Using Green’s Theorem to evaluate

Question

Use Green’s Theorem to evaluate

∮_C(x − 2y²) dx + (y⁴ + 2xy) dy where C consists of the line segment

from (0, 2) to (0, 4), followed by the curve with parametric equations x = 4 cos t, y = 4 sin t from (0, 4) to (−2, 2√3), then the line segment from (−2, 2√3) to (−1, √3), and finally the curve with parametric equations x = 2 sin t, y = 2 cos t from (−1, √3) to (0, 2).

Answer 1

Green's Theorem: ∮C P dx + Q dy = ∬R (∂Q/∂x - ∂P/∂y) dA

Given: P = x - 2y² Q = y⁴ + 2xy

Step 1: Find partial derivatives

∂Q/∂x = ∂(y⁴ + 2xy)/∂x = 2y

∂P/∂y = ∂(x - 2y²)/∂y = -4y

∂Q/∂x - ∂P/∂y = 2y - (-4y) = 6y

Step 2: Identify the region R

Let me trace the path to understand the region:

1. (0, 2) to (0, 4): vertical line segment
2. (0, 4) to (-2, 2√3): arc on circle x² + y² = 16 (radius 4)
3. Angle at (0, 4): θ = π/2
4. Angle at (-2, 2√3): θ = 2π/3
5. (-2, 2√3) to (-1, √3): radial line segment
6. (-1, √3) to (0, 2): arc on circle x² + y² = 4 (radius 2)
7. Both points are at angles π/2 and 2π/3

The region R is an annular sector (ring slice) with:

1. Inner radius: r = 2
2. Outer radius: r = 4
3. Angular range: θ from π/2 to 2π/3

Step 3: Evaluate the double integral using polar coordinates

∬R 6y dA where y = r sin θ and dA = r dr dθ

= ∫[π/2 to 2π/3] ∫[2 to 4] 6(r sin θ) r dr dθ

= ∫[π/2 to 2π/3] ∫[2 to 4] 6r² sin θ dr dθ

= ∫[π/2 to 2π/3] sin θ [2r³]₂⁴ dθ

= ∫[π/2 to 2π/3] sin θ · 2(64 - 8) dθ

= ∫[π/2 to 2π/3] 112 sin θ dθ

= 112[-cos θ]_{π/2}^{2π/3}

= 112[(-cos(2π/3)) - (-cos(π/2))]

= 112[1/2 - 0]

= 56