Question

x²+y²≤1 ∫∫(x²-y)dx dy

∫（x+y²）dx+(x²-y²)dy,已知,A(1,1),B(3,2),C(3,5),用格林公式求曲线积分
∫（x+y²）dx+(x²-y²)dy,L为ABC三角形边界,A(1,1),B(3,2),C(3,5),用格林公式求曲线积分

Answer 1

经过AB的直线：y = x/2 + 1/2
经过BD的直线：x = 3
经过CA的直线：y = 2x - 1
设P = x + y²、∂P/∂y = 2y
设Q = x² - y²、∂Q/²x = 2x
于是∮ (x + y²)dx + (x² - y²)dy
= 2∫∫ (x - y) dxdy
= 2∫(1 → 3) ∫(x/2 + 1/2 → 2x - 1) (x - y) dydx
= 2∫(1 → 3) [xy - y²/2]：(x/2 + 1/2 → 2x - 1) dx
= 2∫(1 → 3) {[x(2x - 1) - (2x - 1)²/2] - [x(x/2 + 1/2) - (x/2 + 1/2)²/2]} dx
= 2∫(1 → 3) (- 3/8)(x - 1)² dx
= (- 3/4) * (x - 1)³/3 ：(1 → 3)
= (- 1/4) * (8 - 0)
= - 2