计算∫(e^xsiny+x)dy-(e^xcosy+y)dx,其中L为从点(-2,0)沿曲线(逆时针)x^2/4+y^2
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计算∫(e^xsiny+x)dy-(e^xcosy+y)dx,其中L为从点(-2,0)沿曲线(逆时针)x^2/4+y^2/2=1到点(2,0)的弧
![计算∫(e^xsiny+x)dy-(e^xcosy+y)dx,其中L为从点(-2,0)沿曲线(逆时针)x^2/4+y^2](/uploads/image/z/533644-52-4.jpg?t=%E8%AE%A1%E7%AE%97%E2%88%AB%28e%5Exsiny%2Bx%29dy-%28e%5Excosy%2By%29dx%2C%E5%85%B6%E4%B8%ADL%E4%B8%BA%E4%BB%8E%E7%82%B9%28-2%2C0%29%E6%B2%BF%E6%9B%B2%E7%BA%BF%28%E9%80%86%E6%97%B6%E9%92%88%29x%5E2%2F4%2By%5E2)
P=-(e^xcosy+y),∂P/∂y=e^xsiny-1
Q=e^xsiny+x,∂Q/∂x=e^xsiny+1
补线段L1:y=0,x从2到-2
则L+L1为封闭曲线,由格林公式
∮(e^xsiny+x)dy-(e^xcosy+y)dx
=∫∫ 2 dxdy
由于半个椭圆的面积为:(√2)π
=2√2π
下面计算L1上的积分:
∫ (e^xsiny+x)dy-(e^xcosy+y)dx
=-∫ [2→-2] e^x dx
=e^x |[-2→2]
=e²-e^(-2)
因此:原式=2√2π-e²+e^(-2)
Q=e^xsiny+x,∂Q/∂x=e^xsiny+1
补线段L1:y=0,x从2到-2
则L+L1为封闭曲线,由格林公式
∮(e^xsiny+x)dy-(e^xcosy+y)dx
=∫∫ 2 dxdy
由于半个椭圆的面积为:(√2)π
=2√2π
下面计算L1上的积分:
∫ (e^xsiny+x)dy-(e^xcosy+y)dx
=-∫ [2→-2] e^x dx
=e^x |[-2→2]
=e²-e^(-2)
因此:原式=2√2π-e²+e^(-2)
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