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08 Line Integrals

Build a practical understanding of line integrals by moving from parametrized curves and arc length to scalar and vector integrals, work, conservative fields, and path independence.

Curves, parametrization, and arc length

A describes a path by assigning a point to each parameter value. In two dimensions or three dimensions, write

r(t)=⟨x(t),y(t)⟩orr(t)=⟨x(t),y(t),z(t)⟩,a≤t≤b.\mathbf r(t)=\langle x(t),y(t)\rangle \quad\text{or}\quad \mathbf r(t)=\langle x(t),y(t),z(t)\rangle, \qquad a\le t\le b.

The parameter determines both position and orientation. The tangent vector is r′(t)\mathbf r'(t), and the differential of arc length is

ds=∥r′(t)∥ dt.ds=\lVert\mathbf r'(t)\rVert\,dt.

Therefore, the length of a smooth curve is

L=∫Cds=∫ab∥r′(t)∥ dt.L=\int_C ds=\int_a^b\lVert\mathbf r'(t)\rVert\,dt.

For a line segment from p\mathbf p to q\mathbf q, a convenient parametrization is

r(t)=(1−t)p+tq,0≤t≤1.\mathbf r(t)=(1-t)\mathbf p+t\mathbf q, \qquad 0\le t\le 1.

For example, the segment from (1,2)(1,2) to (4,6)(4,6) has r(t)=⟨1+3t,2+4t⟩\mathbf r(t)=\langle 1+3t,2+4t\rangle. Since r′(t)=⟨3,4⟩\mathbf r'(t)=\langle 3,4\rangle, its length is

∫0132+42 dt=5.\int_0^1\sqrt{3^2+4^2}\,dt=5.

Takeaway: Parametrization converts geometric information about a curve into functions that can be integrated with respect to one variable.

Accumulating a scalar field along a curve

A weights a scalar field by the amount of path length. For a scalar field ff and a curve parametrized by r(t)\mathbf r(t),

∫Cf ds=∫abf(r(t))∥r′(t)∥ dt.\int_C f\,ds=\int_a^b f\bigl(\mathbf r(t)\bigr)\lVert\mathbf r'(t)\rVert\,dt.

The factor ∥r′(t)∥ dt\lVert\mathbf r'(t)\rVert\,dt is always nonnegative, so reversing the direction of traversal does not change the value when the curve is traversed once.

A physical interpretation is the mass of a wire. If the wire follows CC and has linear density ρ(x,y,z)\rho(x,y,z), then

m=∫Cρ ds.m=\int_C\rho\,ds.

If the density is constant, then m=ρLm=\rho L.

For the circle x2+y2=4x^2+y^2=4, use r(t)=⟨2cos⁡t,2sin⁡t⟩\mathbf r(t)=\langle 2\cos t,2\sin t\rangle for 0≤t≤2π0\le t\le 2\pi. Here ∥r′(t)∥=2\lVert\mathbf r'(t)\rVert=2, and the field f(x,y)=x2+y2f(x,y)=x^2+y^2 has value 44 on the circle. Thus,

∫Cf ds=∫02π4⋅2 dt=16π.\int_C f\,ds=\int_0^{2\pi}4\cdot 2\,dt=16\pi.

Takeaway: For a , substitute the parametrization into the field and multiply by the speed ∥r′(t)∥\lVert\mathbf r'(t)\rVert.

Vector fields and work along paths

A measures how a vector field acts along an oriented path. For a vector field F\mathbf F, use

∫CF⋅dr=∫abF(r(t))⋅r′(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.

In two dimensions, if F=⟨P,Q⟩\mathbf F=\langle P,Q\rangle, the same integral can be written as

∫CP dx+Q dy.\int_C P\,dx+Q\,dy.

