Integration of $2$-form on the sphere using the stereographic projection

Question

Let $\omega$ be the $2$ form $\omega = x \, dy \wedge dz - y \, dx \wedge dz + z \, dx \wedge dy$ on $S^2$. I want to integrate $\int_{S^2} \omega$ using the definition, with the stereographic projection ${\varphi}^{- 1} : {\mathbb{R}}^2 \to S^2 \setminus \{(0 , 0 , 1)\}$ given by $$ {\varphi}^{- 1}(u , v) = \left(x = \frac{2 u}{1 + u^2 + v^2} , y = \frac{2 v}{1 + u^2 + v^2} , z = \frac{u^2 + v^2 - 1}{1 + u^2 + v^2}\right). $$ Then $$ \int_{S^2} \omega = \int_{{\mathbb{R}}^2} {\left({\varphi}^{- 1}\right)}^*(\omega). $$ I proceed to calculate ${({\varphi}^{- 1})}^*(\omega)$. It is $$ {\left({\varphi}^{- 1}\right)}^*(\omega) = x {\left({\varphi}^{- 1}\right)}^*(dy) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) - y {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) + z {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dy). $$ For instance, $$ {\left({\varphi}^{- 1}\right)}^*(dx) = \frac{\partial x}{\partial u} \, du + \frac{\partial x}{\partial v} \, dv = \frac{2 (1 + u^2 + v^2) - 4 u^2}{{(1 + u^2 + v^2)}^2} \, du - \frac{4 u v}{{(1 + u^2 + v^2)}^2} \, dv $$ and similarly $$ {\left({\varphi}^{- 1}\right)}^*(dy) = - \frac{4 u v}{{(1 + u^2 + v^2)}^2} \, du + \frac{2 (1 + u^2 + v^2) - 4 v^2}{{(1 + u^2 + v^2)}^2} \, dv $$ and $$ {\left({\varphi}^{- 1}\right)}^*(dz) = 4 \left(\frac{u}{{(1 + u^2 + v^2)}^2} \, du + \frac{v}{{(1 + u^2 + v^2)}^2} \, dv\right). $$ We calculate now the exterior products: $$ x {\left({\varphi}^{- 1}\right)}^*(dy) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) = - \frac{16 u^2}{{(1 + u^2 + v^2)}^4} \, du \wedge dv, $$ $$ - y {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) = - \frac{16 v^2}{{(1 + u^2 + v^2)}^4} \, du \wedge dv, $$ $$ z {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dy) = - 4 \frac{{(u^2 + v^2)}^2 - 2 (u^2 + v^2) + 1}{{(1 + u^2 + v^2)}^4} \, du \wedge dv. $$ Therefore $$ {\left({\varphi}^{- 1}\right)}^*(\omega) = \frac{4}{{(1 + u^2 + v^2)}^4} (- 2 u^2 - 2 v^2 - 1 - u^4 - 2 u^2 v^2 - v^4) \, du \wedge dv $$ if I did not have any mistake. But how can I proceed with this expression? On the other hand, I know that the integral should be $4 \pi$.

Answer 1

Your results so far are actually all correct. To proceed, you just need to be a little less eager in expanding all the expressions, but opt for factoring more. In particular, the result for $(\phi^{-1})^* (z\, dx\wedge dy)$ can be factored. The numerator is actually just $4(u^2+v^2-1)^2$. When you add up the pullback of the other two terms, you are adding $16(u^2+v^2)$ to the numerator. Thus you get $4(u^2+v^2+1)^2$, which cancels neatly with the denominator.

Alternatively, if you are familiar enough with the expansion of $(x+y+z)^2$, you can recognize straight away that the numerator of your final result is $4(u^2+v^2+1)^2$.

To proceed with the integral, you can either convert the integral into polar coordinates in the $uv$-plane, or do it with trig substitutions. The former method is easier by far.