Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Evaluation of a (parabolic) line integral with respect to arc length

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Evaluate the line integral [tex]\[ \int_c yz\,ds.\][/tex]

    where C is a parabola with z=y^2 , x=1 for 0<=y<=2


    2. Relevant equations
    A hint was given by the teacher to substitute p=t^2 , dp=(2t)dt and use integration by parts.

    I also know from other line integrals with respect to arc length that:

    ds=sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)



    3. The attempt at a solution

    I think that from the information given, the beginning and end points are (1,0,0) to (1,2,4).

    My first guess is:
    x(t) = t
    y(t) = 2t
    z(t) = t^2
    This will be when t goes from 0 to 2.

    So after I have parameterized the curve, I would substitute the functions of t back into the integral to get:

    int((2t)^3*sqrt(1^2+2^2+(2t)^2),t,0,2)

    =8*int(t^3*sqrt(4t^2+5),t,0,2)

    =12032/3

    This doesn't look right to me though. Any help would be appreciated!
     
    Last edited: Mar 10, 2009
  2. jcsd
  3. Mar 11, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi GelatinousFur! Welcome to PF! :smile:
    No, x is constant

    x(t) = 1 :wink:

    (and your y and z don't fit each other)
     
  4. Mar 11, 2009 #3
    Re: Welcome to PF!

    Ah, I see. Thanks for the help (and the welcome)!

    So then...

    X(t) is constant, so x(t)=1
    y(t) =t
    z(t) = t^2

    when t goes from 0 to 2.

    The line integral would then become:

    int(y^3,s) over the curve C, because z=y^2.

    =int(t^3*sqrt(0^2+1^2+(2t)^2),t,0,2)

    =int(t^3*2t,t,0,2)

    =2*int(t^4,t,0,2)

    =64/5

    This answer feels a bit more correct but I still cannot see why the teacher gave us the hint to use integration by parts, as I didn't have to when I just performed that integral.
     
  5. Mar 11, 2009 #4

    lanedance

    User Avatar
    Homework Helper

    Re: Welcome to PF!

    Hi

    in you integrand you have
    [tex] t^3\sqrt{1+4t^2} [/tex]

    I think you simplified away the one from the squareroot

     
  6. Mar 11, 2009 #5
    Re: Welcome to PF!

    Thanks, you are correct.

    The correct integral is:

    int(t^3*sqrt(1+4*t^2),t,0,2)

    So here's where I use integration by parts, but when I integrate sqrt(1+4*t^2) I have to go to an integral table.

    I punched int(t^3*sqrt(1+4*t^2),t,0,2) into MATLAB and it spits this answer out:

    -1/64/pi^(1/2)*(-3128/15*pi^(1/2)*17^(1/2)-8/15*pi^(1/2))

    Is this the wrong answer? Looks a bit weird to me.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook