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Line Integrals

Integrate a scalar function or a vector field along a curve, via parameterization.

The idea. Every integral so far added up a function over a straight domain — an interval, a flat region, a solid. A line integral adds up a function along a curve. Instead of integrating over [a, b] on the x-axis, you integrate along a path that bends through space. The trick that makes it work: parameterize the curve, which collapses the whole thing back to an ordinary single-variable integral in t.

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Parameterize the curve C as x = x(t), y = y(t) for t in [a, b]. The arc-length element is:

ds=(dxdt)2+(dydt)2 dtds = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\, dt

Arc-length element

  • CCThe curve (path) you integrate along.
  • r(t)=⟨x(t),y(t)⟩\mathbf{r}(t) = \langle x(t), y(t) \rangleParameterization of the curve.
  • dsdsArc-length element — a little bit of distance traveled along the curve.

What ds is: a little bit of arc length — the distance traveled along the curve for a little step dt. That square root is speed: how fast the point moves along the path as t ticks. Notice it's the same square-root shape as the surface-area stretch factor — both measure how much a step gets stretched.

The integral becomes a plain integral in t:

∫Cf(x,y) ds=∫abf(x(t),y(t)) (dxdt)2+(dydt)2 dt\int_C f(x,y)\, ds = \int_a^b f\bigl(x(t), y(t)\bigr)\, \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\, dt

Line integral of a scalar function (with respect to arc length)

Line integral of a vector field — the physics-heavy one — work. Here F(x, y) = ⟨P(x, y), Q(x, y)⟩ is a vector field, and you're adding up the component of F along the direction of travel.

With r(t) = ⟨x(t), y(t)⟩:

∫CF⋅dr=∫abF(r(t))⋅r′(t) dt=∫ab[P x′(t)+Q y′(t)]dt\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}\bigl(\mathbf{r}(t)\bigr) \cdot \mathbf{r}'(t)\, dt = \int_a^b \left[P\, x'(t) + Q\, y'(t)\right] dt

Line integral of a vector field (work)

  • r′(t)\mathbf{r}'(t)Velocity vector — tangent to the curve, points in the direction of travel.
  • F=⟨P,Q⟩\mathbf{F} = \langle P, Q \rangleA vector field.
  • ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r}Work done by F along C; depends on the direction (orientation) of travel.

What it means physically: if F is a force, this is the work done moving along C. The dot product picks out how much of the force points the way you're going — force perpendicular to motion does no work, force along it does the most. This is the central object of the chapter.

The one big difference from every prior integral: orientation matters. A line integral of a vector field depends on which direction you traverse the curve. Reverse the path and the integral flips sign. (The scalar ds version does not flip — arc length is arc length regardless of direction. Only the vector/work version cares about orientation.)

∫−CF⋅dr=−∫CF⋅dr\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}

Orientation

Worked example

∫CF⋅dr,F=⟨y, x⟩,r(t)=⟨t, t2⟩,  t∈[0,1]\int_C \mathbf{F} \cdot d\mathbf{r}, \quad \mathbf{F} = \langle y,\ x \rangle, \quad \mathbf{r}(t) = \langle t,\ t^2 \rangle,\; t \in [0,1]
  1. x=t,y=t2x = t, \quad y = t^2

    Parameterize — already given.

  2. r′(t)=⟨1, 2t⟩\mathbf{r}'(t) = \langle 1,\ 2t \rangle

    Find r'(t): differentiate each component.

  3. F(r(t))=⟨t2, t⟩\mathbf{F}(\mathbf{r}(t)) = \langle t^2,\ t \rangle

    Evaluate F along the curve: substitute x = t, y = t^2 into F = ⟨y, x⟩.

  4. F⋅r′=⟨t2,t⟩⋅⟨1,2t⟩=t2(1)+t(2t)=t2+2t2=3t2\mathbf{F} \cdot \mathbf{r}' = \langle t^2, t \rangle \cdot \langle 1, 2t \rangle = t^2(1) + t(2t) = t^2 + 2t^2 = 3t^2

    Take the dot product.

  5. ∫013t2 dt=t3∣01=1\int_0^1 3t^2\, dt = t^3\Big|_0^1 = 1

    Integrate over t. The parameterization turned a curve integral into an ordinary Calc 1 integral — that collapse is the whole method.

Answer:

11

How to solve one of these

  1. 1.

    Parameterize the curve. Write x(t)x(t), y(t)y(t) (and z(t)z(t) in 3D) for tt over some interval. If not given, you construct it — a line segment, a circle (x=cos⁡tx = \cos t, y=sin⁡ty = \sin t), etc.

  2. 2.

    Identify the type. Scalar ∫Cf ds\int_C f\, ds needs the arc-length dsds. Vector ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} needs r′(t)\mathbf{r}'(t) and a dot product.

  3. 3.

    Scalar: compute ds and integrate. Compute ds=x′(t)2+y′(t)2 dtds = \sqrt{x'(t)^2 + y'(t)^2}\, dt, substitute the curve into ff, integrate.

  4. 4.

    Vector: compute r'(t), evaluate F, and dot. Compute r′(t)\mathbf{r}'(t), evaluate F\mathbf{F} along the curve, take F⋅r′\mathbf{F} \cdot \mathbf{r}', integrate.

  5. 5.

    Integrate over t. It's now a single-variable integral. Substitute the tt-bounds.

  6. 6.

    Mind the orientation. For vector line integrals, confirm you've parameterized in the direction the problem specifies — reversing flips the sign.

∫CF⋅dr=∫ab[P x′(t)+Q y′(t)]dt\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \left[P\, x'(t) + Q\, y'(t)\right] dt

Take the component of F along the direction of travel at each point, and add it all up along the curve.

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