Line integrals

Setting up the object. Let $C$ be a curve given by a parametrization $\vec r(t) = (x(t), y(t), z(t))$ for $t \in [a,b]$, and let $\vec F(x,y,z) = F_1\hat\imath + F_2\hat\jmath + F_3\hat k$ be a vector field defined at points of $C$. The line integral of $\vec F$ along $C$ is

where: $\vec r\,'(t) = (x'(t), y'(t), z'(t))$ is the tangent vector to $C$ (its direction is the direction of travel), and the dot product $\vec F\cdot \vec r\,'$ picks out only the part of $\vec F$ pointing along that tangent at each instant. The same integral is often written in the equivalent differential form $\int_C F_1\,dx + F_2\,dy + F_3\,dz$, which is just the dot-product form with $d\vec r = (dx,dy,dz)$ spelled out.

Why the dot product. A vector field can point in any direction relative to the curve — along it, against it, or sideways to it. Only the component along the curve contributes to the running total; a field perpendicular to the path at some instant contributes nothing there. The dot product $\vec F\cdot \vec r\,'$ is exactly the algebraic device that extracts this "along the path" component, instant by instant, before the ordinary integral adds them all up.

Work interpretation. If $\vec F$ is a force field (units: newton, and lengths in metres), then $\displaystyle\int_C \vec F\cdot d\vec r$ is the work (units: joule) that the force does on a particle moving along $C$ from its start point to its end point. This is by far the most common physical reading of a line integral in this course, and almost every exam problem is dressed up as "find the work done."

Orientation matters. A curve traversed in one direction and the same curve traversed in the opposite direction give line integrals that are negatives of each other, because reversing direction flips the sign of the tangent vector $\vec r\,'(t)$. Always check that the chosen parametrization starts and ends where the problem says it should.

Piecewise curves. If $C = C_1 \cup C_2 \cup \cdots$ is made of several joined pieces, the line integral over $C$ is simply the sum of the line integrals over each piece, each evaluated with its own parametrization.

Problem. Evaluate $\displaystyle\int_C \vec F\cdot d\vec r$ where $\vec F(x,y) = (x^2-y)\,\hat\imath + (x+y^2)\,\hat\jmath$ (force in newtons, $x,y$ in metres) and $C$ is the straight-line segment from $(0,0)$ to $(1,2)$.

Step 1. Parametrize $C$. A straight segment from $(0,0)$ to $(1,2)$ is given by $x = t$, $y = 2t$, $t \in [0,1]$: at $t=0$ this is $(0,0)$, and at $t=1$ it is $(1,2)$, matching the required start and end points.

Step 2. Differentiate: $x'(t) = 1$, $y'(t) = 2$, so $\vec r\,'(t) = (1,2)$.

Step 3. Substitute into $\vec F$: $F_1 = x^2 - y = t^2 - 2t$ and $F_2 = x + y^2 = t + 4t^2$. Form the dot product,

The $-2t$ and $+2t$ cancel, a useful consistency check that the algebra is on track.

Step 4. Integrate: $\displaystyle\int_0^1 9t^2\,dt = 9\left[\frac{t^3}{3}\right]_0^1 = 3$. So $\displaystyle\int_C \vec F\cdot d\vec r = \mathbf{3\ J}$: the force does $3$ joules of work moving the particle from $(0,0)$ to $(1,2)$ along the straight line.

Competitive stretch

Now change only the path: let $C'$ be the two-piece path from $(0,0)$ straight to $(1,0)$, then straight up to $(1,2)$, using the same $\vec F$. On $C_1$: $x=t,\,y=0,\,t\in[0,1]$, so $\vec r\,'=(1,0)$, $F_1=t^2$, $F_2=t$, and the integrand is $t^2(1)+t(0)=t^2$, giving $\int_0^1 t^2\,dt = \tfrac13$. On $C_2$: $x=1,\,y=t,\,t\in[0,2]$, so $\vec r\,'=(0,1)$, $F_1=1-t$, $F_2=1+t^2$, and the integrand is $(1-t)(0)+(1+t^2)(1)=1+t^2$, giving $\int_0^2(1+t^2)\,dt = 2+\tfrac83=\tfrac{14}{3}$. Total: $\tfrac13+\tfrac{14}{3}=5$ J — different from the straight-line value of $3$ J for the same endpoints, confirming this $\vec F$ is path-dependent (not conservative), the exact question the next topic in the syllabus takes up systematically.

• Forgetting orientation: reversing the direction of travel along the same curve flips the sign of $\int_C \vec F\cdot d\vec r$, since $\vec r\,'(t)$ reverses. Always check the parametrization matches the stated start and end points, or stated direction around a closed curve.
• Confusing this with the scalar line integral $\int_C f\,ds$ (which uses the arc-length differential $ds = |\vec r\,'(t)|\,dt$) and using $ds$ where the problem actually wants the vector differential $d\vec r$ — these are two different integrals with two different setups.
• Choosing a parametrization for a straight segment that does not actually land on both endpoints at the stated parameter values — always verify $t=a$ and $t=b$ give the correct start and end points before integrating.
• Dropping the dot product and integrating $|\vec F|$ or the field's components separately — the line integral of a vector field intrinsically depends on the angle between $\vec F$ and the tangent direction, not just the field's magnitude.

Ask the tutor: Want to work through a line integral over a helix or an ellipse, or check whether a given field is conservative, with the AI tutor?