Parametrizing and Evaluating a Line Integral on a Given Curve

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
meson0731
Messages
13
Reaction score
0

Homework Statement



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


Homework Equations





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 because 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?
 
Physics news on Phys.org
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?
 
meson0731 said:
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?

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