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Calculus IV Unit 18 Review

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18.1 Line integrals of scalar fields

18.1 Line integrals of scalar fields

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Calculus IV
Unit & Topic Study Guides

Scalar Fields and Parametric Curves

Scalar Fields

A scalar field f(x,y,z)f(x, y, z) assigns a single real number to each point (x,y,z)(x, y, z) in some subset of R3\mathbb{R}^3. Think of temperature at every point in a room, or electric potential throughout a region of space.

You can visualize scalar fields using level curves (in 2D) or level surfaces (in 3D), which are the sets of points where the field takes a constant value.

Parametric Curves and Arc Length

A parametric curve r(t)=(x(t),y(t),z(t))\mathbf{r}(t) = (x(t), y(t), z(t)) traces out a path in space as the parameter tt varies over an interval [a,b][a, b]. Common examples:

  • A helix: (acost,  asint,  bt)(a\cos t,\; a\sin t,\; bt)
  • A circle in the xyxy-plane: (acost,  asint,  0)(a\cos t,\; a\sin t,\; 0)

The arc length of r(t)\mathbf{r}(t) on [a,b][a, b] measures the total distance traveled along the curve:

L=abr(t)dt=ab(x(t))2+(y(t))2+(z(t))2dtL = \int_a^b \|\mathbf{r}'(t)\|\, dt = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\, dt

The quantity r(t)dt\|\mathbf{r}'(t)\|\, dt is the infinitesimal arc length element dsds. This is exactly what shows up inside line integrals.

Scalar Fields, Quadric Surfaces · Calculus

Line Integrals and Their Properties

Definition and Evaluation

The line integral of a scalar field ff along a curve CC is written Cfds\int_C f\, ds. It accumulates the values of ff along the curve, weighted by arc length. If CC is parameterized by r(t)\mathbf{r}(t) for t[a,b]t \in [a, b], the evaluation formula is:

Cf(x,y,z)ds=abf(x(t),y(t),z(t))r(t)dt\int_C f(x, y, z)\, ds = \int_a^b f\bigl(x(t),\, y(t),\, z(t)\bigr)\, \|\mathbf{r}'(t)\|\, dt

Here's the step-by-step process:

  1. Parameterize the curve. Write r(t)=(x(t),y(t),z(t))\mathbf{r}(t) = (x(t), y(t), z(t)) on [a,b][a, b].
  2. Compute r(t)\mathbf{r}'(t) and then its magnitude r(t)\|\mathbf{r}'(t)\|.
  3. Substitute x(t),y(t),z(t)x(t), y(t), z(t) into ff to get f(r(t))f(\mathbf{r}(t)).
  4. Multiply f(r(t))f(\mathbf{r}(t)) by r(t)\|\mathbf{r}'(t)\| and integrate from aa to bb.

A classic application: if a wire has shape CC and density function ρ(x,y,z)\rho(x,y,z), its total mass is Cρds\int_C \rho\, ds.

Scalar Fields, Quadric Surfaces · Calculus

Properties of Scalar Line Integrals

A crucial property of scalar line integrals is independence of orientation. Unlike line integrals of vector fields, reversing the direction you traverse CC does not change the value of Cfds\int_C f\, ds. The scalar line integral depends on the geometric curve itself, not on which direction you walk along it.

Why? The arc length element dsds is always positive regardless of traversal direction, so the integral Cfds\int_C f\, ds stays the same.

The integral is also independent of parameterization: any smooth reparameterization of the same curve gives the same result, as long as the curve is traversed exactly once.

Applications and Theorems

Work and Vector Line Integrals

The section above covers scalar line integrals (integrating fdsf\, ds). There's a closely related but distinct object: the line integral of a vector field, which computes work. The work done by a force F(x,y,z)=(P,Q,R)\mathbf{F}(x,y,z) = (P, Q, R) along CC is:

W=CFdr=abF(r(t))r(t)dtW = \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt

This integral does depend on orientation: reversing the curve flips the sign of WW. That makes physical sense, since walking against a force versus with it should give opposite work values.

Examples include the work done by gravity on an object moving along a curved path, or the work done by an electric field on a charged particle.

Fundamental Theorem of Line Integrals

If F\mathbf{F} is a conservative vector field with potential function ϕ\phi (meaning F=ϕ\mathbf{F} = \nabla \phi), then:

CFdr=ϕ(r(b))ϕ(r(a))\int_C \mathbf{F} \cdot d\mathbf{r} = \phi(\mathbf{r}(b)) - \phi(\mathbf{r}(a))

This is the multivariable analogue of the Fundamental Theorem of Calculus. Two important consequences:

  • The integral depends only on the endpoints, not on the particular path CC connecting them.
  • Over any closed curve (where r(a)=r(b)\mathbf{r}(a) = \mathbf{r}(b)), the integral is zero.

Note that this theorem applies to vector line integrals (CFdr\int_C \mathbf{F} \cdot d\mathbf{r}), not directly to scalar line integrals (Cfds\int_C f\, ds). The original guide conflated these two types. Keep them straight: scalar line integrals use dsds and measure accumulated value along a curve; vector line integrals use drd\mathbf{r} and measure work-like quantities.