Question about evaluating the line Integral...?

(a) This curve consists of two parts:

(i) C1: y = sin x for x in [0, π], clockwise
So, ∫(c1) F * dr
= ∫(c1) (1+y²) dx + y dy
= ∫(x = 0 to π) (1+(sin x)²) dx + sin x (cos x dx)
= ∫(x = 0 to π) [1 + (1/2)(1 - cos(2x)) + sin x cos x] dx
= (1/2) ∫(x = 0 to π) [3 - cos(2x) + 2 sin x cos x] dx
= (1/2) [3x - sin(2x)/2 + sin² x] {for x = 0 to π}
= 3π/2.

(ii) C2: y = 2 sin x for x in [0, π], counterclockwise (use minus sign!)
So, ∫(c2) F * dr
= ∫(c2) (1+y²) dx + y dy
= -∫(x = 0 to π) (1+(2 sin x)²) dx + 2 sin x (2 cos x dx)
= -∫(x = 0 to π) [1 + (4/2)(1 - cos(2x)) + 4 sin x cos x] dx
= -∫(x = 0 to π) [3 - 2 cos(2x) + 4 sin x cos x] dx
= -[3x - sin(2x) + 2 sin² x] {for x = 0 to π}
= -3π.

So, by (i) and (ii), the line integral in question equals
-3π + 3π/2 = -3π/2.
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(b) ∫∫(D) (-2y) dA
= ∫(x = 0 to π) ∫(y = sin x to 2 sin x) -2y dy dx
= ∫(x = 0 to π) -y^2 {for y = sin x to 2 sin x} dx
= ∫(x = 0 to π) -3 sin^2(x) dx
= ∫(x = 0 to π) (-3/2) (1 - cos(2x)) dx
= (-3/2) (x - sin(2x)/2) {for x = 0 to π}
= -3π/2.
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I hope this helps!