- 2076 Chaitra · 5 marks
Verify Green's theorem
Short answer
∂Q/∂x − ∂P/∂y = 2. The double integral over the region between y=√x and y=x² is 2∫₀¹(√x − x²)dx = 2/3. The line integral also equals 2/3, confirming Green's theorem.
Verify Green's theorem in plane for ∮_C [(x − y)dx + (x + y)dy] where C is the boundary of the region enclosed by y² = x and x² = y.
Green's theorem: ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy, with P = x − y, Q = x + y.
The curves y² = x and x² = y meet at O(0,0) and A(1,1). In 0 ≤ x ≤ 1 the curve y = √x lies above y = x².
Double integral
∂Q/∂x − ∂P/∂y = 1 − (−1) = 2
∬_R 2 dx dy = 2 ∫₀¹ (√x − x²) dx = 2 (2/3 − 1/3) = 2/3
Line integral (anticlockwise: O to A along y = x², then A to O along y² = x)
On C₁: y = x², dy = 2x dx, x from 0 to 1. Then (x − y) dx + (x + y) dy = (x − x²) + (x + x²)(2x) = x + x² + 2x³ per dx.
∫₀¹ (x + x² + 2x³) dx = 1/2 + 1/3 + 1/2 = 4/3
C₂: x = y², dx = 2y dy, y: 1→0
(y² − y)(2y) + (y² + y) = 2y³ − 2y² + y² + y = 2y³ − y² + y
∫₁⁰ (2y³ − y² + y) dy = −(1/2 − 1/3 + 1/2) = −2/3
Total = 4/3 − 2/3 = 2/3
Both sides equal 2/3, so Green's theorem is verified.
✓ Answer checked by Claude Opus 5.5 (high effort)
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