Work done by F = (y, x) around the unit circle

∫C F·dr where F = (y, x, 0) along r(t) = (cos(t), sin(t), 0) equals 0 ≈ 0. A worked vector field line integral with an editable calculator.

∫C F·dr where F = (y, x, 0) = 0 ≈ 0. This field is conservative (it is the gradient of xy), so the work around any closed loop is zero.

Vector field F = (P, Q, R)

Each component in terms of x, y and z

Curve C: r(t) = (x(t), y(t), z(t))

Leave z(t) as 0 for a plane curve

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The setup

∫C F·dr where F = (y, x, 0) = 0 ≈ 0

  • Curve: r(t) = (cos(t), sin(t), 0) for t from 0 to 2*pi
  • Arc length of the curve: 6.283185
  • Type: vector field line integral

This field is conservative (it is the gradient of xy), so the work around any closed loop is zero.

More worked line integrals

Circulation of F = (−y, x) around a circle of radius 3 Green's theorem predicts 2 × area = 2π(3²) = 18π; the calculator confirms it. Arc length of a helix Setting f = 1 turns the scalar line integral into arc length: 2π√2 for one turn. ∫(x + y) ds along a line segment For a straight parametrisation the speed |r′(t)| is constant, so it factors out of the integral. ∫x² ds along a parabola Here |r′(t)| = √(1 + 4t²) varies along the curve, which is why the answer is not a round number. Work done by F = (x, y, z) along a straight path A radial field along a straight line: the work is simply the change in ½|r|². Circumference of a circle as a line integral A radius-5 circle has circumference 10π — a good sanity check on the parametrisation. Work against a constant force along a curve A constant downward force does work that depends only on the height change, not the path taken.

Frequently asked questions

What is the value of this line integral? ▾

∫C F·dr where F = (y, x, 0) = 0 ≈ 0. This field is conservative (it is the gradient of xy), so the work around any closed loop is zero.

What curve is being integrated along? ▾

The curve is parametrised as r(t) = (cos(t), sin(t), 0) for t from 0 to 2*pi, and it has arc length 6.283185.

Does the direction of travel matter? ▾

Yes. Reversing the orientation of the curve flips the sign of a vector line integral, because the direction of dr reverses.

About this Calculator

∫C F·dr where F = (y, x, 0) along r(t) = (cos(t), sin(t), 0) equals 0 ≈ 0. A worked vector field line integral with an editable calculator.

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