Evaluate a line integral F=(x−y)i+(x+y)j

Short answer

Using Green's theorem, ∂Q/∂x − ∂P/∂y = 2. The area between y²=x and x²=y from x=0 to 1 is 1/3. The line integral equals 2 × 1/3 = 2/3.

Evaluate ∮_C F·dr, where F = (x − y) i + (x + y) j along the closed curve C bounded by y² = x and x² = y.

F = (x − y) i + (x + y) j, so P = x − y and Q = x + y. The closed curve C is bounded by y² = x and x² = y, which meet at O(0, 0) and A(1, 1).

∮_C F·dr = ∬_R (∂Q/∂x − ∂P/∂y) dx dy = ∬_R (1 − (−1)) dx dy = 2 · Area(R)

Area between the curves (for 0 ≤ x ≤ 1, y runs from x² to √x):

Area = ∫₀¹ (√x − x²) dx = 2/3 − 1/3 = 1/3

∮_C F·dr = 2 × 1/3 = 2/3

Along y = x² (x: 0 → 1), dy = 2x dx: ∫ [(x − x²) + (x + x²)2x] dx = ∫₀¹ (x + x² + 2x³) dx = 1/2 + 1/3 + 1/2 = 4/3.
Along x = y² (y: 1 → 0), dx = 2y dy: ∫ [(y² − y)2y + (y² + y)] dy = ∫₁⁰ (2y³ − y² + y) dy = −(1/2 − 1/3 + 1/2) = −2/3.

Total = 4/3 − 2/3 = 2/3 ✓.

Answer: ∮_C F·dr = 2/3 (counter-clockwise).

✓ Answer checked by Claude Opus 5.5 (high effort)

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