Let $x_0:\Sigma^n\to\mathbb{R}^{n+1}$ be an immersion of an orientable compact $n$-dimensional smooth manifold into the Euclidean space.
It is well-known that given a smooth function $\varphi:\Sigma\to\mathbb{R}$ satisfying the zero-mean condition: \begin{align} \int_{\Sigma}\varphi d\mu=0 \end{align} (where $d\mu$ is the area element in the induced metric), there always exist a volume-preserving normal variation $X:\Sigma\times(-\epsilon,\epsilon)\to\mathbb{R}^{n+1}$ whose variation vector field is \begin{align} Y:=\frac{\partial X}{\partial t}\bigg|_{t=0}=\varphi\nu \end{align} where $\nu$ is the unit normal.
In Barbosa-do Carmo, the proof begins by considering a two-parameter family of immersion \begin{align} x(t,u)=x_0+(t\varphi+ug)\nu & & (1) \end{align} (where $g:\Sigma\to\mathbb{R}$ is any smooth function with $g\equiv 0$ on the boundary $\partial\Sigma$ and $\int_{\Sigma}gd\mu\neq 0$), then considers the equation \begin{align} V(t,u)=const \end{align} (where $V(t,u)$ is the volume of $x(t,u)$) and use the implicit function theorem to deduce that $u$ can be expressed as a function of $t$ in a small open neighbourhood of $0\in\mathbb{R}$. Finally, one checks that $X(p,t)=x(t,u(t))(p)$ is the desired volume-preserving normal variation of $x_0$.
My question is the following: Instead of considering (1), why not the proof considers the one-parameter family \begin{align} x(t)=x_0+t\varphi\nu & & (2) \end{align} of immersions?
This seems to directly give the desired variation (by setting $X(p,t)=x(t)(p)$) and it looks simpler than (1). For one thing, a big theorem like implicit function theorem will not be needed. Is there any reason that I've failed to observe which makes it necessary to consider (2) instead of (1)?
Any comment or answer is welcomed and greatly appreciated.
The variation you suggest will only be volume preserving to first order. Consider the unit circle in the plane, coordinatized with angle $\theta$, the function $f(\theta)=\sin \theta$, for which your variation produces curves which in polar coordinates are given by $r_t(\theta)=1+t\sin \theta$. The area enclosed is
$$\int_0^{2\pi} \left(\int_0^{1+t\sin \theta} r dr \right)d\theta=\int_0^{2\pi}\frac{1}{2} (1+t\sin \theta)^2 d\theta=\pi+\frac{\pi}{2} t^2.$$