What is the line integral along C?
Choose a vector line integral (∫ F · dr) or a scalar one (∫ f ds), type the field or function and the curve r(t), and the range of t. The page gives the value, exact when the algebra finds it.
- Line integral
- 0.9357142857
The line integral along C from t = 0 to t = 1 is 0.9357142857.
- Exact value
- 131/140
- Integrand in t
- 2t^6 + 4t^9 + t^3
Line integral: 0.9357142857. The line integral along C from t = 0 to t = 1 is 0.9357142857.
How to calculate
Finds the line integral ∫ F · dr of a vector field, or ∫ f ds of a function, along a curve r(t) from t = a to t = b.
Example with the default inputs (Integral ∫ F · dr, P (i component of F) y z, Q (j component of F) x y, R (k component, optional) x z, x(t) t^2, y(t) t, z(t) (empty for a plane curve) t^4, From t = a 0, To t = b 1): The line integral along C from t = 0 to t = 1 is 0.9357142857.
Method: ∫ F · dr = ∫ₐᵇ F(r(t)) · r′(t) dt and ∫ f ds = ∫ₐᵇ f(r(t)) ‖r′(t)‖ dt, each derivative and integral from a computer algebra system, checked.
- C runs from r(a) to r(b) as t grows; reversing C changes the sign of ∫ F · dr, not of ∫ f ds.
- Angles in radians. When the algebra finds no antiderivative, adaptive Simpson’s rule gives the value.
Worked examples
Each example is checked against the calculator on every build.
- Integral ∫ F · dr, P (i component of F) y z, Q (j component of F) x y, R (k component, optional) x z, x(t) t^2, y(t) t, z(t) (empty for a plane curve) t^4, From t = a 0, To t = b 1 gives Line integral 0.935714, Exact value 131/140.Source: OpenStax, Calculus Volume 3, section 6.2 Line Integrals (https://openstax.org/books/calculus-volume-3/pages/6-2-line-integrals), Example 6.23 (F = ⟨yz, xy, xz⟩, r(t) = ⟨t², t, t⁴⟩, 0 ≤ t ≤ 1)
- Integral ∫ F · dr, P (i component of F) -y, Q (j component of F) x, x(t) cos(t), y(t) sin(t), From t = a 0, To t = b pi gives Line integral 3.141593.Source: OpenStax, Calculus Volume 3, section 6.2 Line Integrals (https://openstax.org/books/calculus-volume-3/pages/6-2-line-integrals), Example 6.18 (F = ⟨−y, x⟩ on the upper half of the unit circle)
- Integral ∫ f ds, f(x, y, z) x^2 + y^2 + z, x(t) cos(t), y(t) sin(t), z(t) (empty for a plane curve) t, From t = a 0, To t = b 2pi gives Line integral 36.801223.Source: OpenStax, Calculus Volume 3, section 6.2 Line Integrals (https://openstax.org/books/calculus-volume-3/pages/6-2-line-integrals), Example 6.15 (helix): 2√2π + 2√2π²
How it works
The curve C is r(t) = (x(t), y(t), z(t)) for a ≤ t ≤ b, or (x(t), y(t)) when z(t) is empty. a and b are numbers or constants such as pi, with a < b.
- A computer algebra system (nerdamer, open source) finds x′(t), y′(t) and z′(t), each checked against a numeric difference quotient.
- The integrand g(t) is built by putting x(t), y(t), z(t) in place of x, y, z (z is 0 for a plane curve):
- ∫ F · dr: g(t) = P(r(t)) x′(t) + Q(r(t)) y′(t) + R(r(t)) z′(t). An empty R is 0.
- ∫ f ds: g(t) = f(r(t)) √(x′(t)² + y′(t)² + z′(t)²).
- g is shown in the shortest of three forms: as built, simplified, and expanded by the algebra (each checked equal at test points).
- The value is ∫ₐᵇ g(t) dt from the algebra’s definite integral, checked numerically, with its exact form when it has no rounded numbers. When the algebra finds none, the value comes from adaptive Simpson’s rule: each piece is split in two until its two halves agree with it to 1.5 × 10⁻¹² of the size of the integral, with the Richardson correction (L + R − S)/15 added; there is no answer if a piece would need more than 40 halvings, more than 20,000 pieces are needed, or g is not a real number at a point used.
Assumptions
- C is traced once from r(a) to r(b) as t grows.
- Angles are in radians; ln is the natural logarithm.
- An answer that fails its check is not shown.
Worked examples by hand
F = ⟨yz, xy, xz⟩ on r(t) = ⟨t², t, t⁴⟩, 0 ≤ t ≤ 1 (OpenStax Calculus Volume 3, Example 6.23). F(r(t)) = ⟨t⁵, t³, t⁶⟩ and r′(t) = ⟨2t, 1, 4t³⟩, so g(t) = 2t⁶ + t³ + 4t⁹. ∫₀¹ g dt = 2/7 + 1/4 + 2/5 = 131/140 ≈ 0.9357142857.
F = ⟨−y, x⟩ on r(t) = ⟨cos t, sin t⟩, 0 ≤ t ≤ π (Example 6.18). F(r(t)) · r′(t) = sin²t + cos²t = 1, so the integral is π.
f = x² + y² + z on the helix r(t) = ⟨cos t, sin t, t⟩, 0 ≤ t ≤ 2π (Example 6.15). f(r(t)) = 1 + t and ‖r′(t)‖ = √2, so the integral of (1 + t)√2 from 0 to 2π = 2√2π + 2√2π² ≈ 36.80122267.
Other questions people ask
What is a line integral?
An integral along a curve C instead of along an interval. The scalar line integral ∫ f ds adds up f times arc length, for example the mass of a wire with density f. The vector line integral ∫ F · dr adds up the part of F along the curve, for example the work a force F does on an object that moves along C.
How do I compute ∫ F · dr?
Write C as r(t) for a ≤ t ≤ b. Put r(t) into F, take the dot product with r′(t), and integrate from a to b: ∫ F · dr = ∫ₐᵇ F(r(t)) · r′(t) dt. For F = ⟨−y, x⟩ on r(t) = ⟨cos t, sin t⟩, 0 ≤ t ≤ π, the integrand is sin²t + cos²t = 1, so the integral is π.
How do I compute ∫ f ds?
Use ds = ‖r′(t)‖ dt: ∫ f ds = ∫ₐᵇ f(r(t)) ‖r′(t)‖ dt. With f = 1 this is the arc length of C.
Does the direction of the curve matter?
For ∫ F · dr, yes: running C the other way changes the sign of the answer. For ∫ f ds, no: arc length is always positive, so the answer stays the same.
What does a line integral of 0 around a closed curve mean?
If F is conservative (the gradient of a function), ∫ F · dr around every closed curve is 0, and the integral between two points does not depend on the path. A field with curl F ≠ 0 is not conservative.
What is the notation ∫ P dx + Q dy + R dz?
It is ∫ F · dr for F = ⟨P, Q, R⟩. Type P, Q and R as the three components; dx = x′(t) dt, dy = y′(t) dt and dz = z′(t) dt.
How is the answer checked?
Each derivative of the curve comes from a computer algebra system and is checked against a numeric difference quotient. The definite integral from the algebra is checked against a numeric integral. When the algebra finds no antiderivative (often with a square root in ‖r′(t)‖), the page uses adaptive Simpson’s rule to 10 figures and shows no exact value.