Arc length problem

Question

Find the arc length of the curve y = x^2/3 between x = 1 and x = 8

Answer 1

mainly, you have to evaluate the integral (1+((2/3)*x^(-1/3))²)^(1/2)dx.

using 9x^(2/3)as common denominator the integral becomes

(9x^(2/3)+4)^(1/2)*(x^(-1/3))/3dx

yay, the part after (9x^(2/3)+4)^(1/2) is differential of (9x^(2/3)+4) absent a constant

My computation is arc length is (1/27)*(9x^(2/3)+4)^(3/2) evaluated from 1 to 8

or (1/27)*(40^(3/2) -13^(3/2))

Answer 2

The component of the arc length is √[(dx)² +(dy)²]=√(1+(x')²)dy

since y = x^2/3 then x = y^3/2 so x' = (3/2)y^1/2

when x = 1, y = 1^2/3 = 1 and when x = 8, y = 8^2/3 = 4 so change the bounds of integration

Arc length = ₁ ∫⁴√(1 +(9/4)y)dy =(2/3)(4/9) [(9/4)y +1]^3/2¦₁⁴ =(8/27)[(10)^3/2 -(13/4)^3/2]

Inverse function is much easier to deal with.