In the proof of the mean value property for Harmonic functions, one often sees steps such as $$ E(r) := \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u \ dy = \frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} u(x+rz) \ dz $$ which is justified by the change of coordinates $y = x+rz$. The second equality allows one to compute the derivative of the integral w.r.t. $r$, and after applying inverse change of variables we obtain $$ E'(r) = \frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} z \cdot Du(x+rz) \ dz = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \frac{x-y}{r} \cdot Du(y) \ dy \; .$$ From which the mean value property follows easily.
On the other hand, I have attempted to obtain the the mean value property through a similar change of variable, namely $y = rz$. This changes the integral to $$ E(r) = \frac{1}{|\partial B(x,1)|}\int_{\partial B(x,1)} u(rz) \ dz $$ which differentiates to yield $$ E'(r) = \frac{1}{|\partial B(x,1)|}\int_{\partial B(0,1)} z \cdot Du(rz) \ dz = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \frac{y}{r} \cdot Du(y) \ dy \; .$$ (where in the second equality I have again used the inverse change of variables). From this, it follows that we should have $$ \int_{\partial B(x,r)} \frac{x-y}{r} \cdot Du(y) \ dy = \int_{\partial B(x,r)} \frac{y}{r} \cdot Du(y) \ dy \; .$$ Which does not seem correct. I was looking for any guidance on where I have made a mistake.