# Line Integral by two methods

1. Nov 22, 2008

### schroder

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. Nov 22, 2008

### HallsofIvy

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 $\vec{r}(u)= (3-3u)\vec{i}+ (2-u)\vec{j}+ (1-u)\vec{k}$[/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 by a moderator: Nov 23, 2008
3. Nov 22, 2008

### schroder

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.

4. Nov 23, 2008

### HallsofIvy

I have gone back and editted what I wrote before.

5. Nov 23, 2008

### schroder

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.