How to change variables in a surface integral without parametrizing

I know this is an old question, but I thought this explanation might be helpful to some.

By definition (in $\mathbb R^3$):

$$\int_{\partial B(\boldsymbol x,r)}f(\boldsymbol y)dS(\boldsymbol y)= \int_U f(\boldsymbol y(s,t))\left\|\frac{\partial\boldsymbol y}{\partial s}\times\frac{\partial\boldsymbol y}{\partial t}\right\|dsdt$$

Now, observe that $f(\boldsymbol y)=f(\boldsymbol x+r(\frac{\boldsymbol y-\boldsymbol x}{r}))$, and that if $\boldsymbol y(s,t)$ is a parametrization of $\partial B(\boldsymbol x,r)$ for $(s,t)\in U$, then $\frac{\boldsymbol y(s,t)-\boldsymbol x}{r}$ is a parametrization of $\partial B(\boldsymbol 0,1)$ for $(s,t)\in U$. Finally we observe that

$$\left\|\frac{\partial\boldsymbol y}{\partial s}\times\frac{\partial\boldsymbol y}{\partial t}\right\|= r^2\left\|\frac{\partial}{\partial s} \left (\frac{\boldsymbol y-\boldsymbol x}{r} \right )\times\frac{\partial }{\partial t} \left (\frac{\boldsymbol y-\boldsymbol x}{r} \right )\right\|$$

So if we let $\boldsymbol z(s,t)=\frac{\boldsymbol y(s,t)-\boldsymbol x}{r}$, then we have

$$\int_U f(\boldsymbol y(s,t))\left\|\frac{\partial\boldsymbol y}{\partial s}\times\frac{\partial\boldsymbol y}{\partial t}\right\|dsdt= r^2\int_U f(\boldsymbol x +r\boldsymbol z(s,t))\left\|\frac{\partial\boldsymbol z}{\partial s}\times\frac{\partial\boldsymbol z}{\partial t}\right\|dsdt\\= r^2\int_{\partial B(\boldsymbol 0,1)}f(\boldsymbol x+r\boldsymbol z)dS(\boldsymbol z)$$

Edit by OP

As @user5753974 commented, you can generalize this if you use the fact that in $\mathbb R^n$ $$∫_{∂B(\boldsymbol x,r)}f(\boldsymbol y)dS(\boldsymbol y)=∫_{U}f(\boldsymbol y(\boldsymbol z)) \left \|\det\left (\frac{∂\boldsymbol y}{∂z_1},…,\frac{∂\boldsymbol y}{∂z_{n−1}},\boldsymbol n\right) \right \| d^{n−1}\boldsymbol z,$$ where $\boldsymbol n$ is the normal vector to the surface, and that $\boldsymbol n$ does not change when the surface is scaled and translated.

No parametrization is needed, just some careful uncovering of definitions. Maybe the following can help:

Let $S^n(r)=\{x\in\mathbb R^{n+1}|\|x\|=r\}\subset\mathbb R^{n+1}$ (the sphere of radius $r>0$ centered at the origin).

There is a natural volume form $\mu_{r}$ on $S^n(r)$, i.e. a nowhere-vanishing $n$-form, induced by the volume form $\mu=dx_1\wedge\ldots \wedge dx_{n+1}$ on $\mathbb R^{n+1}$ and the "outward" unit normal $N:S^n(r)\to\mathbb R^{n+1}$, $N(x)=x/r$, and is defined as follows:

$$\mu_r(x)(v_1,\ldots,v_n):=\mu(x)(N(x), v_1, \ldots, v_n)$$ for all $v_1, \ldots, v_n\in T_x S^n(r).$

(More informally, we write $\mu_r=\mu/dr$. Note that $dr(N)=1$).

Next let $\Phi:S^n(1)\to S^n(r)$ be defined by $\Phi(x)=rx$. Then $$\Phi^*\mu_r=r^n\mu_1. $$ This you can prove without any parametrization, just from the definitions above and generalities like the chain rule, pull back, etc.

It then follows, by the change of variables formula, that for any continuous function $f:S^n(r)\to\mathbb R$, $$\int_{S^n(r)}f\mu_r=\int_{S^n(1)} \Phi^*(f\mu_r)=r^n\int_{S^n(1)} (f\circ\Phi) \mu_1.$$

Makes sense?

I think the measure theoretic approach works fine, note that the surface measure is n-1 dimensional Hausdorff measure, in general for the s-dimensional Hausdorff measure we have $H^{s} (rA)=r^{s} H^{s}(A)$ and this measure is translation invariant. Now use the measure theoretic change of variable formula.