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Homework Help: Curve Integrals

  1. Jul 6, 2008 #1

    I am evaluating the curve integral below and I am getting an answer of 0. I have looked at my solution many times and cannot see that I have done anything wrong.

    My concern is that a value of 0 for a curve integral does not make sense - a curve integral measures the distance from point A to point B on a curve so how can it be 0.

    Problem Statement:

    Consider the parametrized curve [tex]\varsigma[/tex]: [0:2[tex]\pi[/tex]] -> R[tex]^{s}[/tex], defined by [tex]\varsigma(t) [/tex] = (e[tex]^{t}[/tex]cos(t), e[tex]^{t}[/tex]sin(t)).
    Evaluate the curve integral:
    integral of ( (x / (x[tex]^{2}[/tex] + y[tex]^{2}[/tex]) ) dx + (y / (x[tex]^{2}[/tex] + y[tex]^{2}[/tex]) ) dy )

    Problem Solution:
    Step1: Calculate the norm of parametrized function

    f(t) = e[tex]^{t}[/tex]cos(t) ; f'(t) = e[tex]^{t}[/tex]cos(t) - e[tex]^{t}[/tex]sin(t)
    g(t) = e[tex]^{t}[/tex]sin(t) ; g'(t) = e[tex]^{t}[/tex]sin(t) + e[tex]^{t}[/tex]cos(t)

    Therefore norm of || f'(t) + g'(t) || = sqrt(2) * e[tex]^{t}[/tex]

    (I am not showing the intermediate steps)

    Evaluate f(x) at f(t), g(t) and g(x) at f(t), g(t)

    f( f(t), g(t) ) = cos(t) / e[tex]^{t}[/tex]
    g( f(t), g(t) ) = sin(t) / e[tex]^{t}[/tex]


    Multiply step2 and step 3 = cos(t) + sin(t)


    Evaluate integral of Step3

    integral (0-2*pi) of (cos(t) + sin(t) dt ) = 0

    Plz advise what I am doing incorrectly


    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 7, 2008 #2


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    Homework Helper

    Hi asif! :smile:

    I'm not really following what you've done, but …

    i] shouldn't it be √(x² + y²)?

    ii] Hint: xdx + ydy = d(x² + y²)/2 :smile:
  4. Jul 8, 2008 #3


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    Science Advisor

    No, the curve integral of a given differential does NOT " measure the distance from point A to point B". That is true only for the path integral [itex]\int ds[/itex], a very specific differential.

    In fact, I note that, because of the "symmetry"
    [tex]\frac{\partial \frac{y}{x^2+ y^2}}{\partial x}= \frac{\partial \frac{x}{x^2+ y^2}}{\partial y}[/itex]
    so this is an "exact" differential. It's integral along any closed path is 0.
    Do you recognize that this is a closed path?
    Last edited by a moderator: Jul 8, 2008
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