高等数学几道题目
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29. x = 3 交 y = x 于点 P(3, 3), x = 3 交 y = 2x 于点 Q(3, 6).
Vx = π∫<0, 3> [(2x)^2 - x^2]dx = π∫<0, 3>3x^2dx = π[x^3]<0, 3> = 27π
Vy = π∫<0, 3> [y^2 - (y/2)^2]dy + π∫<3, 6> [3^2 - (y/2)^2]dy
= π∫<0, 3> (3y^2/4)dy + π∫<3, 6> (9 - y^2/4)dy
= π[y^3/4]<0, 3> + π[9y-y^3/12]<3, 6> = 27π/4 + (9+9/4)π = 18π.
30. y = √x - x^(3/2)/3, y' = (1/2)(1/√x - √x),
√(1+y'^2) = √[1+(1/4)(1/x-2+x) = (1/2)√(1/x+2+x) = (1/2)(1/√x+√x)
L = ∫<1, 3> √(1+y'^2)dx = (1/2) ∫<1, 3> (1/√x+√x)dx
= (1/2) [2√x+(2/3)x^(3/2)]<1, 3> = (1/2)(4√3-8/3) = 2√3-4/3
Vx = π∫<0, 3> [(2x)^2 - x^2]dx = π∫<0, 3>3x^2dx = π[x^3]<0, 3> = 27π
Vy = π∫<0, 3> [y^2 - (y/2)^2]dy + π∫<3, 6> [3^2 - (y/2)^2]dy
= π∫<0, 3> (3y^2/4)dy + π∫<3, 6> (9 - y^2/4)dy
= π[y^3/4]<0, 3> + π[9y-y^3/12]<3, 6> = 27π/4 + (9+9/4)π = 18π.
30. y = √x - x^(3/2)/3, y' = (1/2)(1/√x - √x),
√(1+y'^2) = √[1+(1/4)(1/x-2+x) = (1/2)√(1/x+2+x) = (1/2)(1/√x+√x)
L = ∫<1, 3> √(1+y'^2)dx = (1/2) ∫<1, 3> (1/√x+√x)dx
= (1/2) [2√x+(2/3)x^(3/2)]<1, 3> = (1/2)(4√3-8/3) = 2√3-4/3
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