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Find arclength of curve; stuck trying to integrate radical

  1. Feb 3, 2016 #1
    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. jcsd
  3. Feb 3, 2016 #2

    PeroK

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    You may want to double check your differentiation.
     
  4. Feb 3, 2016 #3

    Ray Vickson

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    Your differentiation/arclength formula is incorrect; go back and check your algebra.
     
  5. Feb 3, 2016 #4
    Aha! Thank you both for pointing out that error in computing the derivative - getting a nice factorizable expression now :)
     
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