look.
\operatorname\langle R_y - Q_z,\ P_z - R_x,\ Q_x - P_y\rangle\mathbf___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Each component gets differentiated by its own variable; the result is a scalar.
Pattern
Divergence: sum the matched partials
Why: Each component gets differentiated by its own variable; the result is a scalar.
\[ \operatorname{div}\mathbf{F} = P_x + Q_y + R_z \]
Curl: apply the cross-product pattern
Why: Each component is a difference of two cross partials; the result is a vector.
\[ \operatorname{curl}\mathbf{F} = \langle R_y - Q_z,\ P_z - R_x,\ Q_x - P_y\rangle \]
Quick sanity check: divergence gives a number, curl gives a vector. If you produced the wrong kind of object, you used the wrong operation.
Notation
Annotate
From Recipe: divergence and curl — read this one piece at a time. What is each part doing?
On: \( \operatorname{curl}\mathbf{F} = \langle R_y - Q_z,\ P_z - R_x,\ Q_x - P_y\rangle \)
Elimination
Eliminate the wrong options
What is the value of this scalar line integral?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The speed is the magnitude of the derivative, which is the square root of 3 squared plus 4 squared, equal to 5. With x equal to 3t, the integral is 3t times 5 dt from 0 to 1, giving 15 times one-half, which is 15/2.
Check
Set up carefully and remember the speed factor.
\[ \int_C x\, ds,\qquad \mathbf{r}(t) = \langle 3t,\, 4t\rangle,\ \ 0 \le t \le 1 \]
Check your understanding
What is the value of this scalar line integral?
Answer: A
Why: The speed is the magnitude of the derivative, which is the square root of 3 squared plus 4 squared, equal to 5. With x equal to 3t, the integral is 3t times 5 dt from 0 to 1, giving 15 times one-half, which is 15/2.
Check
Parameterize, dot with the velocity, and integrate.
\[ \mathbf{F} = \langle y,\, 2x\rangle,\qquad \mathbf{r}(t) = \langle t,\, t^2\rangle,\ \ 0 \le t \le 1 \]
Check your understanding
What is the work done by F along this curve?
Answer: A
Why: The velocity is the vector 1 and 2t. Along the curve F is t squared and 2t. The dot product is t squared plus 4 t squared, which is 5 t squared, and its integral from 0 to 1 is 5/3.
Check
Differentiate each component by its own variable, then add.
\[ \mathbf{F} = \langle xy,\, yz,\, xz\rangle,\qquad \text{evaluate at } (1,2,3) \]
Check your understanding
What is the divergence of F at the point (1, 2, 3)?
Answer: A
Why: The divergence is the x-partial of xy plus the y-partial of yz plus the z-partial of xz, which is y plus z plus x. At (1, 2, 3) that is 2 plus 3 plus 1, equal to 6.
Check
Suppose you already found the work along a curve C.
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = 7 \]
Check your understanding
What is the value of the same integral taken along the reversed curve, negative C?
Answer: A
Why: Reversing orientation flips the tangent direction, so the dot product changes sign and the vector line integral becomes its negative. The value is negative 7.
Elimination
Eliminate the wrong options
What is the value of the scalar line integral along the reversed curve, negative C?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Arc length ds is positive regardless of the direction of travel, so a scalar line integral does not change when the curve is reversed. The value stays 12.
Check
Now the scalar flavor over the same reversed curve.
\[ \int_C f\, ds = 12 \]
Check your understanding
What is the value of the scalar line integral along the reversed curve, negative C?
Answer: A
Why: Arc length ds is positive regardless of the direction of travel, so a scalar line integral does not change when the curve is reversed. The value stays 12.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: scalar line integral · Recipe: vector line integral (work) · Recipe: divergence and curl · A vector field attaches a vector to every point · Think wind, water, and force. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A vector field attaches an arrow to every point. Divergence measures net outflow and returns a number; curl measures rotation and returns a vector.
A scalar line integral sums a function over arc length; the key is to convert ds to the speed times dt and never drop that factor.
A vector line integral is the work done by a field along a path: dot the field with the velocity and integrate. Reversing orientation flips its sign, but leaves a scalar line integral unchanged.
| Quantity | Output | Key formula |
|---|---|---|
| Divergence | scalar | P_x + Q_y + R_z |
| Curl | vector | del cross F |
| Scalar integral | number | integral of f times speed dt |
| Work integral | number | integral of F dot velocity dt |
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