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Finding integral of dot product of F and dl

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data
    1. For the vector field
    F = yz ˆx + zx ˆy + xy ˆz
    (^x means the unit vector of x)
    find the integral of F • dl from (0, 0, 0) to (1, 2, 3) in Cartesian coordinates
    in each of the following ways:
    (a) along a straight line path from (0, 0, 0,) to (1, 2, 3)
    (b) along straight line paths from (0, 0, 0,) to (1, 0, 0), then from (1, 0, 0)
    to (1, 2, 0), then from (1, 2, 0) to (1, 2, 3)
    (c) without choosing a particular path


    for part a), i used the dl as (^x + 2 ^y + 3 ^z)dx since vector r= ^x + 2 ^y + 3 ^z
    thus, integral of F dot dl = yzx + z(x^2) + 3/2 (x^2)y evaluated from x=0 to x=1, and get yz+z+3/2y

    for part b), i added the three different solutions from path 1, path 2, and path 3.
    path1: using dl1=dx ^x, integral of path 1 of F*dl1 = yz
    path2: using dl2=dy ^y, integral of path 2 of F*dl2 = 2zx
    path3: using dl3=dz ^z, integral of path 3 of F*dl3 = 3xy
    so my solution to part b) is yz+2zx+3xy

    i was wondering, are my solutions to part a) and part b) correct? If not, please guide me through on how to get the solutions to part a,b, and c. thanks alot!
     
  2. jcsd
  3. Oct 9, 2008 #2

    Defennder

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    Homework Helper

    No, dl is always [tex] \left(\begin{array}{c}dx\\dy\\dz \end{array} \right )[/tex]. Your final answer is supposed to be a numerical value anyway. A routine method would be to parametrise the path taken to evaluate the line integral along the straight line path. However, in this case since the path taken is relatively simple, we can perform the line integral without the parametric method. So firstly find a way to express x,y,z in terms of each other along that straight line path. The vector path representation of that straight line is [tex]\mathbf{r}(t) = t \left(\begin{array}{c}1\\2\\3 \end{array} \right )[/tex] as can be seen just by visualising the path.

    So from the above you can deduce that along the straight line path, x=t, y=2t,z=3t and from this, x=1/2y=1/3z. Now that you've expressed x,y,z in terms of each other you can perform the line integral.

    So we have [tex]\int_{(0,0,0)}^{(1,2,3)} \left(\begin{array}{c}yz\\zx\\xy \end{array} \right ) \cdot \left(\begin{array}{c}dx\\dy\\dz \end{array} \right ) [/tex]. So now, you just need to evaluate each of the dot products, perform the integral for each dx,dy,dz, taking care to express the Fx,Fy,Fz components in terms of x when integrating with respect to dx, in terms of y when integrating wrt to dy and so on.

    As above, your answer here is incorrect because you're supposed to get a numerical answer. For b) it's just performing multiple line integrals by the method explained above. For c) the way the question is phrased serves as a strong hint as to how one can perform a line integral without specifying any path at all. What kind of vector field do you think F is such that the path taken for line integrals doesn't matter? How can you verify that?
     
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