Let $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary$^1$ and $(\Omega,\phi)$ be a $k$-dimensional boundary $C^1$-chart$^2$ of $M$. We know that$^3$ $(\tilde\Omega,\tilde\phi):=(\Omega\cap\partial M,\pi\circ\left.\phi\right|_{\tilde\Omega})$ is a $(k-1)$-dimensional $C^1$-chart of $\partial M$. Let $(U,\psi):=(\phi(\Omega),\phi^{-1})$ and $(\tilde U,\tilde\psi):=(\tilde\phi(\tilde\Omega),\tilde\phi^{-1})$. Then $${\rm D}\tilde\psi(u)={\rm D}\psi(\iota u)\circ\iota\;\;\;\text{for all }u\in\tilde U=\{u\in\mathbb R^{k-1}:(u,0)\in U\}\tag1.$$ By definition, the surface measure on $\mathcal B(\tilde\Omega)$ is given by $$\sigma_{\tilde\Omega}:=\sqrt{g_{\tilde\psi}}\left.\lambda^{\otimes(k-1)}\right|_{\tilde U}\circ\tilde\psi^{-1},$$ where $g_{\tilde\psi}:=\det G_{\tilde\psi}$, $G_{\tilde\psi}(u):={\rm D}\tilde\psi(u)^\ast{\rm D}\tilde\psi(u)$ for $u\in\tilde U$ and $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R)$.
If $k=d$, I would like to show that $$\int f\:{\rm d}\sigma_{\tilde\Omega}=\int_{B_0}f(\psi(u))|\det{\rm D}\phi(u)|\left\|{\rm D}\phi(u)^\ast e_k\right\|\:\lambda^{\otimes k}({\rm d}u)\tag2$$ for all $\sigma_{\tilde\Omega}$-integrable $f:\tilde\Omega\to\mathbb R$, where $B_0:=\{u\in\mathbb R^k:\left\|u\right\|\le1\text{ and }u_k=0\}$ and $(e_1,\ldots,e_k)$ denotes the standard basis of $\mathbb R^k$. How can we do that?
It's easy to see that $\pi$ is the adjoint $\iota^\ast$ of $\iota$, i.e. $\iota^\ast=\pi$. So, if $A\in\mathbb R^{d\times k}$ and $\tilde A:=A\circ\iota$, then $$(\tilde A^\ast\tilde A)_{ij}=\langle Ae_i,Ae_j\rangle=B_{ij}\;\;\;\text{for all }i,j\in\{1,\ldots,k-1\},\tag3$$ where $B:=A^\ast A$. If we denote by $B^{kk}$ the submatrix of $B$ formed by deleting the $k$th row and $k$th column, we have shown that $\tilde A^\ast\tilde A=B^{kk}$. If $\operatorname{cof}(M)$ denotes the cofactor matrix of a square-matrix $M$ and $k=d$, then $$\operatorname{cof}(B)=\operatorname{cof}(A)^\ast\operatorname{cof}(A)\tag4$$ and hence $$\det\tilde A^\ast\tilde A=\operatorname{cof}(B)_{kk}=\left\|\operatorname{cof}(A)e_k\right\|^2\tag5.$$ If $A$ is regular, then $$\operatorname{cof}(A)=\det A(A^{-1})^\ast\tag6$$ and hence $$\sqrt{\det\tilde A^\ast\tilde A}=|\det A|\left\|(A^{-1})^\ast e_k\right\|\tag7.$$ If we apply this to $A={\rm D}\psi(u)$, $u\in\phi(\Omega)$, we obtain $$\sqrt{g_{\tilde\psi}(u)}=|\det{\rm D}\psi(\iota u)|\left\|({\rm D}\psi(\iota u)^{-1})^\ast e_k\right\|\;\;\;\text{for all }u\in\tilde U\tag8.$$ Now, by definition, $$\int f\:{\rm d}\sigma_{\tilde\Omega}=\int\sqrt{g_{\tilde\psi}}(f\circ\tilde\psi)\:{\rm d}\left.\lambda^{\otimes(k-1)}\right|_{\tilde U}.\tag9$$
How do we need to proceed? If $B$ denotes the closed unit ball in $\mathbb R^k$, $B_+:=B\cap(\mathbb H^k)^\circ$ and $B_0=B\cap\partial\mathbb H^k$, I've read that we can assume without loss of generality that $\Omega^\circ=\phi^{-1}(B_+)$ and $\partial\Omega=\phi^{-1}(B_0)$, but I don't get why this is possible.
EDIT: What I could imagine is that we can replace $\phi$ by $$x\mapsto\begin{cases}\frac{\phi(x)}{\max(1,\left\|\phi(x)\right\|)}&\text{, if }\phi(x)\ne0\\0&\text{, otherwise}\end{cases}\tag{10}.$$ This should (am I missing something?) still be a $C^1$-diffeomorphism and it clearly takes values in $B$. Is this the right approach?
$^1$ i.e. each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^k$.
If $E_i$ is a $\mathbb R$-Banach space and $B_i\subseteq E_i$, then $f:B_1\to E_2$ is called $C^1$-differentiable if $f=\left.\tilde f\right|_{B_1}$ for some $E_1$-open neighborhood $\Omega_1$ of $B_1$ and some $\tilde f\in C^1(\Omega_1,E_2)$ and $g:B_1\to B_2$ is called $C^1$-diffeomorphism if $g$ is a homeomorphism from $B_1$ onto $B_2$ and $g$ and $g^{-1}$ are $C^1$-differentiable.
$^2$ A $k$-dimensional $C^1$-chart of $M$ is a $C^1$-diffeomorphism from an open subset of $M$ onto an open subset of $\mathbb H^k$.
$^3$ Let $\iota$ denote the canonical embedding of $\mathbb R^{k-1}$ into $\mathbb R^k$ with $\iota\mathbb R^{k-1}=\mathbb R^{k-1}\times\{0\}$ and $\pi$ denote the canonical projection of $\mathbb R^k$ onto $\mathbb R^{k-1}$ with $\pi(\mathbb R^{k-1}\times\{0\})=\mathbb R^{k-1}$.