For $m\in\mathbb N$ define the ball $B^m \overset{\text{Def.}}=\{ p\in \mathbb R^m:\|p\|_2<1\}$ and the sphere $S^{m-1} \overset{\text{Def.}}= \partial B^m\subset\mathbb R^m$. I want to prove the following statement using the change of variables formula for differential forms in order to strengthen my understanding of the Theorem:
Statement. (Polar coordinates) We have $$\tag{*}\int_{B^m} 1 = \int_0^1 r^{m-1} \int_{S^{m-1}} 1\,\mathrm dx\,\mathrm dr.$$
My idea: Consider the smooth manifold $M=[0,1)\times S^{m-1}$ and the smooth function $f:M\to B^m, (r,x)\mapsto r x$. Then change of variables gives $$\int_{B^m} 1=\int_{B^m} \mathrm dx^1\land\dots\land\mathrm dx^m=\int_M f^*(dx^1\land\dots\land\mathrm dx^m ),$$ so I would just have to prove that the pullback $f^*(dx^1\land\dots\land\mathrm dx^m )$ is $r^{m-1}\mathrm dr\land\dots$. How can I do this?
Also: How can I extend this method when I replace $1$ in (*) by a, say smooth, function $J:B^m\to\mathbb R$ ?
Note that any differential form is, by definition, an alternating tensor field. Therefore, the pullback $$f^*(J\, \mathrm dx^1\land\dots\land\mathrm dx^m),$$ by [LeeSM13; Corollary 12.28], is equal to, with the notation $(r,y)=(r,y^1,\dots, y^{m-1})$, $$J\circ f\, \mathrm d(x^1\circ f)\land\dots\land\mathrm d(x^m\circ f).$$
(Henceforth, as is usual in differential geometry, we will also use $r,y^1,\dots, y^{m-1}$ to denote the respective projection operators from $[0,1[\times S^{m-1}$ into $\mathbb R$.)
Now for each $k\in\{1,\dots, m\}$, we have $$x^k\circ f:M\to[0,1[, \ (r, y)\mapsto r y^k.$$ By [LeeSM13; pages 62, 63 or formula (11.11)], we thus have ("formula for the total derivative") $$\mathrm d(x^k\circ f)=\frac{\partial(r y^k)}{\partial r}\,\mathrm dr+\sum_{i=1}^{m}\frac{\partial (r y^k)}{\partial y^i}\,\mathrm dy^i=y^k\,\mathrm dr+r\,\mathrm dy^k.$$
Therefore, $$f^*(\mathrm dx^1\land\dots\land\mathrm dx^m)=\bigwedge_{k=1}^m (y^k\,\mathrm dr+r\,\mathrm dy^k).$$
Claim. In the above setting, for any $m\in\mathbb N\setminus\{1\}$, we have $$\bigwedge_{k=1}^m (y\,\mathrm dr+r\,\mathrm dy^k)=r^{m-1}\mathrm dr\wedge\omega,$$ where (for notation see [LeeSM13; Pages 315 and following]) $$\omega=\omega(m)=\sum_{j=1}^{m}(-1)^{j-1} y^j\,\mathrm dy^1\wedge\dots\wedge \mathrm dy^{j-1}\wedge\mathrm dy^{j+1}\wedge\dots\wedge\mathrm dy^{m}\in\Gamma(\Lambda^{m-1}(S^{m-1})).$$
In particular, once we have chosen the claim, we have proven the desired statement in the question, since $\omega$ is the volume form of $S^{m-1}$ by [Kö02; Page 436].
Proof of Claim. Note first that $\mathrm dr\wedge\mathrm dr=0$. Now consider $$\bigwedge_{k=1}^m (y\,\mathrm dr+r\,\mathrm dy^k).$$ When fully expanding the wedge, if we take any product which contains $\mathrm dr$ at least twice, then that term will be equal to $0$. If the term contains $\mathrm dr$ not at all, then it must be equal to some function times $\mathrm dy^1\wedge\dots\wedge\mathrm dy^{m}$, which is also equal to $0$, as $$\sum_{k=1}^m y^k\mathrm dy^k = 0,$$ as can be derived from $$\sum_{k=1}^m (y^k)^2=1.$$
Therefore, we are just left with $$\bigwedge_{k=1}^m (y\,\mathrm dr+r\,\mathrm dy^k)=\sum_{k=1}^m r^{m-1} \,\mathrm dy^1\wedge\dots\wedge\mathrm d y^{k-1}\wedge\mathrm dr\wedge\mathrm dy^{k+1}\wedge\dots\wedge \mathrm dy^m=\mathrm dr\wedge\omega.$$ $\square$
Literature
Note: I lost a lot of time in the proof since I first wanted to go by induction, but there is the nasty detail (lost in my notation) that $$\mathrm dy^1\vert_{S^m}\wedge\dots\wedge \mathrm dy^m\vert_{S^m}=0$$ but $$\mathrm dy^1\vert_{S^{m+1}}\wedge\dots\wedge \mathrm dy^m\vert_{S^{m+1}}\neq 0.$$