The dot product selects the component of the field in the direction of motion. Reversing the orientation changes the sign:

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

For example, let F(x,y)=⟨y,x⟩\mathbf F(x,y)=\langle y,x\rangle, and let the path from (0,0)(0,0) to (2,1)(2,1) be r(t)=⟨2t,t⟩\mathbf r(t)=\langle 2t,t\rangle, with 0≤t≤10\le t\le 1. Then r′(t)=⟨2,1⟩\mathbf r'(t)=\langle 2,1\rangle and

F(r(t))=⟨t,2t⟩.\mathbf F\bigl(\mathbf r(t)\bigr)=\langle t,2t\rangle.

Consequently,

∫CF⋅dr=∫01⟨t,2t⟩⋅⟨2,1⟩ dt=∫014t dt=2.\int_C\mathbf F\cdot d\mathbf r =\int_0^1\langle t,2t\rangle\cdot\langle 2,1\rangle\,dt =\int_0^1 4t\,dt =2.

Takeaway: Orientation matters for vector line integrals because the integral records the field's directional effect along the path.

The endpoint method for gradient fields

The applies when a vector field is the gradient of a scalar function. If F=∇f\mathbf F=\nabla f, then a curve from r(a)\mathbf r(a) to r(b)\mathbf r(b) satisfies

∫CF⋅dr=∫C∇f⋅dr=f(r(b))−f(r(a)).\int_C\mathbf F\cdot d\mathbf r =\int_C\nabla f\cdot d\mathbf r =f\bigl(\mathbf r(b)\bigr)-f\bigl(\mathbf r(a)\bigr).

The chain rule explains the result. Along r(t)\mathbf r(t),

ddtf(r(t))=∇f(r(t))⋅r′(t).\frac{d}{dt}f\bigl(\mathbf r(t)\bigr) =\nabla f\bigl(\mathbf r(t)\bigr)\cdot\mathbf r'(t).

Integrating this derivative gives the endpoint difference.

For F(x,y)=⟨2x,4y⟩\mathbf F(x,y)=\langle 2x,4y\rangle, choose f(x,y)=x2+2y2f(x,y)=x^2+2y^2, since ∇f=F\nabla f=\mathbf F. Any curve from (0,0)(0,0) to (2,2)(2,2) therefore gives

∫CF⋅dr=f(2,2)−f(0,0)=(22+2⋅22)−0=12.\int_C\mathbf F\cdot d\mathbf r =f(2,2)-f(0,0) =\bigl(2^2+2\cdot 2^2\bigr)-0 =12.

No parametrization of the particular path is needed.

Takeaway: Before performing a direct , check whether the field has a .

Conservative fields and an efficient workflow

A has a , so its vector line integrals are : they depend only on the starting and ending points. Every closed curve has zero integral,

∮CF⋅dr=0.\oint_C\mathbf F\cdot d\mathbf r=0.

In the plane, for F=⟨P,Q⟩\mathbf F=\langle P,Q\rangle with continuous first partial derivatives, a standard test is

∂P∂y=∂Q∂x.\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}.

On a simply connected region, this condition is also sufficient. The domain is important: a field can satisfy the derivative equality away from a missing point or curve and still fail to be globally conservative.

For direct evaluation, use this sequence:

  1. Choose a parametrization r(t)\mathbf r(t) and interval [a,b][a,b].

  2. Compute r′(t)\mathbf r'(t).

  3. Substitute r(t)\mathbf r(t) into the scalar or vector field.

  4. Use ∫Cf ds=∫abf(r(t))∥r′(t)∥ dt\int_C f\,ds=\int_a^b f(\mathbf r(t))\lVert\mathbf r'(t)\rVert\,dt for a scalar field.

  5. Use ∫CF⋅dr=∫abF(r(t))⋅r′(t) dt\int_C\mathbf F\cdot d\mathbf r=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt for a vector field.

  6. Check orientation.

  7. Look for a before integrating directly.

Takeaway: Recognizing a conservative field can replace a path calculation with a simple endpoint calculation, but the domain must support the conclusion.