Let $x:\Sigma^n\to\mathbb{R}^{n+1}$ be an immersion of an orientable smooth manifold. For simplicity, let us assume $\Sigma$ is compact and $x$ is an embedding. The volume of $\Sigma$ in $x$ is defined to be the integral \begin{align} V=V_{\Sigma}(x)=\frac{1}{n+1}\int_{\Sigma}\langle x,\nu\rangle d\mu & & (1) \end{align} where $\nu$ is the Gauss map while $d\mu$ is the area element in the induced metric. This definition can be found in, e.g., Barbosa-do Carmo. In case where $\Sigma$ is closed, by divergence theorem this is essentially the volume of the region bounded by $x(\Sigma)$.
Now, let $x:\Sigma\times(-\epsilon,\epsilon)\to\mathbb{R}^{n+1}$ be a variation of $x$. For convenience, we also denote $x(\cdot,t)$ by $x_t$. Thus each $x_t:\Sigma\to\mathbb{R}^{n+1}$ is also an immersion of $\Sigma$. In Barbosa-do Carmo-Eschenburg, the volume function $V:(-\epsilon,\epsilon)\to\mathbb{R}$ is defined by \begin{align} V(t):=\int_{\Sigma\times[0,t]}x^*(dV) & & (2) \end{align} where $dV$ is the volume element in $\mathbb{R}^{n+1}$ and the superscript $^*$ denotes the pullback.
My question is as follow: By viewing $\{x_t\}$ as a family of immersions, I would expect $V(t)$ to be given by \begin{align} V(t)=V_{\Sigma}(x_t)=\frac{1}{n+1}\int_{\Sigma}\langle x_t,\nu\rangle d\mu & & (3) \end{align} using (1). Is it true that (3) is equivalent to (2) in general, or at least under some nice conditions? If so, how would us prove it? If not, what motivates us to take (2) as the definition instead of (3)?
Any comment or answer is welcomed and greatly appreciated.