设L为抛物线y^2=x上从A（1,-1）到B（4,-2)的一段弧.求∫xydx 设l是抛物线y^2=x从点(1，1)到点(4，2)的一段弧的...
\u8bbeL\u4e3a\u629b\u7269\u7ebfy^2=x\u4e0a\u4eceA\uff081,-1\uff09\u5230B\uff081,1\uff09\u7684\u4e00\u6bb5\u5f27.\u6c42\u222bxydx\u6cd51.
I =\u222bxydx =\u222bx(-\u221ax)dx + \u222bx(\u221ax)dx
= 2\u222bx(\u221ax)dx = 2(2/5)[x^(5/2)] = 4/5.
\u6cd52. \u8865\u5145\u7ebf\u6bb5 BA\uff0c \u6210\u5c01\u95ed\u56fe\u5f62\u3002
I = \u222bxydx = \u222exydx + \u222bxydx
\u524d\u8005\u7528\u683c\u6797\u516c\u5f0f\uff0c\u540e\u8005 x = 0\uff0c dx = 0
I = -\u222b\u222b (-x)dxdy = \u222b\u222b xdxdy
= \u222b xdx \u222b dy
= 2\u222b x^(3/2)dx = (4/5)[x^(5/2)] = 4/5
y² = x\uff0c2y dy = dx
\u222b_C (y - x)dy + (x + y)dx
= \u222b(1\u21922) [(y - y²) + (y² + y)(2y)] dy
= \u222b(1\u21922) (y - y² + 2y³ + 2y²) dy
= \u222b(1\u21922) (2y³ + y² + y) dy
= [(2/4)y⁴ + (1/3)y³ + (1/2)y²]\uff1a(1\u21922)
= [(1/2)(16) + (1/3)(8) + (1/2)(4)] - [1/2 + 1/3 + 1/2]
= 34/3
简单计算一下即可，答案如图所示
x = y^2, 第 4 象限部分 y = -√x
∫<L>xydx = ∫<1, 4>x(-√x)dx = -∫<1, 4>x^(3/2)dx = -(2/5)[x^(5/2)]<1, 4> = -62/5
y² = x，2y dy = dx
∫_C (y - x)dy + (x + y)dx
= ∫(1→2) [(y - y²) + (y² + y)(2y)] dy
= ∫(1→2) (y - y² + 2y³ + 2y²) dy
= ∫(1→2) (2y³ + y² + y) dy
= [(2/4)y⁴ + (1/3)y³ + (1/2)y²]：(1→2)
= [(1/2)(16) + (1/3)(8) + (1/2)(4)] - [1/2 + 1/3 + 1/2]
= 34/3
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