Find the arc length of the curve y^3=8x^2 fr0m x=1 to x=8.

Answer 1

since y^3 = 8x^2,
y = 2x^(2/3)
y' = 4/3 x^(-1/3)
y'^2 = 16/(9x^(2/3))

arc length is thus
∫[1,8] √(1+16/(9x^(2/3))) dx

u=x^-2/3
then let v = u^1/2
then let 4w = 3tanv

and you finally wind up with

1/27 √(9+16/x^2/3) (9x + 16x^1/3) [1,8]
= 1/27 ((√(9+4)(72+32))-(√(16+9)(9+16)))
= 1/27 (104√13 - 125)