Line integrals

(1)




W = ⌠ ⌡ (1,2) (0,0)  F·dr,

where



F ≡ (x y, x2), (1) dr = (dx, dy). (2)




W = ⌠ ⌡ x y dx + x2 dy = ⌠ ⌡ x (2x) dx + x2 2 dx = ⌠ ⌡ 1 0  (2x2 + 2x2) dx = ⌠ ⌡ 1 0  4x2 dx = 4 3 . (3)


(2)




W = ⌠ ⌡ x y dx + x2 dy = ⌠ ⌡ x (2 x2) dx + x2 (4x dx) = ⌠ ⌡ 1 0  (2 x3 + 4 x3) dx = ⌠ ⌡ 1 0  6 x3 dx = 3 2 . (4)

It can be seen that the values of line integrals generally depend on the integral paths.
(3)




F ≡ (2 x y, x2 − 1).




W = ⌠ ⌡ 2 x y dx + (x2 − 1) dy = ⌠ ⌡ 5/2π 0  { 2 (rcosθ)(rsinθ) (dr cosθ− r sinθdθ) + ((rcosθ)2 − 1) (dr sin θ+ r cos θdθ) } = … = −5/2 π, (5)

where


x = r cos θ, (6) y = r sin θ, (7) dx = dr cos θ− r sin θdθ, (8) dy = dr sin θ+ r cos θdθ, (9)

was used.
Alternatively, if one defines


G(x, y) ≡ x2 y − y,1




⌠ ⌡ F·dr = ⌠ ⌡ dG = [G](x,y)=(0, 0)(x,y)=(0, 5/2π) = − 5 2 π.

While the line integrals of Examples 1 and 2 depend on the integral path, the line integral of Example 3 is independent of the path. This difference will be clarified in the next section.

Stokes' theorem

Irrotational Field

(1)


v = ∇ϕ.


(2)


∇×v = 0.


(3)


⌠ (⎜) ⌡ v·dr = 0.


(4)


⌠ ⌡ v·dr,

is independent of the integral path.

1. 1 → 2.
   

   ∇×(∇ϕ)=0
2. 2 → 3
   

   ⌠ (⎜) ⌡ ∂S  v·dr = ⌠ ⌡ ⌠ ⌡ S  m·(∇×v) dA = 0
3. 3 → 4 (shown in class)
4. 4 → 1
   Define ϕ as
   

   ϕ(r) ≡ ⌠ ⌡ r a  v·dr,

   then
   
   

   ϕ(r+dr)−ϕ(r) ≈ ∇ϕ·dr2

   (15)

   On the other hand,
   
   

   ϕ(r+dr)−ϕ(r) = ⌠ ⌡ r+dr a  v·dr − ⌠ ⌡ r a  v·dr ≈ v ·dr, (16)

   so
   

   v = ∇ϕ.

Footnotes: