【242dy】求曲线积分∫L(x y)^2dx-(x^2 y^2)dy其中L是以A(11....

发布时间:2021-04-03 16:04:30

求曲线积分∫L (x+y)^2dx-(x^2+y^2)dy其中L是以A(1,1),B(3,2),C(2,5)为顶点的三角形ABC 数学

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【答案】 围成D的三条直线分别是
  AB:y = x/2 + 1/2
  BC:y = - 3x + 11
  CA:y = 4x - 3
  假设L是逆时针方向行走的.用格林公式:
  P = (x + y)²、P'y = 2(x + y) = 2x + 2y
  Q = (- x² - y²)、Q'x = - 2x
  Q'x - P'y = - 2x - 2x - 2y = - 4x - 2y = - 2(2x + y)
  ∮L (x + y)² dx - (x² + y²) dy
  = - 2∫∫ (2x + y) dxdy
  = - 2[∫(1→2) dx ∫(x/2 + 1/2→4x - 3) (2x + y) dy + ∫(2→3) dx ∫(x/2 + 1/2→- 3x + 11) (2x + y) dy]
  = - 2[(7/8)∫(1→2) (17x² - 22x + 5) dx - (7/8)∫(2→3) (3x² + 14x - 69) dx]
  = - 2[(7/8)(35/3) - (7/8)(- 15)]
  = - 140/3
  普通方法:
  AB:y = x/2 + 1/2、dy = 1/2 dx、x:1→3
  BC:y = - 3x + 11、dy = - 3 dx、x:3→2
  CA:y = 4x - 3、dy = 4 dx、x:2→1
  ∫AB = ∫(1→3) {(x + x/2 + 1/2)² - [x² + (x/2 + 1/2)²](1/2)} dx
  = ∫(1→3) (13x²/8 + 5x/4 + 1/8) dx
  = 58/3
  ∫BC = ∫(3→2) {(x - 3x + 11)² - [x² + (- 3x + 11)²](- 3)} dx
  = ∫(3→2) (34x² - 242x + 484) dx
  = - 283/3
  ∫CA = ∫(2→1) {(x + 4x - 3)² - [x² + (4x - 3)²](4)} dx
  = ∫(2→1) (- 43x² + 66x - 27) dx
  = 85/3
  所以∮L = ∫AB + ∫BC + ∫CA = - 140/3
  逆时针是- 140/3
  顺时针是140/3
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