1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Line Integral of a parametric curve

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the line integral over the curve C
    [tex]\int_{C}^{}e^xdx[/tex]
    where C is the arc of the curve
    [tex]x=y^3[/tex]
    from (-1,-1) to (1,1)
    2. Relevant equations
    [tex]\int_{C}^{}f(x,y)ds=\int_{a}^{b}f(x(t),y(t))\sqrt((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)dt[/tex]
    3. The attempt at a solution
    I tried parametrizing the curve to y=t and x=t^3
    therefore dy/dt = 1 and dx/dt = 3t^2
    plug this back into the formula, we get
    ∫ from -1 to 1 (e^t^3)sqrt(3t^2+1)dt
    but this is an insolvable integral, anything I did wrong or is there another way?
     
  2. jcsd
  3. May 11, 2012 #2

    sharks

    User Avatar
    Gold Member

    Your line integral is obviously wrong. You might have typed it incorrectly. It should be:
    [tex]\int_{C}^{}e^xds[/tex]
    [tex]\frac{dy}{dx}=\frac{1}{3y}[/tex]
    [tex]\int_{C}^{}f(x,y)ds=\int_{x=a}^{x=b}f(x,y)\sqrt {1+(\frac{dy}{dx})^2}\,.dx[/tex]
    Afterwards, use the substitution [itex]\frac{1}{3x^{1/3}}=\tan \theta[/itex] and evaluate your line integral.
     
    Last edited: May 11, 2012
  4. May 11, 2012 #3
    Keep in mind there are three basic ways of integrating over a curve (line integral), you can integrate over the arc-length (ds), but also over the shadow of the curve along the x and y axis (by dx or dy). Now you wrote it as dx, so that's just a regular integral:

    [tex]\int_C f(x,y)dx=\int_C f(x,y(x))dx=\int_{-1}^1 e^x dx[/tex]

    If it were ds, then you'd need to use the formula you posted.
     
  5. May 11, 2012 #4
    But I am suppose to integrate over the arclength(ds). I'm suppose to parametrize the curve with respect to t so that the curve imoves along 1 unit of length per unit of time. That's how I got the bounds for the integral
     
  6. May 11, 2012 #5

    sharks

    User Avatar
    Gold Member

    Post the entire question correctly.
     
  7. May 11, 2012 #6
    The entire question is posted correctly.
     
  8. May 11, 2012 #7

    sharks

    User Avatar
    Gold Member

    In that case, you won't have any use for the relevant equations that you've posted. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook