Show that integral of $f$ over a manifold $M$ is independent of coordinate chart 
Definition: Suppose $M\subseteq \mathbb{R}^n$ is a compact smooth k-manifold and $f:M\to \mathbb{R}$ is a continuous function. Since $S=\text{support}(f)$ is compact suppose $\phi:U\to V$ is a chart such that $S\subseteq V$ and $U$ is bounded. We define the integral of $f$ over $M$ to be $$\int_M fdV=\int_{U^{int}} (f\circ \phi)\text{Vol}(D\phi)$$

Suppose $x_1,...x_k\in\mathbb{R}^n$ form the vertices of a k-parallelepiped $P$, and $X$ be the matrix with columns $x_i$ then the volume of $P$ is $\sqrt{\det(X^TX)}$
Prove this definition is independent of coordinate chart $\phi$
Suppose $\psi:\widetilde U\to V$ is coordinate chart, then $\phi^{-1}\circ\psi:\widetilde U\to U$ is a change of variables.
Applying change of variables to the integral $$\int_{U^{int}} (f\circ \phi) \text{Vol}(D\phi)$$
$$\int_{\widetilde U^{\text{int}}} (f\circ\phi\circ(\phi^{-1}\circ\psi))\vert \det D(\phi^{-1}\circ\psi)\vert \text{Vol}(D\phi)$$
$$=\int_{\widetilde U^{\text{int}}} (f\circ\psi)\vert \det D\phi^{-1} D\psi\vert\text{Vol}(D\phi)$$
I'm not sure how to deal with the jacobians here. Since I don't really understand how you can take the determinant of $D\phi$ when $\phi$ is a map from $\mathbb{R}^k$ to $\mathbb{R}^n$ so it isn't a square matrix.
 A: Applying the change of variables we get
$$\int_{U^{int}} (f\circ \phi) \text{Vol}(D\phi)=\int_{\widetilde U^{\text{int}}} (f\circ\psi)\cdot\text{Vol}(D\phi)\circ\phi^{-1}\circ\psi\cdot\vert \det D(\phi^{-1}\circ\psi)\vert$$
So we should show 
$$\text{Vol}(D\psi)=\text{Vol}(D\phi)\circ\phi^{-1}\circ\psi\cdot\vert \det D(\phi^{-1}\circ\psi)\vert$$
Writing this out, taking squares and using $|\det A|^2=\det (A\cdot A^T)$ and the multiplicativity of the determinant   wee see that this is equivalent to
$$\det (D\psi^T\cdot D\psi)
=\det\left(((D\phi)\circ\phi^{-1}\circ\psi)^T\cdot ((D\phi)\circ\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi)^T\right).$$
Since $\psi=\phi\circ(\phi^{-1}\circ\psi)$ by applying the chain rule the right hand side becomes
$$\det\left(((D\phi)\circ\phi^{-1}\circ\psi)^T\cdot D\psi\cdot D(\phi^{-1}\circ\psi)^T\right).$$
Using the commutativity of the determinantant this is equal to 
$$
\det\left((D(\phi^{-1}\circ\psi)^T\cdot ((D\phi)\circ\phi^{-1}\circ\psi))^T\cdot D\psi\right)$$
which due to $(A\cdot B)^T=B^T\cdot A^T$ is
$$\det\left((((D\phi)\circ\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi))^T\cdot D\psi\right)$$
and so again by using the chain rule this is just
$$
\det\left(D\psi^T\cdot D\psi\right)$$
which is what we wanted to show.
