急 急 此題數學題目🙏?

Area = ∫_[0,2] 2πy√[1+(dy/dx)^2] dx
    =  ∫_[0,2] 2π(x^3/9)√[1+(x^2/3)^2] dx
    =  ∫_[0,2] 2π (x/3)(x^2/3) √[1+(x^2/3)^2] dx

取變數變換 x^2/3 = tanθ, 則
   (2x/3) dx = sec^2(θ) dθ
   0 ≦ x ≦ 2  ←→  0 ≦ θ ≦ atan(4/3) = θ0[0,θ0]
∴ secθ > 0

∴ area = ∫_[0,θ0] π tan(θ) sec(θ) sec^2(θ) dθ
           = π (sec^3(θ)/3)_[0,θ0]
sec(atan(4/3)) = √[1+(4/3)^2] = 5/3
sec(0) = 1
∴ area = π[(5/3)^3/3 - 1/3]  = (98/81)π