Evaluate a line integral F=sinyi+x(1+cosy)j

Here F = sin y i + x(1 + cos y) j, so with F = P i + Q j:

P = sin y ,    Q = x(1 + cos y)

The path is the circle C: x² + y² = a², z = 0, which is a simple closed curve (taken counter-clockwise). The integral is ∮_C F·dr = ∮ (P dx + Q dy).

Use Green's theorem

∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy

Compute the derivatives:

• ∂Q/∂x = 1 + cos y
• ∂P/∂y = cos y

∂Q/∂x − ∂P/∂y = (1 + cos y) − cos y = 1

So

∮_C F·dr = ∬_R 1 dx dy = area of the disc x² + y² ≤ a² = πa²

Answer: ∮_C F·dr = πa² (counter-clockwise; −πa² if the circle is traversed clockwise).

Direct check (numerical)

Putting x = a cos t, y = a sin t, 0 ≤ t ≤ 2π, the integral becomes ∫₀^{2π} [ −a sin t · sin(a sin t) + a² cos²t (1 + cos(a sin t)) ] dt. For a = 1 this evaluates to 3.14159 = π, and for a = 1.5 to 7.0686 = π(1.5)², agreeing with πa².