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Homework Help: Curve Integration problem

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data
    Well i got the formula down, im having trouble with the integration though:
    Q: {(x^3 + y) ds , x = 3t, y = t^3, 0<=t<=1

    = {[ (27t^3 + t^3) sqrt(9t^2 + t^6)] dt

    2. Relevant equations
    f(x,y)ds = f(x(t), y(t)) sqrt[ (x'(t))^2 + (y'(t))^2 ]

    3. The attempt at a solution
    I simplified it down to: 28 {[ (t^4) sqrt(9 + t^4) ] dtAny help would be appreciated.

    t^2 = 3cosu
    = 28{ [(9cos^2u) sqrt(9 + 9cos^2u)] du
    =756{ [(cos^2u) sqrt(1 + cos^2u)] du
    no helpful identity..
    dt^2 = -3sinu du
    Last edited: Nov 22, 2007
  2. jcsd
  3. Nov 23, 2007 #2


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    Science Advisor

    WHY would you let t^2= 3 cos u? Since sin^2(x)+ cos^2(x)= 1, you have sin^2(x)= 1- cos^2(x) so "v= cos(x)" is useful for things like 1- cos^2(x), not 1+ cos^2(x).

    Rather, dividing sin^2(x)+ cos^2(x)= 1 by cos^2(x) you have tan^2(x)+ 1= sec^2(x) so t^2= 3tan(u) would be much better.
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