(40) It looks like the line integral is

where C is the boundary of the region D,

(a) To evaluate the line integral directly, split up C into two paths C₁ and C₂, parameterized by
• C₁ : x = t ²/8 and y = -t, where -4 ≤ t ≤ 4
(the y component is negative to make this path have a positive/counterclockwise orientation)
• C₂ : x = 2 and y = t, where -4 ≤ t ≤ 4
Then the line integral over C is the sum of the line integrals over C₁ and C₂ :

(b) Using Green's theorem, we have

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(41) I assume you meant to use θ in the parameterization, and not 6, so that C is parameterized by x = θ - sin(θ) and y = 1 - cos(θ). There's no range given for θ, so I'll just assume 0 ≤ θ ≤ π.
Then ... (for some reason, the math text won't render properly. I've attached the computation as an image)
(where s = sin(θ) and c = cos(θ))