1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluate the line integral

  1. Jul 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫(x^3 + y^3)ds where C : r(t)=<e^t , e^(-t)>, 0 <= t <= ln2
    c


    2. Relevant equations



    3. The attempt at a solution

    I tried to parametrize the integral and change ds to sqrt(e^(2t) + e^(-2t)) dt.

    I then change (x^3 + y^3) to (e^(3t) + e^(-3t)

    so i ended up with


    ln2
    ∫(e^(3t) + e^(-3t)) * sqrt(e^(2t) + e^(-2t))dt
    0

    I feel like i set the integral up wrong becuase I would have no idea of how to do this integral. Even wolframalpha gives me a crazy answer. Is there another way to do this or did i make a mistake?
     
  2. jcsd
  3. Jul 25, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It looks correct to me. For what it's worth, Maple gives:$$
    \frac 1 8\ln \left( \frac{13\sqrt{17}+51}{13\sqrt{17}-51}\right)+\frac{63\sqrt{17}}{64}$$
     
  4. Jul 25, 2012 #3
    So i have to actually do that integral? Is there a way to write it in differential form or another form that would be easier?
     
  5. Jul 25, 2012 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Beats me. I don't right off see a simple way to work it myself. I thought about expressing the integrand in terms of ##\cosh(3t)## and ##\cosh(2t)## and I still didn't see anything obvious. But then again, I haven't been losing any sleep over it and maybe someone else will see something clever.
     
  6. Jul 25, 2012 #5
    Yeah i tried changing it into hyperbolic but it just got even more messy....
     
  7. Jul 25, 2012 #6
    It may help to rewrite the curve that it travels over.
    Instead of integrating $$r(t)=<e^t,e^{-t}>$$ it may be easier to integrate $$r(t)=<t,\frac{1}{t}>$$

    It haven't tried it though, so it may not be any easier.
     
  8. Jul 25, 2012 #7
    I'm working on this same problem. I did indeed rewrite it as r(t)=<t,1/t>, but this integral is no easier to solve. The solution given by Wolfram Alpha for this integral was the same numerically as that given by Maple in the above post.
     
  9. Jul 26, 2012 #8
    I haven't evaluated many line integrals, but if we are on the curve y = 1/x, can't we skip using the parameter t and integrate [itex](x^3 + \frac{1}{x^3})\sqrt{1 + \frac{1}{x^4}} dx[/itex]
    from 1 to 2?
    I'm having success evaluating it using the substitution [itex]x^2 = tan\theta[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Evaluate the line integral
  1. Evaluate line integral (Replies: 3)

  2. Evaluate line integral (Replies: 3)

  3. Line Integral Evaluation (Replies: 25)

Loading...