How to calculate differential forms on $S^2$ parameterized by stereographic projection? Suppose that we have the stereographic mapping $\varphi: \mathbb{R}^2\to M$ where $M=S^2-\{(0,0,1)\}$.
I've already found that the stereographic parametrization of $S^2-\{(0,0,1)\}$ is given by:
$$(x,y) \mapsto (\dfrac{2x}{1+x^2+y^2},\dfrac{2y}{1+x^2+y^2},\dfrac{x^2+y^2-1}{1+x^2+y^2})$$
Normally, if I want to find a basis for $T_p(M)$, I can find $\dfrac{\partial{\varphi} }{\partial x}$ and $\dfrac{\partial{\varphi} }{\partial y}$ and then consider the plane that is spanned by $<\dfrac{\partial{\varphi} }{\partial x},\dfrac{\partial{\varphi} }{\partial y}>$ at each point $p$. I've already found these two vectors to be:
$$\dfrac{\partial{\varphi} }{\partial x}=<\dfrac{2(1-x^2+y^2)}{(1+x^2+y^2)^2}, \dfrac{-4xy}{(1+x^2+y^2)^2}, \dfrac{4x}{(1+x^2+y^2)^2}>$$
$$\dfrac{\partial{\varphi} }{\partial y}=<\dfrac{-4xy}{(1+x^2+y^2)^2}, \dfrac{2(1+x^2-y^2)}{(1+x^2+y^2)^2}, \dfrac{4y}{(1+x^2+y^2)^2}>$$
The vectors are orthogonal at all points and therefore create a basis for $T_p(M)$. The problem is that calculation with this basis looks very very hard, almost impossible to be done by hand. 
Now the exercise asks me to show that the surface element on $M$, i.e. $dS$ is given by the differential form $\omega = x \space dy \wedge dz + y \space dz \wedge dx + z \space dx \wedge dy $.
Apparently, the surface element on $M$ at the point $p$ should be given by $\|\dfrac{\partial{\varphi} }{\partial x} \times \dfrac{\partial{\varphi} }{\partial y} \|_p $. So I should show that when we input $<\dfrac{\partial{\varphi} }{\partial x},\dfrac{\partial{\varphi} }{\partial y}>$ into $w$ it gives the same result. Am I right?
But how can we do that? The calculations seem to be crazily hard. Even wolfram alpha couldn't do it. Is there another way to approach this problem?
 A: Using the stereographic coordinates for this problem is a trap, especially since the desired final answer is in the $(x,y,z)$ coordinates anyway. The easiest way to approach this problem is to remember that the metric on $S^2$ comes from the embedding $S^2 \to \mathbb R^3$. 
For any isometric immersion like this we have (up to sign/after choosing orientations appropriately) the formula $dS = \iota_\nu dV$ where 


*

*$dV$ is the volume form of $\mathbb R^3$

*$\nu$ is a smooth unit normal vector field to $S^2$

*$\iota$ is contraction/the interior product. 


This formula should be somewhat intuitive: whenever you have an orthonormal basis for $T_pS^2$ you can add $\nu|_p$ to get an orthonormal basis for $T_p \mathbb R^3$.
In this case we can use $\nu = (x,y,z)$, and thus the linearity of $\iota$ gives
$$ dS = x\ \iota_{\partial_x} dV + y\ \iota_{\partial_y} dV + z\ \iota_{\partial_z} dV.$$
Now remember that $dV = dx \wedge dy \wedge dz = dy \wedge dz \wedge dx = dz \wedge dx \wedge dy$; so contracting $dV$ with $\partial_{x^i}$ just removes the $dx^i$. Thus
$$ dS = x\ dy \wedge dz + y\ dz \wedge dx + z\ dx \wedge dy $$
as desired.
