Prove the sphere is orientable Is there an easy way to show that the sphere
$$\mathbb S^n = \{ x\in \mathbb R^{n+1} : \|x\| =1\}$$
is orientable other then using stereograohic projection? I am preferably looking for something derived from a basic theorem in elementary geometry with respect to the unit normal.
 A: One way to do it is finding a volume form for $\mathbb{S}^n$.
Obviously, using the unit normal assumes the embedding $\iota:\mathbb{S}^n\hookrightarrow \mathbb{R}^{n+1}$.
The idea is to take the usual volume form on $\mathbb{R}^{n+1}$, namely $\eta:=dx_1\wedge...\wedge dx_{n+1}$, contract it with respect to the radial field $X=\sum_{i=1}^{n+1}x_i\,\frac{\partial}{\partial x_i}$ and then restrict it to $\mathbb{S}^n$ (notice that $X|_{\mathbb{S}^n}$ is the unit normal).
More precisely, define $\omega:=\iota^*(i_X\eta)$. This is an $n$-form on $\mathbb{S}^n$. To prove that $\omega$ is a volume form, we take a basis $\{v_1,...,v_n\}$ for $T_p\mathbb{S}^n$ and prove that $\omega(v_1,...,v_n)\neq 0$.
Since $T_p\mathbb{S}^n=\text{span}(X_p)^\perp$, then $\{X_p,v_1,...,v_n\}$ is a basis for $\mathbb{R}^{n+1}$. But $\eta$ is a volume form on $\Bbb{R}^{n+1}$, hence
$$\omega(v_1,...,v_n)=\eta(X_p,v_1,...,v_n)\neq 0.\,\,_\blacksquare$$
A: Yes, any level set $X=\{f=0\}$ of a smooth function $f\colon U \to \mathbb{R}$ with $\nabla f \ne 0$ on $\{f=0\}$ is an orientable manifold ( a submanifold of $U$). Here $U$ is an open subset of $\mathbb{R}^n$, or, more generally, an orientable manifold. 
To get the orientation: Consider the gradient field $\text{grad} f$ (we  need a Riemannian structure on $U$, the standard one if $U$ is a subset of a numeric space). At each point $x$ in $\{f=0\}$ we have $\text{grad}f(x) \perp T_x(X)$. Choose the orientation on $T_x(X)$ as follows: a basis in $(e_1, \ldots , e_{n-1})$ of $T_x(X)$ is positively oriented if $(e_1, \ldots, e_n, \text{grad} f(x))$ is positively oriented in $T_x(U)$. 
A: Here's an orientable atlas with three charts obtained using ordinary spherical coordinates.
Use spherical coordinates $0<\phi<\pi$, $0<\theta<2\pi$, where $x=\sin\phi \cos\theta$, $y=\sin\phi \sin\theta$, $z=\cos\phi$ as a coordinate chart for the complement of the prime meridian (where $\theta=0$). 
Use the exact same spherical coordinate formulas but with domains $0<\phi<\pi$, $-\pi<\theta<\pi$ as a coordinate chart for the complement of the meridian $\theta=\pi$.
So far this gives charts that cover all but the north pole $(0,0,1)$ and the south pole $(0,0,-1)$.
Finally, to cover the two poles, change your Euclidean coordinates to $x'=y$, $y'=z$, $z'=x$, and rewrite the spherical coordinate formulas as $x'=\sin\phi \cos\theta$, $y'=\sin\phi \sin\theta$, $z' = \cos\phi$.
