设曲线L;y=x,从点a(0,0)到点b(1,1),则积分 ∫(y²-x²)ds=?
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曲线 L 的方程为 y = x,所以曲线 L 的速度为 v = sqrt(x'² + y'²) = sqrt(1 + 1) = sqrt(2)
所以积分 ∫(y² - x²)ds = ∫(x² - x² + y²)ds = ∫(y²)ds = ∫(x²)ds
因为从点 a 到点 b,所以积分区间为 [0,1]
∫(x²)ds = ∫x²sqrt(2)dx = ∫x²sqrt(2)dx = (1/3)x³sqrt(2) |0,1 = (1/3)1³sqrt(2) - (1/3)0³sqrt(2) = (1/3)*sqrt(2)
∫(y²)ds = ∫y²sqrt(2)dy = ∫y²sqrt(2)dy = (1/3)y³sqrt(2) |0,1 = (1/3)1³sqrt(2) - (1/3)0³sqrt(2) = (1/3)*sqrt(2)
∫(y² - x²)ds = ∫(y²)ds - ∫(x²)ds = (1/3)*sqrt(2) - (1/3)*sqrt(2) = 0
因此积分 ∫(y² - x²)ds = 0
所以积分 ∫(y² - x²)ds = ∫(x² - x² + y²)ds = ∫(y²)ds = ∫(x²)ds
因为从点 a 到点 b,所以积分区间为 [0,1]
∫(x²)ds = ∫x²sqrt(2)dx = ∫x²sqrt(2)dx = (1/3)x³sqrt(2) |0,1 = (1/3)1³sqrt(2) - (1/3)0³sqrt(2) = (1/3)*sqrt(2)
∫(y²)ds = ∫y²sqrt(2)dy = ∫y²sqrt(2)dy = (1/3)y³sqrt(2) |0,1 = (1/3)1³sqrt(2) - (1/3)0³sqrt(2) = (1/3)*sqrt(2)
∫(y² - x²)ds = ∫(y²)ds - ∫(x²)ds = (1/3)*sqrt(2) - (1/3)*sqrt(2) = 0
因此积分 ∫(y² - x²)ds = 0
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