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Line Integrals - Cartesian and Parametric

  1. Aug 13, 2006 #1
    Hello

    Im working on some line integral problems at the moment. The first one is really only a check - I think ive worked it out...

    Compute the line integral of the vector field B(r) = x^2 e(sub 1) + y^2 e(sub 2) along a straight line from the origin to the point e(sub 1) + 2 e(sub 2) + 4 e(sub 3). Explain why your answer should be independant of the path of integration.

    So quickly - z = 2y = 4x
    y = 2x so y^2 = 4x^2 -> dy = 8x dx

    thus B(r) = x^2 e + 4x^2
    so we integrate this over 0->1 (as we converted all the bits to x) and thus I end with an answer of 11.

    The second part I think reads something like "because the line integral depends only on the arc length", but really thats something I read off wikipedia or somewhere so am not particularly sure about :)

    The other question is proving more tricky...

    Compute the line integral of a vector field F = xe(sub1) + ye(sub 2) + 2ze(sub 3) along a circular helix between the two points (a,0,0) and (a,0,b) parametrized by r = (a cos phi, a sin phi, (b/2*pi) phi). Can you check your result using a differant path and why? Do so for example using a direct line.

    So as before I might go ahead and say x = y = 1/2 z, but then I wouldnt know where to go with the parametric component to the question. Any pointers in this would be great - I think I have to differentiate the parametric part, but then dont know where to go from there (suspect some cross-producting in there, such as int(r x r') but im really not very confident any hypothesis I have!
     
  2. jcsd
  3. Aug 13, 2006 #2

    HallsofIvy

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    What you say you did seems right but the answer is not "11"! Yes, on the line through (0,0,0) and (1, 2, 4), y= 2x and z= 4x as x goes from 0 to 2. Another way of putting that is that r= te1+ 2te2+ 4t3 where I have more or less arbitrarily take "t" as the parmeter (since x= t, I could as easily have used x). Of course, dr= (e1+ 2e2+ 4e3)dt since this is a straight line. Yes, the function to be integrated is, in your terms, x2e1+ 4x2e2 or, in terms of my t, t2e1+ 4t2e2. The dot product is then t2+ 8t2= 9t2. Since x (or t) ranges between 0 and 1, the integral is
    [tex]\int_0^1 9t^2 dt[/tex]
    That integral is not 11!

    There are two ways to show that the integral is independent of the path. We can always write dr as dxe1+ dye2+ dze3. Since the first component of the function to be integrated involves only x, the second only y, and the third is 0, we can write the dot product as x2dx+ y2dy, no matter what the path is and do those as two separate integrals:
    [tex]\int_0^1 x^2 dx+\int_0^2 y^2dy[/tex]
    That can be integrated without regard for the path and is the same as the integral above.

    The other method is more general. The integral will be independent of the path if and only if the integrand is an "exact differential". That is, if there exist a function F(x,y,z) having the integrand as its differential:
    [tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z} dz[/tex]
    which can be integrated by evaluating F at the endpoints no matter what the path is.
    Since, for a simple function with continuous partials, the "mixed second derivatives" must be equal, that is
    [tex]\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial^2 F}{\partial y\partial x}[/tex]
    f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz is an "exact differential" if and only if
    fy= gx, fzz= hx, and gz= hy. In this problem it is easy to see that all of those partial derivatives are 0 and so equal.

    No, x= y= 1/2 z is incorrect. I guess you get that from the formula for F= xe1+ ye2+ ze3 but that is NOT the equation for the path. And, no, there is no cross product. The integral of F on a path r is the dot product [itex]F\cdot r[/itex] Since you are given that
    [tex]r= (a cos\phi, a sin\phi, \frac{b}{2\pi}\phi)[/tex]
    you are given that
    [tex]x= a cos\phi, y= a sin\phi, z= \frac{b}{2\pi}\phi[/tex]
    so that
    [tex]dr= (dx, dy, dz)= (-a sin\phi, a cos \phi, \frac{b}{2\pi})d\phi[/tex]
    The function to be integrated is
    [tex]F= (x, y, z)= (a cos\phi, a sin\phi, \frac{b}{2\pi}\phi)[/tex]
    and so the integral is of the dot product:
    [tex]\int_0^{2\pi}(-a^2sin\phi cos\phi+ a^2sin\phi cos\phi+ \frac{b^2}{4\pi^2}\phi d\phi= \frac{b^2}{4\pi^2}\int_0^{2\pi}\phi d\phi[/tex]

    Of course, the straight line from (a, 0, 0) to (a, 0, b) is given by
    x= a, y= 0, z= bt with t going from 0 to 1. That is r= (a, 0, bt) so
    dr= (0, 0, b)dt. The function is given by F= (x, y, z)= (a, 0, bt}. The integral is [itex]\int_0^1 b^2 t dt[/itex]. Check to see if those two integrals are the same. (For exactly the same reasons, the answer should be "yes".)
     
  4. Aug 14, 2006 #3
    In general, it would be good to destinguish between "exact" and "closed" forms.
     
  5. Aug 15, 2006 #4
    just a few bits I dont understand still, bear with me if you will :)

    the first bit you say we need to get the intergral(9t^2), but I could have done that right from the start, AFTER saying B(r) = this and r = that. You later go on to say "The integral of F on a path r is the dot product F.R" - so I do all that but dont really need the dr part, the y = 2x part, or even to integrate x^2 + 4x^2.

    Is what your suggesting here to just get B (or F!) and r and dot them to get the line integral ?

    For the second part I think you have a few mistakes but im basically following...

    F . dr is what I want to calculate where...

    F = (x,y,2z) = (a cos phi, a sin phi, 1/pi phi)
    dr = (-asin phi, a cos phi, b / 2 pi) d phi

    Then I integrate this - only I dont understand where you have got the limits from!

    Sorry for my slow reply to this thread, I work quite alot and this is the first chance ive had to look back in at it :) I really appreciate the work you have put into it!
     
  6. Aug 15, 2006 #5
    Hey, if you are who I think you are... good luck tomorrow I'll be there too :-)
     
  7. Aug 16, 2006 #6
    Im scared how all through my life everybody knows me and I know nobody... Maybe im the guy from the truman show *shifty eyes*

    Also, it seems /everybody/ who takes it will be there! :S Evil maths for the lose.
     
    Last edited: Aug 16, 2006
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