Evaluate LHS of Green's Theorem

the question is to evaluate the left hand side of the theorem

where P(x,y) = tan^-1 (y/x)
Q(x,y) = log(base e)(x^2 + y^2)

and C is the boundary of the region (given in polar coordinates) 1<=r<=2
0<=theta<=pi, oriented clockwise

Parameterize C one segment at a time.

(i) r = 2 with θ in [0, π] (clockwise).
Parameterize this by x = 2 cos θ, y = 2 sin θ with opposite orientation.

So, ∫c₁ (arctan(y/x) dx + ln(x^2 + y^2) dy)
= -∫(θ = 0 to π) [θ * (-2 sin θ) + ln(2^2) * (2 cos θ)] dθ
= -∫(θ = 0 to π) [2θ * -sin θ + (4 ln 4) cos θ] dθ
= -[(2θ cos θ - 2 sin θ) + (4 ln 4) sin θ] {for θ = 0 to π}
= 2π.

(ii) r = 1 with θ in [0, π] (counterclockwise).
Parameterize this by x = 1 cos θ, y = 1 sin θ.

So, ∫c₂ (arctan(y/x) dx + ln(x^2 + y^2) dy)
= ∫(θ = 0 to π) [θ * (-sin θ) + ln(1^2) * (cos θ)] dθ
= ∫(θ = 0 to π) θ * -sin θ dθ
= (θ cos θ - sin θ) {for θ = 0 to π}
= -π.

(iii) x-axis from x = 2 to x = 1.
Parameterize via x = t, y = 0 on [1, 2] with opposite orientation

So, ∫c₃ (arctan(y/x) dx + ln(x^2 + y^2) dy)
= -∫(t = 1 to 2) (0 + 0) dt
= 0.

(iv) x-axis from x = -1 to x = -2.
Parameterize via x = t, y = 0 on [-2, -1] with opposite orientation

So, ∫c₄ (arctan(y/x) dx + ln(x^2 + y^2) dy)
= -∫(t = -2 to -1) (π + 0) dt, since arctan(y/x) = π on negative x-axis
= -π.
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From (i)-(iv), we have
∫c (arctan(y/x) dx + ln(x^2 + y^2) dy) = 2π + (-π) + 0 + (-π) = 0.
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Double check:
∫∫ [(∂/∂x) ln(x^2 + y^2) - (∂/∂y) arctan(y/x)] dA
= ∫∫ [2x/(x^2 + y^2) - (1/x)/(1 + (y/x)^2)] dA
= ∫∫ [2x/(x^2 + y^2) - x/(x^2 + y^2)] dA
= ∫∫ x dA/(x^2 + y^2)
= ∫(r = 1 to 2) ∫(θ = 0 to π) (r cos θ) * (r dθ dr)/r^2, converting to polar coordinates
= ∫(r = 1 to 2) dr * ∫(θ = 0 to π) cos θ dθ
= (2 - 1) * 0
= 0.
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I hope this helps!