Let $u \in C^2(U)$ with $U \subset \mathbb{R}^{n}$ and $\alpha(n)$ as the volume of unit n-dimensional ball. I want to show that
$$\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)dS(y)=\frac{1}{n\alpha(n)}\int_{\partial B(0,1)}u(x+rz)dS(z)$$
So, I have attempted to do pure calculation but somehow my result is different from this one. This result is actually found in L.C. Evans (PDE). I wonder where do I make mistakes.
So, this is my result
Observe that $y_{i} = x_{i} + rz_{i}$ and therefore $\frac{\partial y_{i}}{\partial z_{k}}= 0$ whenever $k\neq i$ and $\frac{\partial y_{i}}{\partial z_{i}}= r$ for any $i,k \in \{1,2,...,n\}$. Therefore we can deduce that the Jacobian matrix $J = rI_{n\times n}$ given $I_{n\times n}$ as an identity matrix of size $n \times n$. Furthermore, $\det(J) = r^{n}$.
Now, we will have:
$$\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)dS(y)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(0,1)}u(x+rz)|\det(J)|dS(z)$$
$$\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)dS(y)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(0,1)}u(x+rz)r^{n}dS(z)$$
$$\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)dS(y)=\frac{r}{n\alpha(n)}\int_{\partial B(0,1)}u(x+rz)dS(z)$$
This is as far as I go and I have different result. Where do I make mistakes here?
Please help and any advice or suggestion is very much welcomed!