Alternate notation for line integrals

We initially denoted line integrals using the notation

∫ CF ⋅ ds = ∫ abF(c(t)) ⋅ c'(t)dt (1)
where c(t) = (x(t),y(t),z(t)). We can derive a new notation by “multiplying out” the dot products on both sides of equation (1).

Let’s assume we are in three dimensions to F = (F1,F2,F3). If we multiply out the left hand side, thinking of ds being the “vector” ds = (dx,dy,dz), it becomes

∫ CF ⋅ ds = ∫ CF1dx + F2dy + F3dz.
Also, since
c'(t) = ( ) dx dy dz dt-,dt, dt-,
we could “multiply out” the right hand side of of equation (1) to write it as
∫ abF(c(t)) ⋅ c'(t)dt = ∫ ab( ) dx- dy- dz- F1 (c(t)) dt + F2(c(t))dt + F3 (c (t))dtdt.

Frequently, line integral problems are presented in this notation. In this case, one can compute them directly using the formula

∫ CF1dx + F2dy + F3dz = ∫ ab( ) F (c(t))dx- + F (c(t))dy-+ F (c(t))dz- 1 dt 2 dt 3 dtdt. (2)

Example

Evaluate

∫ Cydx + (x + y)dy + dz
where thee curve C is parameterized by c(t) = (t, 1 - t, 1), 0 ≤ t ≤ 1.

Solution: If we write c(t) = (t, 1 -t, 1) = (x(t),y(t),z(t)), Then dx dt = 1, dy dt = -1, and dz dt = 0. Since F1(x,y,z) = y, F2(x,y,z) = x + y, and F3(x,y,z) = 1, we can use formula (2) to calculate the integral.

∫ Cydx + (x + y)dy
= ∫ 01( dx dy dz ) y(t)---+ (x(t) + y(t)) ---+ --- dt dt dtdt
= ∫ 01[(1 - t)(1) + (t + (1 - t))(- 1)] dt
= ∫ 01[(1 - t) + (-1)]dt
= ∫ 01(-t)dt = t2||1 - --|| 2 0 = -1 -- 2