# Substitution of integral - pdes

$$\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits}$$

One line in a proof I'm looking at is $$\varphi(r) = \avint_{\partial B_r(x)} u(y) d\sigma(y) = \avint_{\partial B_1(0)} u(x+rz) d\sigma(z)$$

I am not sure how they have transformed the integral from $$B_r(x)$$ to $$B_1(0)$$.

I think it might have something to do with the fact that $$\partial B_R(x) = R\partial B_1(0) + x$$ which suggest a substitution like $$y=rz + x$$. But then not sure how to get the area element...

Any help is appreciated.

$$\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits}$$
Your initial suspicions is correct: the integral follows from the change of variables $$y = x + rz$$.
Let's support this last paragraph by a calculation. Let $$|S| = |\partial B_1(0)|$$ be the area ($$(n-1)$$-dimensional Hausdorff measure) of the unit ball. Then, $$\avint_{\partial B_r(x)} u(y) d\sigma(y) = \frac{1}{r^{n-1} |S|}\int_{\partial B_r(x)} u(y)\;d\sigma(y)$$ If we take the change of variables $$y = rz + x$$, then the new area element is $$d\sigma(y) = r^{n-1}d\sigma(z)$$ (since we are working with an $$(n-1)$$-dimensional area, the differential $$r^{n-1}$$). Thus, after doing the change of variables, we get $$\avint_{\partial B_r(x)} u(y) d\sigma(y) = \frac{r^{n-1}}{r^{n-1}|S|} \int_{\partial B_1(0)} u(rz +_ x)\;d\sigma(z) = \avint_{\partial B_1(0)} u(rz + x) \;d\sigma(z)$$