Final answer:
To evaluate the work integral ∮C F · ds, where F = (-y, x), and C is the path counterclockwise along the circle x² + y² = 16 from (4,0) to (-4,0), we can use Green's theorem to calculate the double integral of the vector field components.
Step-by-step explanation:
To evaluate the work integral ∮C F · ds, where F = (-y, x), and C is the path counterclockwise along the circle x² + y² = 16 from (4,0) to (-4,0), we can use Green's theorem. Green's theorem states that if F = (P, Q) is a vector field whose components have continuous first partial derivatives on an open region in the plane, and C is a positively oriented, piecewise smooth, simple closed curve in the plane, then:
∮C F · ds = ∬D (Q_x - P_y) dA
In this case, P = -y and Q = x. Therefore, P_y = -1 and Q_x = 1. Plugging these values into the formula, we get:
∮C F · ds = ∬D (1 - -1) dA = ∬D 2 dA
The circle x² + y² = 16 can be parametrized as x = 4cos(t) and y = 4sin(t), where 0 ≤ t ≤ π. Plugging these values into the formula, we get:
∮C F · ds = ∫₀^π∫₀⁴ (2)(r) dr dt = 2∫₀^π (1/2)(4²) dt = 2∫₀^π 8 dt = 2(8π) = 16π
Therefore, the correct answer is D) None of the above.