Green’s Theorem (example continued)

Strategy: In calculating the line integral on the left hand side of Green’s Theorem

∫ F . dr Strategy: Break the path C into four parts,

C

C₁, C₂, C₃, and C₄ proceeding counterclockwise starting at (1,1).

i.e. C₁ goes from (1,1) to (2,1), C₂ is from (2,1) to (2,2), C₃ is from (2,2) to (1,2),

and finally C₄ is from (1,2) returning to (1,1).

F . dr = [(y i - x j)/ (x² + y²)] . [ d x i + dy j ] = (y dx – x dy) / (x² + y²)

∫ F . dr = ∫ F . dr + ∫ F . dr + ∫ F . dr + ∫ F . dr

C C₁ C₂ C₃ C₄

Note: On C₁ y = 1, dy = 0, and 1 ≤ x ≤ 2.

2 2

So ∫ F . dr = ∫ (dx / (x² + 1) = tan^-1 (x) | = tan^-1(2) - tan^-1(1)

C₁ 1 1

The other three line integrals are similar in that they involve the inverse tangent

function and when added together sum to zero. You should verify this result

by completing the remaining three line integrals and summing them up.

Click here for an example “extending” the use of Green’s Theorem for the case

where the vector field, F(x,y) = P(x,y) i + Q(x,y) j is not continuous at all points

in region R.

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