Foundations to FrontierPart VI: Calculus IIIBezhig

Module 42: Vector fields, scalar line integrals, work, and circulation

8 exercises, each with a complete solution. 1 figure.

On this page
  1. A. Prerequisites and observable goals
  2. B. The central idea: integrate along a path, not across an interval of space
  3. C. Two line integrals with different orientation behavior
  4. D. Why work uses a dot product
  5. E. A parameterization procedure
  6. F. Worked examples
  7. G. Independent practice
  8. H. Complete solutions: Module 42
  9. I. Retention and readiness

A. Prerequisites and observable goals#

Before

Vector curves, dot products, arc length, and single-variable integration.

After

Distinguish scalar fields from vector fields, choose between  ds\dd s and  dr\dd\vv r, compute work or wire mass, and track orientation.

B. The central idea: integrate along a path, not across an interval of space#

A scalar field assigns one number to each point, such as temperature or density. A vector field assigns a vector, such as force or fluid velocity. A path samples the field along a curve. Scalar line integrals weight length; vector line integrals weight directed displacement. They answer different questions even when both involve the same curve.

C. Two line integrals with different orientation behavior#

For a piecewise smooth parameterized curve r(t)\vv r(t), a≤t≤ba\le t\le b,

∫Cf ds=∫abf(r(t))∥r′(t)∥ dt,\int_C f\dd s=\int_a^b f(\vv r(t))\|\vv r'(t)\|\dd t,

while

∫CF⋅ dr=∫abF(r(t))⋅r′(t) dt.\int_C\vv F\cdot\dd\vv r=\int_a^b\vv F(\vv r(t))\cdot\vv r'(t)\dd t.

The first is unchanged by reversing a once-traced curve; the second changes sign. A nonnegative linear density makes the first a wire mass. A force field makes the second work, because only the force component along displacement contributes. Notation P dx+Q dy+R dzP\dd x+Q\dd y+R\dd z denotes the same vector work integral for F=⟨P,Q,R⟩\vv F=\langle P,Q,R\rangle. [O3Line]

For a closed path, the work-type integral is also called circulation in a velocity-field context. In a counterclockwise oriented simple closed plane curve, the outward normal is to the right of travel. The planar outward flux is

∮CF⋅n ds=∮CP dy−Q dx.\oint_C\vv F\cdot\vv n\dd s=\oint_C P\dd y-Q\dd x.

This measures crossing of the boundary, not motion along it. The next module relates circulation and flux to area integrals.

D. Why work uses a dot product#

A small displacement Δr\Delta\vv r receives work approximately F⋅Δr\vv F\cdot\Delta\vv r: a force perpendicular to the motion does no work, while a force opposed to motion contributes negatively. Summing and limiting gives the integral. Wire mass instead uses positive segment lengths, so it contains ∥r′∥\|\vv r'\|, not a signed coordinate derivative.

A rotating field F=⟨−y,x⟩\vv F=\langle-y,x\rangle is tangent to circles centered at the origin: it has circulation but zero normal flux through those circles. A radial field G=⟨x,y⟩\vv G=\langle x,y\rangle has positive outward flux but zero circular work. This distinction previews curl versus divergence.

E. A parameterization procedure#

State the curve and orientation, choose a parameter interval tracing it as intended, and differentiate the parameterization. Substitute every coordinate in the field. Then choose the final factor: speed for  ds\dd s, tangent vector for  dr\dd\vv r, or an oriented normal for flux. Compute and check units and signs. Repeated tracing counts repeatedly, so a geometric picture alone does not fix the integral without its parameter information.

F. Worked examples#

Work along a segment. For F=⟨y,x⟩\vv F=\langle y,x\rangle from (0,0)(0,0) to (2,1)(2,1), use r(t)=⟨2t,t⟩\vv r(t)=\langle2t,t\rangle, 0≤t≤10\le t\le1. Then F(r)=⟨t,2t⟩\vv F(\vv r)=\langle t,2t\rangle and r′=⟨2,1⟩\vv r'=\langle2,1\rangle, giving work ∫01(2t+2t) dt=2\int_0^1(2t+2t)\dd t=2.

Flux versus circulation. On the counterclockwise unit circle, r=(cos⁡t,sin⁡t)\vv r=(\cos t,\sin t), r′=(−sin⁡t,cos⁡t)\vv r'=(-\sin t,\cos t), and outward normal n=(cos⁡t,sin⁡t)\vv n=(\cos t,\sin t). For the radial field G=r\vv G=\vv r, the work integrand is zero while flux integrand is one. Total circulation is zero; outward flux is 2π2\pi.

Exact sampled rotating field F=(−y,x)\vv F=(-y,x) with a counterclockwise unit circle. Arrows show tangent flow, not outward flow. Sampling visualizes the field; the line integral is computed from its formula.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M42.01FSolution

Evaluate ∫C(x+y) ds\int_C(x+y)\dd s along the segment from (0,0)(0,0) to (3,4)(3,4).

M42.02CSolution

Evaluate ∫C(x+y) ds\int_C(x+y)\dd s along the first-quadrant unit-circle arc from (1,0)(1,0) to (0,1)(0,1).

M42.03CSolution

Find work by F=⟨2x,3y⟩\vv F=\langle2x,3y\rangle along the segment from (0,0)(0,0) to (1,2)(1,2).

M42.04TSolution

For F=⟨−y,x⟩\vv F=\langle-y,x\rangle, find circulation counterclockwise around the circle of radius two and its outward flux through that circle.

M42.05TSolution

For F=⟨y,0⟩\vv F=\langle y,0\rangle, compare work from (0,0)(0,0) to (1,1)(1,1) along the straight segment and along the broken path first to (1,0)(1,0) and then to (1,1)(1,1).

M42.06CSolution

Evaluate ∫C⟨x,y,z⟩⋅ dr\int_C\langle x,y,z\rangle\cdot\dd\vv r for r(t)=⟨cos⁡t,sin⁡t,t⟩\vv r(t)=\langle\cos t,\sin t,t\rangle, 0≤t≤π0\le t\le\pi.

M42.07TSolution

A wire is r(t)=⟨t,t2⟩\vv r(t)=\langle t,t^2\rangle, 0≤t≤10\le t\le1, with linear density equal to its xx coordinate in consistent units. Find its mass.

M42.08SSolution

Explain by a substitution why a one-to-one smooth orientation-preserving reparameterization leaves both line-integral values unchanged, while orientation reversal changes only the work integral’s sign.

Optional small hints

For M42.01, the segment speed on [0,1][0,1] is five. For M42.05, work is ∫y dx\int y\dd x; a vertical segment has  dx=0\dd x=0. For M42.08, contrast the factor ϕ′\phi' with ∣ϕ′∣|\phi'|.

H. Complete solutions: Module 42#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M42.01Show solution

A scalar field weights positive arc length

Use r(t)=(3t,4t)\vv r(t)=(3t,4t), 0≤t≤10\le t\le1. Then x+y=7tx+y=7t and speed is 55. Hence ∫C(x+y) ds=∫0135t dt=35/2\int_C(x+y)\dd s=\int_0^135t\dd t=\boxed{35/2}. Reversing the segment would not change this scalar line integral.

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M42.02Show solution

Unit-circle speed equals one for its angle parameter

Use (x,y)=(cos⁡t,sin⁡t)(x,y)=(\cos t,\sin t) for 0≤t≤π/20\le t\le\pi/2. Speed is one, so the integral is ∫0π/2(cos⁡t+sin⁡t) dt=[sin⁡t−cos⁡t]0π/2=2\int_0^{\pi/2}(\cos t+\sin t)\dd t=[\sin t-\cos t]_0^{\pi/2}=\boxed2. The parameter interval, not a full circle, matches the specified arc.

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M42.03Show solution

Substitute the field and dot with velocity

With r=(t,2t)\vv r=(t,2t), F(r)=(2t,6t)\vv F(\vv r)=(2t,6t) and r′=(1,2)\vv r'=(1,2). The work integrand is 2t+12t=14t2t+12t=14t, so work is ∫0114t dt=7\int_0^114t\dd t=\boxed7. Integrating the field’s magnitude instead would calculate a different quantity.

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M42.04Show solution

A tangential field circulates without crossing the circle

Parameterize r=(2cos⁡t,2sin⁡t)\vv r=(2\cos t,2\sin t), 0≤t≤2π0\le t\le2\pi. Then F=(−2sin⁡t,2cos⁡t)=r′\vv F=(-2\sin t,2\cos t)=\vv r', so F⋅r′=4\vv F\cdot\vv r'=4 and circulation is 8π\boxed{8\pi}. The outward unit normal is (cos⁡t,sin⁡t)(\cos t,\sin t) and its dot product with the field is zero, so outward flux is 0\boxed0.

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M42.05Show solution

Two paths reveal path dependence

Along (t,t)(t,t), work is ∫01t dt=1/2\int_0^1t\dd t=\boxed{1/2}. On the first broken segment, y=0y=0, so y dx=0y\dd x=0; on the second, xx is constant, so  dx=0\dd x=0. The broken-path work is 0\boxed0. Because the same endpoints give different values, the field’s work is path-dependent; an endpoint-only rule cannot apply to this field globally.

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M42.06Show solution

Orthogonal horizontal motion cancels in the dot product

The substituted field is (cos⁡t,sin⁡t,t)(\cos t,\sin t,t) and velocity is (−sin⁡t,cos⁡t,1)(-\sin t,\cos t,1). Their dot product is −cos⁡tsin⁡t+sin⁡tcos⁡t+t=t-\cos t\sin t+\sin t\cos t+t=t. Thus the work is ∫0πt dt=π2/2\int_0^\pi t\dd t=\boxed{\pi^2/2}. The horizontal circular motion contributes no radial-field work; the vertical motion does.

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M42.07Show solution

Density must be multiplied by speed

Density along the path is t≥0t\ge0, and speed is 1+4t2\sqrt{1+4t^2}. Thus mass is

m=∫01t1+4t2 dt=[(1+4t2)3/212]01=55−112.m=\int_0^1t\sqrt{1+4t^2}\dd t =\left[\frac{(1+4t^2)^{3/2}}{12}\right]_0^1 =\boxed{\frac{5\sqrt5-1}{12}}.

The substitution u=1+4t2u=1+4t^2 contributes the factor  du=8t dt\dd u=8t\dd t. Positivity of the density and length factor makes the mass positive.

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M42.08Show solution

The chain rule explains reparameterization invariance

Let t=ϕ(u)t=\phi(u) be a smooth one-to-one monotone reparameterization. The new velocity is r′(ϕ(u))ϕ′(u)\vv r'(\phi(u))\phi'(u). For work, the integrand acquires the signed factor ϕ′\phi', exactly the ordinary substitution differential; reversing endpoint order reverses the integral. For a scalar line integral, the speed acquires ∣ϕ′∣|\phi'|, so orientation reversal changes the bounds and the sign of ϕ′\phi' together, leaving the positive length weighting unchanged. This argument assumes the same curve is traced once; a multiple-covering map is not the same reparameterization and can multiply the integral.

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I. Retention and readiness#

Retrieve M31.05 and M34.02. Say whether the problem is asking for density times length, force along motion, circulation, or flux across a boundary. Advance when you can predict orientation behavior before performing the integral.