# Find arclength of curve; stuck trying to integrate radical

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1. Feb 3, 2016

### Ryaners

1. The problem statement, all variables and given/known data
Find the arclength of the curve x = ⅓(y2+2)3/2 from y=0 to y=1.

2. Relevant equations
(Please forgive the crazy definite integral symbols - I'm taking a LaTex class tomorrow so hopefully I'll be able to communicate more clearly from then on..!)

arclength of curve = ∫ab √ (1 + f'(x)2) dx

3. The attempt at a solution
*Note: We haven't covered trig substitution in class yet & so I'm looking for a way to do this using u-substitution or another method.*

This is where I've gotten so far:

dx/dy = y√(y+2) = √(y3 + 2y2)

arclength = ∫01 √(1 + [√(y3 + 2y2)]2) dy
= ∫01 √(1 + y3 + 2y2) dy

Then I get stuck. I've tried the following substitution:
u = 1 + y3 + 2y2
du = 3y2 + 2y

but that doesn't seem to help. I've also tried getting y in terms of x & integrating that but I reach a similar stumbling block with that approach too. Wolfram Alpha gives a nice clean answer of 4/3 which makes me think I'm missing something fairly obvious. Any pointers much appreciated!

2. Feb 3, 2016

### PeroK

You may want to double check your differentiation.

3. Feb 3, 2016