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Homework Help: A cumbersome line integral

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \int_{C}(xy+\ln{x})\mathrm{d}s
    [/tex]

    where C is the arc of the parabola [itex]y=x^2[/itex] from (1,1) to (3,9)

    2. Relevant equations

    [tex]
    \int_{C}f(x,y)\mathrm{d}s = \int_{a}^{b}f(x(t),y(t))\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\mathrm{d}t
    [/tex]

    3. The attempt at a solution

    Ok, so I take x as the parameter t, so the parametric equations of the line [itex]x=t[/itex] and [itex]y=t^2[/itex] where [itex]t\in [1,3][/itex]. I can then rewrite this line integral as a regular definite integral by using the formula mentioned above (pardon the bad formatting). However, when I do that, I come up with

    [tex]
    \int_{1}^{3}(t^3+\ln{t})\sqrt{1+4t^2}\mathrm{d}t
    [/tex]

    I can't seem to find a way to integrate this function. It is integrable, at least according to the Wolfram Integrator. A solutions manual for the book I'm studying from comes up with

    [tex]
    \int_{1}^{3}(t^3+\ln{t})\sqrt{4t^2}\mathrm{d}t = \int_{1}^{3}(t^3+\ln{t})2t\mathrm{d}t
    [/tex]

    which is quite straightforward, however, it seems to me that it's a mistake since [itex]\frac{dx}{dt}=1[/itex] thus there is a 1 missing under the square root.

    So can someone please help me find a way to integrate the integral I've come up with or am I missing something simple, i.e. it has been set up correctly in the book?
     
  2. jcsd
  3. Aug 31, 2010 #2
  4. Aug 31, 2010 #3
    Oh yeah, I had incorrectly input it in the wolfram integrator. The integral indeed cannot be expressed in terms of elementary functions.

    So how do I go about the integral I have?
     
  5. Aug 31, 2010 #4
  6. Aug 31, 2010 #5
    Ok, cheers man.
     
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