Calculus III:

Fundamental theorem for line integrals

1. Definition and method

In Calculus I we had used the fundamental theorem of calculus that show us how to evaluate definite integrals. That is : ∫_a^b f'(x) dx = f(b) - f(a)

There is a version of this theorem for line integrals over certain kinds of vector fields:

Theorem

Suppose that C is a smooth curve given by (t), a ≤ t ≤ b. Also suppose that f is a function of multiple variable whose gradient vector ∇f is continuous on C. Then:

∫_C∇f . = f((b)) - f((a))

Note that (a) represents the initial point on C while (b) represents the final point on C.

Indeed,

∫_C∇f . = ∫_a^b (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz =
∫_a^b (∂f/∂x)(dx/dt)dt + (∂f/∂y)(dy/dt)dt + (∂f/∂z)(dz/dt)dt =
∫_a^b [(∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt) + (∂f/∂z)(dz/dt)] dt =
∫_a^b (d[f((t))]/dt) dt = ∫_a^b df((t))

Using the Fundamental theorem of calculus for single integrals, we obtain:

∫_C∇f . = f( (b)) - f((a))

Example 1

Evaluate ∫_C∇f . where f(x, y, z) = xy + xz + yz, and C is any path from (- 1 , 0, 1) and (- 2, 1, 2).

We have then:

(a) = 〈 - 1 , 0, 1 〉 , and (b) = 〈 - 2, 1, 2 〉

f((a)) = (- 1)(0) + (- 1)(1) + (0)(1) = - 1

f((b)) = (- 2 )(1) + (- 2)(2) + (1)(2) = - 4

Therefore

f((b)) - f((a)) = - 4 - (- 1) = - 3

∫_C ∇f . = - 3