Fourier transform as an integral on a surface

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Let $v\in C_c^\infty(\mathbb{R}^n)$ with support $B\subset \mathbb{R}^n.$ Let $h$ be real valued smooth function on a neighborhood of $B$ and $\xi \in \mathbb{R}^n, t\in \mathbb{R}$ and consider the integral

$$I(\xi, t)= \int_{B}{\rm e}^{{\rm i}(\xi\cdot x + t h(x))}v(x)\,{\rm d}x,$$ where $x\cdot \xi$ denotes the euclidean scalar product in $\mathbb{R}^n.$

Define now in $\mathbb{R}^{n+1}$the surface $$S:= \{(x_1, x_2,\cdots, x_n, h(x))\,|\, (x_1, x_2,\cdots, x_n) =:x \in B\}.$$

How can the integral $I$ be written in the following form $$J = \int_{X\in S} {\rm e}^{{\rm i}Y\cdot X }u(X)\,{\rm d}S(X),$$ where ${\rm d}S$ is the surface measure and $u$ is to be chosen properly.

As far as I know, the surface measure on $S$ is given by $${\rm d}S = \sqrt{1+|\nabla h(x)|^2}\, {\rm d}x.$$ Then we can write $$I(\xi, t)= \int_{B}{\rm e}^{{\rm i}(\xi\cdot x + t h(x))}\,\frac{v(x)}{\sqrt{1+|\nabla h(x)|^2}} \, \sqrt{1+|\nabla h(x)|^2}\,{\rm d}x = J,$$ where $X = (x,h(x)), \, Y = (\xi, t)$ and $u(X) = \frac{v(x)}{\sqrt{1+|\nabla h(x)|^2}}.$

Thank you in advance