In the last post, we showed the properties of vector line integrals. Let's use them to compute a vector line integral.
Solution
Let's draw the vector field and represent the domain curve
As seen in the figure, the domain curve is the sum of the four sides of a rectangle: C₁ from (0,0) to (2,0), C₂ from (2,0) to (2,1), C₃ from (2,1) to (0,1), C₄ from (1,0) to (0,0).
The pattern
- Start at the first vertex of the side, in the counterclockwise order (0,0) → (2,0) → (2,1) → (0,1) → (0,0).
- Keep one coordinate fixed, since the side is horizontal or vertical.
- Let the other coordinate change at speed 1, increasing or decreasing depending on the direction of travel.
- Let t run from 0 to the length of the side, so that t measures distance traveled.
The four sides
| Side | From → to | Parametrization | Range of t | r′(t) |
|---|---|---|---|---|
| C₁ (bottom) | (0,0) → (2,0) | ⟨t, 0⟩ | 0 ≤ t ≤ 2 | ⟨1, 0⟩ |
| C₂ (right) | (2,0) → (2,1) | ⟨2, t⟩ | 0 ≤ t ≤ 1 | ⟨0, 1⟩ |
| C₃ (top) | (2,1) → (0,1) | ⟨2 − t, 1⟩ | 0 ≤ t ≤ 2 | ⟨−1, 0⟩ |
| C₄ (left) | (0,1) → (0,0) | ⟨0, 1 − t⟩ | 0 ≤ t ≤ 1 | ⟨0, −1⟩ |
Interpretation
In the table above we find the parametrization for each side r(t), the range of the parameter and r'(t)
- C₁: y stays at 0 and x grows from 0 to 2. At t = 0 you are at (0,0), and at t = 2 you are at (2,0).
- C₂: x stays at 2 and y grows from 0 to 1. The side has length 1, so t stops at 1.
- C₃: y stays at 1, but x must go from 2 down to 0. Writing 2 − t does this, since at t = 0 it gives 2 and at t = 2 it gives 0. Using ⟨t, 1⟩ would trace the same segment in the wrong direction.
- C₄: x stays at 0 and y goes from 1 down to 0, so 1 − t with t from 0 to 1.
According to the 4th property of vector line integral, we have:
Let's calculate F(r(t)) and substitute r'(t):
Let's do the same for the other integrals:
Adding all the 4 integrals, we finally have:
