Evaluate the line integral

meson0731

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?

LCKurtz

It looks correct to me. For what it's worth, Maple gives:$$
\frac 1 8\ln \left( \frac{13\sqrt{17}+51}{13\sqrt{17}-51}\right)+\frac{63\sqrt{17}}{64}$$

meson0731

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?

LCKurtz

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.

meson0731

Yeah i tried changing it into hyperbolic but it just got even more messy....

Villyer

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.

nileszoso

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.

Villyer

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].