∫(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
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![∫(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](/uploads/image/z/1064255-23-5.jpg?t=%E2%88%AB%EF%BC%88x%2By%26%23178%3B%EF%BC%89dx%2B%28x%26%23178%3B-y%26%23178%3B%29dy%2C%E5%B7%B2%E7%9F%A5%2CA%281%2C1%29%2CB%283%2C2%29%2CC%283%2C5%29%2C%E7%94%A8%E6%A0%BC%E6%9E%97%E5%85%AC%E5%BC%8F%E6%B1%82%E6%9B%B2%E7%BA%BF%E7%A7%AF%E5%88%86%E2%88%AB%EF%BC%88x%2By%26%23178%3B%EF%BC%89dx%2B%28x%26%23178%3B-y%26%23178%3B%29dy%2CL%E4%B8%BAABC%E4%B8%89%E8%A7%92%E5%BD%A2%E8%BE%B9%E7%95%8C%2CA%281%2C1%29%2CB%283%2C2%29%2CC%283%2C5%29%2C%E7%94%A8%E6%A0%BC%E6%9E%97%E5%85%AC%E5%BC%8F%E6%B1%82%E6%9B%B2%E7%BA%BF%E7%A7%AF%E5%88%86)
∫(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
∫(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
∫(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
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