# Line Integral Parameterization

#### dgiroux48

##### New member
1. Consider the curve c= (x(t),y(t),z(t)) in space as t varies over [0, T ]. We could also parameterize this curve by
c= x(τ^2 ),y(τ^2 ),z(τ^2) τ ∈ [0, sqrt(T)].
Show that one obtains the same value for the line integral using either parameterization.

The line integral is just the integral for the arc legnth of the parameterized curve. I understand the concept intuitively, I just don't really know how to derive it.

(I dont know how to write out integral signs and all here but the formula can be found at this site Pauls Online Notes : Calculus III - Line Integrals - Part I)

Thanks!

#### HallsofIvy

##### Well-known member
MHB Math Helper
Looks to me like it is just a matter of using the chain rule.

We are saying that (x(t), y(t), z(t)), for $$0\le t\le T$$ and $$(x(\tau^2), y(\tau^2), z(\tau^2))$$ are parameterizations of the same curve: $$t= \tau^2$$ so that $$dt= 2\tau d\tau$$.

#### ModusPonens

##### Well-known member
1- $\int_{C_1} f(x,y,z)ds=\int_0^T f(x(t),y(t),z(t))||r(t)||dt$.

2- On the other hand, $\int_{C_2}f(x,y,z)ds=\int_0^{\sqrt{T}} f(x(t^2),y(t^2),z(t^2))||r(t^2)|| |2t| dt$, where the $|2t|$ comes from the computation of $||r(t^2)||$, since it's a composition of functions.

So now, all you have to do is to note that the second integral in 1 is the same as the second integral in 2, because it is the result of changing the variables.