Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let
$$\sigma(U)=\iint_U d(x,y)dxdy$$
Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.
My question is how does $\sigma(k U)$ change with respect to $\sigma(U)$. I have doubt with two results. Is $\sigma(kU)=k\cdot\sigma(U)$ or does it depend on its dimension, i.e. $\sigma(kU)=k^n\cdot\sigma(U)$, or it is something else?
I have doubts because every distance in the new shape is scaled by $k$ with respect to the old shape, but the area that gets scaled by a factor of $k^n$. Any help would be appreciated. Thanks.
Suppose $U$ is our shape of dimension $n$ and each of its points will have coordinates $x=(x_1,\ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',\ldots,x_n')$ with the relations
$$x_i'=kx_i$$
$$dx_i'=kdx_i$$
So $dx'=\prod_{i=1}^ndx_i'=\prod_{i=1}^nk dx_i=k^ndx$, and because $x_i=x_i'/k$, the shape $kU$ gets transformed to $\{x\mid \exists x'\in kU:x=x'/k\}=U$, so we have that:
\begin{align} \sigma(kU) &= \iint_{kU}d(x',y')dx'\ dy' \\ &= \iint_U d(kx,ky)k^ndx\ k^ndy \\ &= \iint_U kd(x,y)k^{2n}dx\ dy \\ &= k^{2n+1}\iint_Ud(x,y)dx\ dy \\ \sigma(kU)&= k^{2n+1}\sigma(U) \end{align}