Understanding $\rho^*\delta_0$ the pullback of the delta distribution by a function

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I don't understand the meaning of theorem 6.15 (page 136) from the Hörmander Book "The Analysis of Linear Partial Differential Operators I". It states a connection between the pullback of the delta distribution by a well behaved function:

If $\rho$ is a real valued function in $\mathcal{C}^\infty(X),$ $X \subset \mathbb{R}^n,$ and if $|\rho'| = \left(\sum |\partial \rho / \partial x_j|^2\right)^{1/2} \neq 0$ when $\rho = 0,$ then $\rho^*\delta_0 = dS/|\rho'|$ where $dS$ is the surface measure on the surface $\{ x\in X \mid \rho(x) = 0 \}$.

The main problem I have is understanding the right part of the equation i.e. $dS/|\rho'| .$ I have a general understanding of the surface measure but I have a really hard time understanding it in this context. Is there an example I can look up?

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The set $\Sigma:=\rho^{-1}(\{0\})=\{x\in X \subseteq \mathbb{R}^n \mid \rho(x)=0 \}$ is an $(n-1)$-dimensional hypersurface if $\rho'\neq 0$ on it. The measure $dS$ is the induced measure on $\Sigma$; for $n=3$ it's just the area measure on a 2D-surface. The measure $dS/|\rho'|$ is defined so that $(dS/|\rho'|)(A)=\int_A (1/|\rho'|) \, dS,$ or equivalently $\int f \, (dS/|\rho'|) = \int (f/|\rho'|) \, dS$ for every measurable function $f.$