求高数大神解答
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简单计算一下，答案如图所示
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8. y = x^3, dy = (3x^2)dx
I = ∫<L>(2x+y)dx+(x+y)dy
= ∫<0, 1>(2x+x^3)dx+(x+x^3)(3x^2)dx
= ∫<0, 1>(2x+4x^3+3x^5)dx
= [x^2+x^4+(1/2)x^6]<0, 1> = 5/2.
15. z = u^2lnv, u = x/y, v = x-2y
∂z/∂x = (∂z/∂u)(∂u/∂x) + (∂z/∂v)(∂v/∂x)
= 2ulnv · (1/y) + (u^2/v) · 1
= (2x/y^2)ln(x-2y) + x^2/[y^2(x-2y)] ;
∂z/∂y = (∂z/∂u)(∂u/∂y) + (∂z/∂v)(∂v/∂y)
= 2ulnv · (-x/y^2) + (u^2/v) · (-2)
= (-2x^2/y^3)ln(x-2y) - 2x^2/[y^2(x-2y)].
I = ∫<L>(2x+y)dx+(x+y)dy
= ∫<0, 1>(2x+x^3)dx+(x+x^3)(3x^2)dx
= ∫<0, 1>(2x+4x^3+3x^5)dx
= [x^2+x^4+(1/2)x^6]<0, 1> = 5/2.
15. z = u^2lnv, u = x/y, v = x-2y
∂z/∂x = (∂z/∂u)(∂u/∂x) + (∂z/∂v)(∂v/∂x)
= 2ulnv · (1/y) + (u^2/v) · 1
= (2x/y^2)ln(x-2y) + x^2/[y^2(x-2y)] ;
∂z/∂y = (∂z/∂u)(∂u/∂y) + (∂z/∂v)(∂v/∂y)
= 2ulnv · (-x/y^2) + (u^2/v) · (-2)
= (-2x^2/y^3)ln(x-2y) - 2x^2/[y^2(x-2y)].
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