Having trouble proving change of variables from integral of differential forms. From what I understand, the integral of a differential form $dx^1\wedge\cdots\wedge dx^k$ on $[a^1,b^1]\times\cdots\times [a^k,b^k]$ is defined by
$$
\int_{[a^1,b^1]\times\cdots\times [a^k,b^k]} dx^1\wedge\cdots\wedge dx^k := \int_{a^k}^{b^k}\cdots\int_{a^1}^{b^1} dx^1 \cdots dx^k
$$
Where the right is $k$ repeated Riemann integrals. 
I am having trouble proving the change of variables rule for Riemann integrals using differential forms. I don’t see where the absolute sign would appear. Here’s what I have so far...
Let 
$$
\phi : [p^1,q^1]\times\cdots\times [p^k,q^k] \to [a^1,b^1]\times\cdots\times [a^k,b^k], (t^1, \dots , t^k) \mapsto \left( \phi^1 (t^1,\dots ,t^k), \dots , \phi^k (t^1,\dots ,t^k)\right)
$$
be bijective and its total derivative bijective, too. 
Then 
$$
\int_{a^k}^{b^k}\cdots\int_{a^1}^{b^1} dx^1 \cdots dx^k = \int_{[a^1,b^1]\times\cdots\times [a^k,b^k]} dx^1\wedge\cdots\wedge dx^k 
\\
= \int_{[p^1,q^1]\times\cdots\times [p^k,q^k]} \phi^* \left(dx^1\wedge\cdots\wedge dx^k\right)
= \int_{[p^1,q^1]\times\cdots\times [p^k,q^k]} det( D\phi ) \,\,\, dt^1\wedge\cdots\wedge dt^k
$$
where $D\phi$ denotes the total derivative of $\phi$ to $t^1,\dots , t^k$. 
I feel like the next step is to turn this into a Riemann integral but then there wouldn’t be an absolute sign on that determinant. 
 A: Let $f^1,...,f^k$ be smooth $\mathbb{R}$-valued functions on $M$ corresponding to a diffeomorphism $f$. Then we have,
\begin{align*} (df^1 \wedge \cdots \wedge df^k)(v_1,...,v_k)&= \sum_{\sigma \in S_k} \textbf{sgn}(\sigma) \ df^1(v_{\sigma(1)}) \cdots df^k(v_{\sigma(k)}) \\ & = \textbf{det}[df^i(v_j)]_{i,j} \end{align*}
where the last equality follows from the definition of the determinant. Hence, if we redefine the vectors $v_i = \partial_i:= \frac{\partial}{\partial x^i}$ then $(df^1 \wedge \cdots \wedge df^k)(\partial_1,...,\partial_k) = \textbf{det}[df^i(\partial_j)] = \textbf{det}\left[\frac{\partial f^i}{\partial x^j}\right]_{i,j}$. We will now use this fact.

*

*On charts $(U, \textbf{x})$ and $(V, \textbf{y})$ of $M$, the overlap map $f(\textbf{x}) = \textbf{y} = \begin{pmatrix} f^1(\textbf{x})& \cdots & f^n(\textbf{x}) \end{pmatrix}^T$ is a diffeomorphism. Moreover we have $y^i = f^i$ and the following chain of identities,

\begin{align*} dy^1 \wedge \cdots \wedge dy^n (\partial_{i_1},...,\partial_{i_n}) &= \textbf{sgn}(\sigma) \ dy^1 \wedge \cdots \wedge dy^n(\partial_1,...,\partial_n) \\ \\ &= \textbf{sgn}(\sigma) \cdot  \textbf{det}\left[\frac{\partial f^i}{\partial x^j}\right]_{i,j} \\ \\ &=  \textbf{sgn}(\sigma) \cdot \textbf{det}\left[\frac{\partial y^i}{\partial x^j}\right]_{i,j} dx^1 \wedge \cdots \wedge dx^n( \partial_1,...,\partial_n) \end{align*}
Consider now the problem in finding the area of $X = U \cap V$. Recalling that the signature of permutation is $\pm 1$, then by the above identity and the fact that $dy^1 \wedge \cdots \wedge dy^n$ and $dx^1 \wedge \cdots dx^n$ are volume forms on $X$ we have,
\begin{align*} \int_X \textbf{sgn}(\sigma) \ dy^1 \wedge \cdots \wedge dy^n  &= \int_X \textbf{sgn}(\sigma) \cdot \textbf{det}\left[\frac{\partial y^i}{\partial x^j}\right]_{i,j} dx^1 \wedge \cdots \wedge dx^n \\ \\ & = \int_X \ \left|\textbf{det}\left[\frac{\partial y^i}{\partial x^j}\right]_{i,j}\right| \ dx^1 \wedge \cdots \wedge dx^n \end{align*}
However, by the change of variables formula $\int_X dy^1  \cdots dy^n=\int_X  \left|\textbf{det}\left[\frac{\partial y^i}{\partial x^j}\right]_{i,j}\right| \ dx^1  \cdots  dx^n $

*

*Hence if the Jacobian of the transition charts are always positive then $M$ is Riemann Integrable. If you know the definition of orientable i.e smoothly varying frame then one can show that this property of the transition charts is equivalent to $M$ being orientable.

Geometrically one can reason this using the fact that the determinant gives the oriented volume of the parallelepiped spanned by the columns vectors.

*

*In this case the two sets of column vectors are the ones in the coordinates of the cotangent basis $dy^1,...,dy^n$ and the other in $dx^1,...,dx^n$. Let's call these $d \textbf{x}$ and $d \textbf{y}$ for short. Then $d\textbf{x}(v_1,...,v_n)$ and $d\textbf{y}(v_1,...,v_n)$ projects the parallelepiped spanned by $v_1,...,v_n$ onto the $(x_1,...,x_n)$ and $(y_1,...,y_n)$ coordinate planes and computes the volume. Hence, both should return the same signed volume.

It's in this projection that the sign issued is singled out. Why?

*

*Think of projecting the plane $x_3 = 1$ onto the $x_1,x_2$-plane which is $x_3 = 0$. If we orient  the $x_1,x_2$-plane with the counterclockwise orientation then $x_3$ is the usual "up" i.e toward our normal sky. However, if we choose the clockwise orientation, $x_3$ points toward the ground and this is understood to be "up". The signature handles this and recognizes that to project onto this version of the $x_1,x_2$-plane (which is the $y_1,y_2$-plane) we'll have to multiply be a negative since the overlap map is,
$$f(x_1,x_2,x_3) = (y_1(x_1,x_2,x_2),y_2(x_1,x_2,x_3), y_3(x_1,x_2,x_3))$$
where $y_1= x_1,y_2 = x_2$ and $y_3 = -x_3$ i.e in the frame induced by the coordinate system $\textbf{y}$, the plane $x_3 = 1$ is $x_3 = -1$.
