Stokes' theorem in higher dimensions I'm trying to follow along with Stokes's theorem on page 46 and I'm not quite understanding where this comes from: $$ V=\int_{B} dx \wedge dy \wedge dz \wedge dt = - \int_{S} t dx \wedge dy \wedge dz $$
I can understand the negative due to the anticommutivity, $\int_{B} dx \wedge dy \wedge dz \wedge dt = - \int_{B} dt \wedge dx \wedge dy \wedge dz$ due to how there would be three interchanges
But where does the $d$ for the $t$ go? Is it as simple as integrating with respect to $t$ but leaving all the rest the same? (So far, I've only applied $d$ to various forms and so I'm not sure how it works in reverse.
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I can understand the first line is Stokes' theorem, but how would you know that α is that? I'd know that the volume is $\int_{B} dx \wedge dy \wedge dz \wedge dt = \int_{domain} d\alpha $ and then I'd need α itself to integrate over the surface like $\int_{surface} \alpha$. But how would we get $\alpha$ if given $d\alpha$?

This is exactly where you applied the Stokes theorem
$$ \int _B d\alpha = \int_S \alpha$$
to $\alpha = -t  dx \wedge dy \wedge dz$, since
\begin{align}
d\alpha &= -d (t  dx \wedge dy \wedge dz) \\
&= -dt \wedge  dx \wedge dy \wedge dz.  
\end{align}
Note that we have the last equality: essentially this is how we define the exterior derivative, but you can also think of it as the application of Leibniz rules
$$ d (t  dx \wedge dy \wedge dz) = dt  \wedge dx \wedge dy \wedge dz \pm t d ( dx \wedge dy \wedge dz)$$
and that $d^2 = 0$, which gives (by some Leibniz rule again)
$$d ( dx \wedge dy \wedge dz) = 0.$$
