- #1
Ryaners
- 50
- 2
Homework Statement
Find the arclength of the curve x = ⅓(y2+2)3/2 from y=0 to y=1.
Homework 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
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!