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Math Help - Line Intergal

  1. #1
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    Line Intergal

    I need help with creating suitable parameters to solve the integral

    thanks
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  2. #2
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    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
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  3. #3
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    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]
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