An example of a conservative vector field

If a vector field F is a gradient field, meaning F = $\nabla$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

$$\int_{{C}}^{}$$

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) = $\nabla$f (x, y) = (y², 2xy). We know that F must be conservative.

What is $\int_{{C}}^{}$F ^. ds where C is a path c(t) = (t², t³) for 1$\le$t$\le$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

$$\int_{{C}}^{}$$ = f (4, 8) - f (1, 1) = 4(8²) - 1(1²) = 256 - 1 = 255

We could also compute $\int_{{C}}^{}$F ^. ds the direct way using the parametrization c(t):

$$\int_{{C}}^{} \int_{{a}}^{{b}}$$

$$\int_{1}^{2}$$

$$\int_{1}^{2}$$

$$\int_{1}^{2} \Big\vert _{1}^{2}$$

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$\le$t$\le$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

$$\int_{{B}}^{} \int_{{0}}^{{1}}$$

$$\int_{0}^{1}$$

$$\int_{0}^{1} \left(\vphantom{(1+7t)^2, 2(1+3t)(1+7t)}\right. \left.\vphantom{(1+7t)^2, 2(1+3t)(1+7t)}\right)$$

$$\int_{0}^{1}$$

$$\int_{0}^{1}$$

$$\Big\vert _{0}^{1}$$

Indeed, we got the same answer again.