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Line Integrals

  1. Sep 24, 2007 #1

    This comes from Mathematical Methods for the Physicist from Susan Lea, Chapter 1 Question 25 Part B (Incase anyone is familar with the book). The question asks to evaluate the line integral
    integral (u*dl)

    where the vector u is:
    u = x*y^2 i + y*x^2 j

    along the path of a semi-circle in the xy plane with a radius of a and the flat side of the semi-circle along the x-axis

    2. Relevant equations

    see below

    3. The attempt at a solution

    I am near certain the answer is 0, evaluating this in polar coor. gives me 0 and taking advantage of the del operator to find the surface integral gives me 0, so two methods out of three give me zero.

    however when I evaluate the integral in cartesian cord. I end up with 1/4 * a^4 for the half circle path and 0 for the bottom path. I am almost sure that the bottom path should not be zero for this to work, but I have no idea how to find it.

    My sanity really depends on knowing how I messed this up, because it looks so good on my paper :)

    Anyways, thanks for the help

    PS. Anyone know a good way I can enter equations onto the forums, I don't think typing it this way really makes the situation clear.
    Last edited: Sep 24, 2007
  2. jcsd
  3. Sep 24, 2007 #2
    Check out this for some info on LaTeX.

    As for the problem, note that the vector field is conservative (with the potential function being [itex]F = \frac{1}{2}x^2y^2[/itex]). Since you're going along a closed path, the answer should be zero.
  4. Sep 25, 2007 #3


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    Homework Helper

    As durt points out, this is a "conservative" field, so the line integral on a closed path within it will be zero. An equivalent form of saying this, if you'd have Stokes' Theorem yet, is that since this field has a curl of zero (calculate it or take a look at the symmetry of the field directions among the quadrants), the line integral around the path will be zero.

    I think you'll find that the (u*dl) contributions between the first and second quadrants on the semicircle cancel each other, giving zero on that portion of the closed path. As for the portion along the x-axis, y = 0 there, so you have u = 0 all along that section.
  5. Sep 25, 2007 #4
    Thanks for the info, I'm going to have to look at this again. dynamicsolo, your absolutely correct in that when you look at Stoke's Theorem it pops out right away that the answer is zero. As for looking at the semi-circle part of the integral, I worked on it for a couple of hours today, and I didn't end up with zero, though I have to admit, I believe you more than all my work, just because it makes more sense.
    Thanks for the help, and durt thanks for the info on LaTex, tomorrow will be another great opportunity to bury my head in the books!
  6. Sep 25, 2007 #5


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    Staff Emeritus
    Science Advisor

    You made a mistake in the integration.

    If you insist on staying in xy- coordinates, you have x2+ y2= a2 so [itex]y^2= a^2- x^2[/itex] or [itex]y= \sqrt{a^2- x^2}[\itex] and 2ydy= -2xdx
    [tex]\int xy^2 dx+ \int x^2y dy= \int x(1-x^2)dx+ \int \sqrt{a^2- x^2}(x^2)(-2x)dx[/itex]
    The first integral, since it is of an odd function, will be 0 when integrated between -a and a. Make the change of variable u2= a2- x2 and you will see that the second integral and you will see that it is also an odd function in u. Again, the integral is 0.

    Of course, the easy way to integrate over the semi-circle is to use the parametrization [itex]x= a cos(\theta)[/itex] and [itex]y= a sin(\theta)[/itex]. Then the integral becomes [tex]\int_0^\pi a^3 (cos^3(\theta)sin(\theta)- sin^3(\theta)cos(\theta))d\theta[/tex]

    Since those are odd powers of sine and cosine it is easy to substitute [itex]u= sin(\theta)[/itex] and see that the integral is 0.

    Since I am more math oriented than physics oriented, I would say the integrand is "an exact differential" rather than "conservative" but the result is the same: the integral around any closed path is 0.

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