# Line Integrals

1. Dec 6, 2014

### dwn

1. The problem statement, all variables and given/known data

Find the line integral of f(x,y,z) = x+y+z over the straight-line segment from (1,2,3) to (0,-1,1).

2. Relevant equations

∫ f(x,y,z)ds = ∫ f(g(t), h(t), k(t)) |v(t)| dt

3. The attempt at a solution

I arrived at the correct solution, but I'd like some clarity on the result.

The final answer to this is 3√(14) or -3√(14) depending on which point you choose as your parametric equation.
x = -t
y = -3t-1
z = -2t+1
From using the point (0,-1,1) and (-1,-3,-2) as my direction vector.

What I would like to understand is the meaning of the positive and negative result. Does it matter?
It just seems to me that my result should have been positive since I am moving from a lower position to a higher position, no?

BTW, QUICK SHOUTOUT TO PF --- THE NEW SITE IS AMAZING! GREAT JOB ON THE NEW LAYOUT.

2. Dec 6, 2014

### SteamKing

Staff Emeritus
Which is the correct solution? You show 2 different values below.
How did you arrive at this parameterization for the line segment specified? For example, if t = 0, does your parameterization return the (x,y,z) of the first point on the line segment?

Yes, it matters. Some line integrals are path independent. Is this one?

How did you arrive at this conclusion? The OP states that the line integral is to be taken over the line segment from (1,2,3) to (0,-1,1), not the other way around.

3. Dec 6, 2014

### dwn

Correct Answer: 3√(14) but I got the negative term because I used the wrong point as my starting position.

Why does direction of the integral matter though? Because the value of the integral will remain the same. I see that they are asking us to go from a specified point and not the other, but is it really necessary?

Last edited: Dec 6, 2014
4. Dec 6, 2014

### SteamKing

Staff Emeritus
Is $\int_{a}^{b} f(x) dx = \int_{b}^{a} f(x) dx$ ?

5. Dec 6, 2014

### dwn

No it is not.

I actually have a question about a and b though. How do we determine the interval? I still don't quite understand that.

Last edited: Dec 7, 2014