Find the length of the boundary from parametric equation I have to find the length of the boundary for the parametric equation
$\begin{bmatrix}x \\y \\z \end{bmatrix}=\begin{bmatrix}e^u+e^{-u} \\2u \\v(e^u-e^{-u}) \end{bmatrix}, 0\le u\le1,0\le v\le1$
I know it can be done by finding $\int_{a}^{b}\sqrt{f'(t)^2+g'(t)^2+h'(t)^2}$ but I'm unsure of how to do it, while accounting for both variables.
I hope someone can help me with this?
 A: The parameter domain is the square $[0,1] \times [0,1]$ in the $uv$-plane. The key thing is that the boundary of the parameter domain is mapped to the boundary of the surface, so you can split that boundary of the parametric surface into four (parametric) curves corresponding to the sides of the square in the parameter domain:


*

*for $v=0$, let $u:0 \to 1$;

*for $v=1$, let $u:0 \to 1$;

*for $u=0$, let $v:0 \to 1$;

*for $u=1$, let $v:0 \to 1$.


For example, in the first case with $v=0$, the parametric curve corresponding to one of the four 'sides' is given by:
$$\begin{pmatrix} x(u) \\ y(u) \\ z(u) \end{pmatrix} =
\begin{pmatrix} e^u+e^{-u} \\ 2u \\ 0 \end{pmatrix} \quad,\quad 0 \le u \le 1$$
And the length of the curve is given by the formula you had in mind:
$$\int_{0}^{1} \sqrt{x'(u)^2+y'(u)^2+z'(u)^2} \,\mbox{d}u = \ldots$$
You can do this for each of the four sides.

To get an idea, you can use WolframAlpha for a plot of the surface and (one of) the curve(s):
$\quad\quad$  $\quad\quad\quad$ 
Note that it automatically gives you the arc length as well - so you can check.
