Line Integrals as Integrals of Vectors

From Work, By Way of Riemann

We've already met line integrals as $\int_C P\,dx + Q\,dy$ in §5.2 and as $\int_C f\,ds$ in §5.3. Now we want a third reading - one that shows the integrand is a vector dot product, and explains why the previous two forms are really the same animal in different clothes.

Start with the physical setup. A particle moves along a curve $C$ in the plane or in space, parametrized by $\mathbf{r}(t) = (x(t), y(t), z(t))$ for $t \in [a,b]$. A vector field $\mathbf{u}(\mathbf{r})$ acts on the particle. We want the total work done by $\mathbf{u}$.

For a constant force on a straight segment, work is $\mathbf{u} \cdot \Delta\mathbf{r}$. The trouble is that $\mathbf{u}$ varies along $C$, and $C$ itself isn't straight. So we zoom in. Cut $C$ into $n$ small pieces. On each piece - if it's small enough - the path is nearly straight (replace it by the displacement vector $\Delta\mathbf{r}_k$) and $\mathbf{u}$ is nearly constant (evaluate it at one point $\mathbf{r}_k$ on the piece). The work on that piece is approximately

Sum, then shrink:

That's the vector form. It's nothing new conceptually - it's just the work-from-Riemann-sums story we already used in §5.1, written compactly with a dot product so the structure is visible.

Computing the Integral - Parametric Reduction

The definition is geometric, but to get a number we parametrize. With $\mathbf{r}(t) = (x(t), y(t), z(t))$, the differential displacement is

So the line integral collapses to an ordinary one-variable integral:

This is the workhorse formula. Plug in the parametrization, dot the field with the velocity vector, integrate over the parameter.

Worked example. Take $\mathbf{u}(x,y) = (-y, x)$ - the rotation field - and let $C$ be the unit circle traced counterclockwise: $\mathbf{r}(t) = (\cos t, \sin t)$, $t \in [0, 2\pi]$.

Then $\mathbf{u}(\mathbf{r}(t)) = (-\sin t, \cos t)$ and $\mathbf{r}'(t) = (-\sin t, \cos t)$. The dot product is $\sin^2 t + \cos^2 t = 1$, and

$\displaystyle \oint_C \mathbf{u}\cdot d\mathbf{r} \;=\; \int_0^{2\pi} 1\,dt \;=\; 2\pi.$

The field is everywhere tangent to the circle and aiding the motion - so the work integrates to a positive number. Reverse direction (use $-t$ as parameter) and we'd get $-2\pi$. Run halfway and stop and we'd get $\pi$. Direction and length both matter.

Component Form - The Same Integral, Three Notations

Write $\mathbf{u} = L\,\mathbf{i} + M\,\mathbf{j} + N\,\mathbf{k}$ and $d\mathbf{r} = dx\,\mathbf{i} + dy\,\mathbf{j} + dz\,\mathbf{k}$. The dot product expands:

That's the differential-form notation Kaplan uses throughout §5.2 and §5.5 (Green's). It's not a new integral - it's the same vector line integral with the dot product written out. So when we see $\int_C P\,dx + Q\,dy$ in a 2-D problem, we can mentally replace it with $\int_C \mathbf{u}\cdot d\mathbf{r}$ where $\mathbf{u} = (P, Q)$, or vice versa.

And the §5.3 connection. Let $\mathbf{T}$ be the unit tangent vector along $C$. Then $d\mathbf{r} = \mathbf{T}\,ds$, and

Same number, three readings: $\int \mathbf{u}\cdot d\mathbf{r}$ (vector), $\int L\,dx + M\,dy + N\,dz$ (differential forms), $\int u_T\,ds$ (scalar tangential component along arc length). Pick whichever is computational easiest - usually the parametric reduction in card 2.

Properties That Pay Off

Two structural facts about $\int_C \mathbf{u}\cdot d\mathbf{r}$ keep showing up. Worth memorizing.

One more, conceptually different. Suppose $\mathbf{u} = \nabla f$ for some scalar function $f$ - i.e. the field is conservative. Then

The path drops out entirely - only endpoints matter. This is the line-integral analogue of the Fundamental Theorem of Calculus, and Kaplan develops it fully in §5.6 (Independence of Path). The lesson here: before computing, check if $\mathbf{u}$ is a gradient. If yes, the work is trivial.

Final Review - When You See It, What to Reach For

§5.4 problems usually look like one of these. The right move is almost always the parametric reduction; the question is whether to dodge it via a property first.

Traps to Watch For

The signed nature of $\int_C \mathbf{u}\cdot d\mathbf{r}$ is the source of most mistakes. Reading the next six items is cheap; rediscovering them on the exam isn't.

1. Forgetting that direction matters. Reversing $C$ flips the sign. If we parametrize with $t$ going the wrong way (e.g. circle clockwise when the problem says counterclockwise), our answer is the negative of the right one. Always check the orientation against the problem's statement.
2. Mixing up $d\mathbf{r}$ and $ds$. $d\mathbf{r} = \mathbf{r}'(t)\,dt$ is a vector and integrates with a dot product. $ds = |\mathbf{r}'(t)|\,dt$ is a scalar (the arc length element) and integrates against a scalar function $f$. Using $ds$ where $d\mathbf{r}$ is needed kills all directional information.
3. Pulling a constant vector $\mathbf{u}$ outside. Yes, we can: $\int_C \mathbf{c}\cdot d\mathbf{r} = \mathbf{c}\cdot \int_C d\mathbf{r} = \mathbf{c}\cdot (\mathbf{r}(b) - \mathbf{r}(a))$. So a constant force on any path depends only on endpoints. This is the simplest case of "conservative" - and a common quick-check.
4. Computing the dot product wrong. Component-wise: $(L, M, N)\cdot(dx, dy, dz) = L\,dx + M\,dy + N\,dz$. Each pairing matches indices. Writing $L\,dy$ or $M\,dx$ is a standard slip on this topic.
5. Skipping the gradient check. Before grinding through a parametric integral, test whether curl is zero. If yes, $f(B) - f(A)$ might be a 30-second computation instead of a five-minute one. We'll always pay for not checking.
6. Confusing line integrals with definite integrals. $\int_a^b f(t)\,dt$ runs over a real interval; $\int_C \mathbf{u}\cdot d\mathbf{r}$ runs over a curve in space. The formula $\int_a^b \mathbf{u}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\,dt$ converts one to the other - the parameter interval $[a,b]$ is not the curve, just a label for points on the curve.