Suppose $E \subset \mathbb R^n$ is a set with smooth boundary and let $\phi :E \to \mathbb R^n$ be a $C^\infty$ map. Then by change of variables $$ \int_{\phi(E)} f = \int_{E} (f \circ \phi) |\det d \phi| $$ where $d\phi$ is the matrix of partial derivatives of $\phi$.
How can I express the integral on the boundary? The formula $$ \int_{ \phi(\partial E)} f = \int_{\partial E} (f \circ \phi) |\det d \phi| $$ clearly does not work (for example taking $\phi$ a dilation by $k$, one would obtain that the integral changes like $k^n$, while it should change like $k^{n-1}$). What should I put instead of $|\det d \phi| $?
I think that what you want is the area formula for Lipschitz functions. You can find it in the book of Evans and Gariepy "Measure theory and fine properties of functions". Essentially, if you consider a local chart $\varphi:V\to \mathbb{R}^n$ for $\partial E$, where $V\subset \mathbb{R}^{n-1}$, then (locally) $$\int_{\partial E}g(x)\,dS=\int_Vg(\varphi(y))[[D\varphi(y)]]\,dy,$$ where $[[D\varphi(y)]]=\sqrt{\det((D\varphi(y))^tD\varphi(y))}$. On the other hand, $\phi\circ\varphi:V\to \mathbb{R}^n$ is a local chart for $\phi(\partial E)$ so you get $$\int_{\phi(\partial E)}h(x)\,dS=\int_Vh(\phi(\varphi(y)))[[D(\phi\circ\varphi)(y)]]\,dy.$$ So now you want to make the two right-hand sides equal. I don't know how messy are the computations using the chain rule for $[[D(\phi\circ\varphi)(y)]]$.