Circumference of a circle as a line integral
∫C (1) ds along r(t) = (5*cos(t), 5*sin(t), 0) equals 10π ≈ 31.41592654. A worked scalar field line integral with an editable calculator.
∫C (1) ds = 10π ≈ 31.41592654. A radius-5 circle has circumference 10π — a good sanity check on the parametrisation.
Example: x+y, x^2*y, sqrt(x^2+y^2). Use f = 1 to get arc length.
Curve C: r(t) = (x(t), y(t), z(t))
Leave z(t) as 0 for a plane curve
The setup
∫C (1) ds = 10π ≈ 31.41592654
- Curve: r(t) = (5*cos(t), 5*sin(t), 0) for t from 0 to 2*pi
- Arc length of the curve: 31.415927
- Type: scalar field line integral
A radius-5 circle has circumference 10π — a good sanity check on the parametrisation.
More worked line integrals
Frequently asked questions
What is the value of this line integral? ▾
∫C (1) ds = 10π ≈ 31.41592654. A radius-5 circle has circumference 10π — a good sanity check on the parametrisation.
What curve is being integrated along? ▾
The curve is parametrised as r(t) = (5*cos(t), 5*sin(t), 0) for t from 0 to 2*pi, and it has arc length 31.415927.
Does the direction of travel matter here? ▾
No. A scalar line integral is taken with respect to arc length, so travelling the curve backwards gives exactly the same value.
About this Calculator
∫C (1) ds along r(t) = (5*cos(t), 5*sin(t), 0) equals 10π ≈ 31.41592654. A worked scalar field line integral with an editable calculator.