An example of a conservative vector field

If a vector field F is a gradient field, meaning F = ∇f for some scalar-valued function f, then we can compute the line integral of F along a curve C from some point a to some other point b

∫ CF ⋅ ds = f(b) - f(a).
Note that this integral does not depend on the entire curve C; it depends on only the endpoints a and b. If we replaced C by another curve with the same endpoints, the integral would be unchanged. Hence F is path-independent (or conservative).

As an example, consider f(x,y) = xy2, and let F(x,y) = ∇f(x,y) = (y2, 2xy). We know that F must be conservative.

What is ∫ CF ⋅ ds where C is a path c(t) = (t2,t3) for 1 ≤ t ≤ 2? The starting point is a = c(1) = (1, 1), and the ending pont is b = c(2) = (4, 8). Hence the integral must be

∫ CF ⋅ ds = f(4, 8) - f(1, 1)
= 4(82) - 1(12) = 256 - 1 = 255

We could also compute ∫ CF ⋅ ds the direct way using the parametrization c(t):

∫ CF ⋅ ds = ∫ abF(c(t)) ⋅ c'(t)dt
= ∫ 12F(t2,t3) ⋅ (2t, 3t2)dt
= ∫ 12((t3)2, 2t2t3) ⋅ (2t, 3t2)dt
= ∫ 128t7dt = | t8|2 1 = 256 - 1 = 255,
which agrees with the first answer.

We should get the same answer for any path from (1, 1) to (4, 8). Since (4, 8) - (1, 1) = (3, 7), we let the curve B be the straight line path parametrized by p(t) = (1, 1) + t(3, 7) = (1 + 3t, 1 + 7t) for 0 ≤ t ≤ 1. (Note that this is not a reparametrization of C. The curve C was not a straight line, so B is a completely different curve.)

The integral along B is

∫ BF ⋅ ds = ∫ 01F(p(t)) ⋅ p'(t)dt
= ∫ 01F(1 + 3t, 1 + 7t) ⋅ (3, 7)dt
= ∫ 01( 2 ) (1 + 7t) ,2(1 + 3t)(1 + 7t)⋅ (3, 7)dt
= ∫ 01(1 + 14t + 49t2, 2 + 20t + 42t2) ⋅ (3, 7)dt
= ∫ 01(17 + 182t + 441t2)dt
= 2 3 |1 (17t + 91t + 147t )|0 = 17 + 91 + 147 = 255.
Indeed, we got the same answer again.