let $M$ one $k$- manifold orientable on $\mathbb R^n$ with boundary and $p\in \partial M$ and $\alpha$ one coordinate patch around $p$. i know that the induced orientation on $\partial M$
By Munkres (Analysis on Manifolds, page 318) $\alpha_0=\alpha\circ b$ where $b:\mathbb R^{k-1}\to \mathbb R^k$ such that $b(x_1,\ldots,x_{k-1})=(x_1,\ldots,x_{k-1},0)$, $\alpha_0$ is one coordinate patch on $p\in \partial M$. Munkres say if $k$ is even then $\alpha_0$ belongs on induced orientation of $\partial M$ and if $k$ is odd then belongs the oposite orientation. Hence $\det [(-1)^nn :D\alpha_0]>0$ is say $\partial M$ corresponds with the orientation induced by $M$.
In particular, if $M$ is defined by $\{(x,y,z)\in \mathbb R^3: x^2+ y^2=1, 0\leq z\leq 1\}$. Consider the orientation on $M$ sucht that the coordinate system $\alpha:(0,1)^2\to \mathbb R^3$ defined by $\alpha(u,v)=(\cos 2\pi u,\sin 2\pi u,v)$ belongs to the orientation of $M$. I calculated the unitary normal vector on $M$ where $n=(\cos 2\pi u, \sin 2\pi u, 0)$ , but when i applied $\alpha_0$ to find orientation on $\partial M$ then
$\alpha_0(x)=\alpha\circ b(x,0)=\alpha(x,0)=(\cos 2\pi x,\sin 2\pi x,0)$
but is not possible calculate $\det [(-1)^nn :D\alpha_0]>0$ because the result is not a square matrix
I want to find the orientation of $\partial M$ compatible with $M$, what need to do? does has other way, can somebody help me please!