I am revisiting the following problem:
Let $\iota : S^3 \hookrightarrow \mathbb{R}^4$ be the inclusion and let $\omega : =\iota^\ast \alpha$ where $$\alpha : = \sum_{i = 1}^4 (-1)^{i - 1} x_{i} \, dx_{1} \wedge \cdots \wedge \widehat{dx_{i}} \wedge \cdots \wedge dx_{4} \in \Omega^3(\mathbb{R}^4).$$ Evaluate $\displaystyle \int \limits_{S^3} \omega$.
If we use polar coordinates \begin{eqnarray*} &F : (0,\pi) \times (0, \pi) \times (0, 2\pi) \longrightarrow S^3 \subseteq \mathbb{R}^4 & \\ & (\theta^1 , \theta^2 , \theta^3)\longmapsto \begin{pmatrix}\cos \theta^1 \\ \sin \theta^1 \cos \theta^2 \\ \sin \theta^1 \sin \theta^2 \cos \theta^3 \\ \sin \theta^1 \sin \theta^2 \sin \theta^3\end{pmatrix} & \end{eqnarray*} Then it is not too hard to show $F^\ast \omega = \sin^2 \theta^1 \sin \theta^2 \, d \theta^1 \wedge d \theta^2 \wedge d \theta^3$, which implies $$ \int\limits_{S^3} \omega = \int\limits_{0}^{2 \pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\pi} \sin^2 \theta^1 \sin \theta^2 \, d \theta^1 d \, \theta^2 \, d \theta^3 = 2 \pi^2. $$ Now I did this problem some time ago (so I am trusting that my notes are correct) and it was not too difficult; it was just a matter of taking the differential of relatively simple functions and applying basic properties of the wedge product. However, it was an incredibly long and tedious computation. Is there a quicker way to compute $F^\ast \omega$?
I realize now that I can apply Stokes's Theorem to make things a little bit shorter: If we let $D^4$ denote the unit ball in $\mathbb{R}^4$, then we have $$ \int\limits_{S^3} \omega = \int\limits_{D^4} d \omega = \int\limits_{D^4}\iota^\ast d \alpha. $$ One can show $d \alpha = 4 \, dx_{1} \wedge d x_{2} \wedge dx_{3} \wedge dx_{4}$. We can again convert this to polar so that it is easier to integrate but it is still a somewhat lengthy calculation (though shorter than what is above).
The original problem also mentions that $\alpha = i_{E}(d V)$, where $E$ is "Euler's Vector Field," i.e. $E|_{x} = x$. I never used this in any of my work, so is there a way to use this in order to simplify things?