I've just learned about line integrals, and I need some help understanding an example problem in my textbook.

The question is supposed to be really easy. Integrate $f(x,y,z)=x-3y+z$ over the line segment $C$ joining the origin to the point $(1,1,1)$.

I've been told to parametrize the line segment first, and this is the part I'm confused about. My knowledge on parametrizations is a bit wobbly.... From what I remember, I thought $ \bar{r}(t)$ would be $ \bar{r}(t)=<1,1,1> + t<1,1,1>= <1+t,1+t,1+t>=(1+t) \hat{i}+(1+t) \hat{j}+ (1+t) \hat{k} $

but my textbook says $ \bar{r}(t)=t\hat{i}+t\hat{j}+t\hat{k}$, with $0 \leq t \leq 1$

Why is this? what does this mean? I would be grateful for a bit of help!

  • $\begingroup$ The line segment joins the origin to the point $(1,1,1)$, so the y-intercept is $(0,0,0)$ not $(1,1,1)$. Therefore, the position vector in your example should be $\bar{r}(t)=<0,0,0> + t<1,1,1> = t\hat{i} + t\hat{j} + t\hat{k}$ $\endgroup$
    – Tymric
    Commented Jul 10, 2014 at 10:22
  • $\begingroup$ oh right!!!! gosh I am so stupid. of course it is!! thank you!! @Timmy $\endgroup$
    – user125342
    Commented Jul 10, 2014 at 11:08

1 Answer 1


Basically, when you parametrize the line segment, the form of the vector is $$\vec{r}(t) = <x_0, y_0,z_0> + t<x_1, y_1,z_1>$$ Where $(x_0, y_0, z_0)$ is your initial point, and $(x_1, y_1, z_1)$ is your final point, and $t$ always have to be from $0\le t \le 1$ (do you see why?).

Think back to lines and how you parametrized them. This is no different, except there's a limit of $t$.

EDIT: The vector $<x_0, y_0,z_0>$ is the vector to your initial point, and the vector $<x_1, y_1,z_1>$ is the displacement vector between your two points, and the reason $t$ is bounded by $0\le t \le 1$ is because when $t = 0$, we get our initial point, and then when $t = 1$, we get our full length line segment, and all other values of $t$ are for all the other line segments that are pieces of the full length one.


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