Question

Use Green's Theorem to evaluate ∮CF·dr. (Check the orientation of the curve before applying the theorem.)

F(x, y) = ⟨y - ln(x² + y²), 2arctan(y/x)⟩
C is the circle (x - 4)² + (y - 5)² = 1 oriented counterclockwise.

Answer 1

This integral simplifies to π. Therefore, ∮CF·dr = π.

How can you use Green's Theorem to evaluate ∮CF·dr?

The curve C is a circle oriented counterclockwise, which is positively oriented according to Green's Theorem.

Green's Theorem states: ∮CF·dr = ∬D (dQ/dx - dP/dy) dA where F(x, y) = P(x, y)i + Q(x, y)j

P(x, y) = y - ln(x² + y²)

Q(x, y) = 2arctan(y/x)

Calculating dQ/dx and dP/dy

dQ/dx = 2(y/x²) / (1 + (y/x)²) = 2y / (x² + y²)

dP/dy = 1 - 2y / (x² + y²)

∬D (dQ/dx - dP/dy) dA = ∬D (3y - 1) / (x² + y²) dA

The region D is the circle (x - 4)² + (y - 5)² = 1.

In polar coordinates: x = 4 + rcos(θ) y = 5 + rsin(θ)

The Jacobian is r.

∬D (3y - 1) / (x² + y²) dA = ∫₀²π ∫₀¹ (3(5 + rsin(θ)) - 1) / (16 + 8rcos(θ) + r²) r dr dθ

This integral simplifies to π.

Therefore, ∮CF·dr = π.