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Line Integral by two methods

  1. Nov 22, 2008 #1
    Recently I was working through a problem involving a force field, and came up with a question I could not answer, so I thought I would post it here. I solved the problem using a vector representation and a line integral, and although I am sure the answer is correct, I would like to solve it by a slightly different method. My question is about the method.

    Here is the original problem: I have a vector field x^3 i + 3zy^2 j + -x^2y k and I am calculating the line integral along the straight line segment passing through points (3,2,1) to (0,1,0)
    I set this up in terms of u : (3-3u) i + (2-u) j + (1-u) k
    Now I substituted into the vector field to get:
    [(3 - 3u)^3 (-3)] + [3 (1 - u) (2 – u)^2 (-1) ] + [(3 -3u)^2 (2 – u) (-1)]
    This is now integrated in respect to u between the limits 0 to 1 yielding: -19.25

    So far so good. Now what I would like to do is integrate the same field in terms of (x,y,z) instead of u. To do that I need to derive a path as a function of (x,y,z) from the given points. That is my question! Of course, I know how to derive the function for two points given in (x,y) by using the point-slope formula. But it has somehow escaped my memory on how to derive a three variable function from the given points.

    I’m sure someone here knows how to do this. Can you help?
  2. jcsd
  3. Nov 22, 2008 #2


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    What do you mean by that? You can't write a line or curve as a single equation of 3 variables. A line is one-dimensional. If you were to write a single equation f(x,y,z)= 0 you could choose any two of those variables and solve for the third: that defines a two-dimensional surface.

    This was typed first-then I realized I had misunderstood your question so I went back and changed- but I forgot to erase this:
    You already have that [itex]\vec{r}(u)= (3-3u)\vec{i}+ (2-u)\vec{j}+ (1-u)\vec{k}[/itex][/itex] and now you want to use x itself as parameter? Let x= 3- 3u. Then 3u= 3- x so u= 1- x/3. Replace u by that in the formula: [/itex]\vec{r}(x)= x\vec{i}+ (5- x/3)/vec{j}+ (4- x/3)\vec{k}[/itex]
    I thought when I started that that you were asking how to write the curve as a function of x, not as a function of "x, y, z".
    Last edited: Nov 23, 2008
  4. Nov 22, 2008 #3
    It seems you didn't quite finish what you started to say, but you told me what I needed to know; it can't be done by defining a path in (x,y,z) even though the path is in three dimensional space it is still a straight line. Thanks.
  5. Nov 23, 2008 #4


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    I have gone back and editted what I wrote before.
  6. Nov 23, 2008 #5
    OK. I see it now. Apparently there is still something wrong with the LaTex, but I get the idea. All I need to do is a reverse sunstitution from terms of u to terms of x,y,z. Thank you.

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