Stokes theorem to find integral of differential form 
Let $$\alpha=2w dx\land dy\land dz + z dx\land dy\land dw + ydx\land dz \land dw + xdy\land dz\land dw.$$
Use Stokes theorem to calculate $\int_{\partial S}\alpha$ where $S$ is a domain bounded by $x=1-\left(y^2+z^2+w^2\right)$ and $-x=1-\left(y^2+z^2+w^2\right)$.


My attempt:
By Stokes theorem, I need to calculate $$\int_{\partial S}\alpha= \int_{S}{\rm d}\alpha\ .$$ I found ${\rm d}\alpha= dx\land dy\land dz\land dw$, but I am not sure how to figure out boundaries for $S$.
Do I just put the equations for $x$ and $-x$ as boundaries for $x$? What would boundaries for other variables be?
 A: The region $S$ is a kind of $4$-dimensional parabola. For example, if you set $z=w=0$, then you can clearly see that the intersection of the region $S$ with the $x$-$y$ plane is the region $y^2-1 \leq x \leq 1-y^2$ (i.e one parabola is opening down, the other is opening up, and we're looking at the overlapping region).
I'll trust your calculation that $d \alpha = dx \wedge dy \wedge dz \wedge dw$. So, we have
\begin{align}
\int_{\partial S} \alpha &= \int_Sdx \wedge dy \wedge dz \wedge dw \\
&= \int_S dx \, dy \, dz \, dw,
\end{align}
where in the last equal sign, I dropped the $\wedge$ signs, because $d\alpha$ is the positively oriented volume form, so integrating the $4$-dimensional volume form (almost by definition) is simply the standard Riemann integral.
So, we have $S = \{(x,y,z,w) \in \Bbb{R}^4| \, \, (y^2 + z^2 + w^2) -1 \leq x \leq 1-(y^2 + z^2 + w^2) \}$. One way to calculate this volume integral is to parametrize $S$ as follows: define the map $\xi: \Bbb{R}^4 \to \Bbb{R}^4$ by
\begin{align}
\xi(x,r, \theta, \phi)&= (x, r \sin \theta \cos \phi, r \sin\theta \sin \phi, r \cos \theta)
\end{align}
So, this is a kind of polar coordinate parametrization. If we define 
\begin{align}
U = \{(x,r,\theta, \phi):\, r \in (0,1), \, \, \theta \in (0,\pi), \, \, \phi \in (0,2\pi),\,\, \, |x|< 1-r^2\},
\end{align}
then it is easy to see that $\xi[\overline{U}] = S$, and $\overline{U} \setminus U$ has measure zero, and $\xi$ restricted to $U$ is injective. So, by the change of variables theorem, we have
\begin{align}
\int_S dx \, dy \, dz \, dw &= \int_U \left|\det \xi'(x,r, \theta, \phi)\right|\, dx \, dr \, d \theta\, d \phi \\
&= \int_U r^2 \sin \theta \, dx \, dr \, d \theta\, d \phi\\
&= \int_0^{2\pi} \int_0^{\pi} \int_0^1 \int_{r^2-1}^{1-r^2} \, r^2 \sin \theta \, dx \, dr \, d \theta\, d \phi.
\end{align}
I leave it to you to calculate.
