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**1. Homework Statement**

find the length of the curve

x = 1/3√y(y-3) 1 ≤ y ≤ 9

**2. Homework Equations**

L = ∫ √(1 + (dx/dy)^2)

**3. The Attempt at a Solution**

x = 1/3√y(y-3) 1 ≤ y ≤ 9

x = 1/3 (y^3/2 - 3y^1/2)

dx/dy = 1/2(y^1/2) - 1/2(y^-1/2)

dx/dy = 1/2(y^1/2 - y^-1/2)

(dx/dy)^2 = 1/4( y -2 + 1/y)

L = ∫ √[(1 + 1/4(y - 2 + 1/y)] 1 ≤ y ≤ 9

L = ∫ √[(1 + 1/4(y^2 - 2y + 1)/y] 1 ≤ y ≤ 9

L = ∫ √[(1 + 1/4(y-1)^2/y] 1 ≤ y ≤ 9

Im tempted to go ahead with the trig substitution, but this is getting a little complex, so I am wondering if I am even on the right track.