# Math Help - Line Intergal

1. ## Line Intergal

I need help with creating suitable parameters to solve the integral

thanks

2. ## Re: Line Intergal

Since you are integrating along the path $xy^2=1$ you can use it in your integral (where ever possible). So your integral becomes

$\int_1^4 \left(x^2 - e^{-x}\right) dx + \int_1^{1/2} \left( \frac{2}{y^3} + \cos y\right) dy$

3. ## Re: Line Intergal

Wow, that's very clever! jonnyl, if it had not worked out that nicely (or, like me, you didn't see to the core as quickly as Danny), here's the hard way:
From $xy^2= 1$, you have $x= \frac{1}{y^2}= y^{-2}$ so $dx= -2y^{-3}dy= \frac{-2}{y^3}dy$. Now, $x^4y^4= \frac{y^4}{y^8}= y^{-4}$, $x^2y^2= \frac{y^2}{y^4}= y^{-2}$, $\frac{2x}{y}= \frac{2}{(y^2)(y)}= \frac{2}{y^3}$, and $xy^3= \frac{y^3}{y^2}= y$. That makes the path integral
$\int_1^{1/2}\left[\left(y^{-4}- e^{-y^{-2}}\right)\left(-2y^{-3}dy\right)+ (2y^{-3}+ cos(y))dy\right]$