Question

∫L -2√y³dx+3(x+2)√ydy,其中L为从O（0,0）经y³=4x²到（1/2,1）那一段

Answer 1

Y型：
y³ = 4x² → x = √(y³/4) = (1/2)y^(3/2)、其中x ≥ 0
dx = (1/2)(3/2)y^(1/2) dy = (3/4)√y dy、y：0→1
∫L - 2√y³dx + 3(x + 2)√ydy
= ∫(0→1) {- 2√y³(3/4)√y + 3[(1/2)y^(3/2) + 2]√y} dy
= 6∫(0→1) √y dy
= 6(2/3)y^(3/2) |(0→1)
= 4

X型：
y³ = 4x² → y = ³√(4x²) = 2^(2/3) * x^(2/3)
dy = 2^(2/3) * (2/3)x^(- 1/3) dx = 1/3*2^(5/3)*x^(- 1/3) dx，x：0→1/2
∫L - 2√y³dx + 3(x + 2)√ydy
= ∫(0→1/2) [- 2*2x + 3(x + 2)*2^(1/3)*x^(1/3)*1/3*2^(5/3)*x^(- 1/3)] dx
= 8∫(0→1/2) dx
= 8x |(0→1/2)
= 8(1/2)
= 4

格林公式(参考，好学的话自己理解)：
补上L₁：y = 0、dy = 0、x：0→1/2、逆时针
补上L₂：x = 1/2、dx = 0、y：0→1、逆时针
其中L⁻ + L₁ + L₂ = C、围成闭区域D、逆时针。
P = - 2√y³、Q = 3(x + 2)√y
∂Q/∂x - ∂P/∂y = 3√y - (- 3√y) = 6√y
IC = ∮C - 2√y³dx + 3(x + 2)√ydy = 6∫∫D √y dxdy
= 6∫(0→1/2) ∫(0→³√(4x²)) √y dydx
= 6∫(0→1/2) (2/3)y^(3/2) |(0→³√(4x²)) dx
= 4∫(0→1/2) 2x dx
= 4 * x² |(0→1/2)
= 1
IL⁻ + IL₁ + IL₂ = IC
IL⁻ + 0 + ∫L₂ 3(1/2 + 2)√y dy = 1
IL⁻ + (15/2)∫(0→1) √y dy = 1
IL⁻ + (15/2)(2/3)y^(3/2) |(0→1) = 1
IL⁻ + 5 = 1
IL⁻ = - 4
- IL = - 4
IL = 4

这里L⁻跟L的方向相反，L顺时针、L⁻逆时针，有IL = - IL⁻
IL是原式积分的简写。

Answer 2

题目不全 无解

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老师我们练习册上的题，我又看了一遍没打错