In section 2.4 (wave equation) of Partial Differential Equation (Evans), the author used the descent method to derive the Poisson's formula for two-dimensional wave equation.
For the $n=3$ case, the author has derived the Kirchhoff's formula for the IVP problem of the wave equation: $$\begin{cases} u_{tt}-\Delta u=0 & \text{in}\;\mathbb{R}^n\times (0,+\infty) \\ u=g,\; u_t=h & \text{on}\;\mathbb{R}^n\times\{t=0\} \end{cases}$$ i.e., $$u(x,t) = \frac{\partial}{\partial t}\biggl(\frac{1}{4\pi t}\int_{\partial B(x,t)}g\,\mathrm{d}S\bigg)+\frac{1}{4\pi t}\int_{\partial B(x,t)}h\,\mathrm{d}S.$$
For the $n=2$ case, assume $u\in C^2(\mathbb{R}^2\times [0,+\infty))$ solves the above IVP and write $$\bar{u}(x_1,x_2,x_3,t):=u(x_1,x_2,t),$$ $$\bar{g}(x_1,x_2,x_3):=g(x_1,x_2),\quad \bar{h}(x_1,x_2,x_3):=h(x_1,x_2),$$ $$x=(x_1,x_2),\quad \bar{x}=(x_1,x_2,x_3).$$ Then $\bar{u}$ solves $$\begin{cases} \bar{u}_{tt}-\Delta \bar{u}=0 & \text{in}\;\mathbb{R}^3\times (0,+\infty), \\ \bar{u}=\bar{g},\; \bar{u}_t=\bar{h} & \text{on}\;\mathbb{R}^3\times\{t=0\}, \end{cases}$$ and hence $$u(x,t)=\bar{u}(\bar{x},t)=\frac{\partial}{\partial t}\biggl(\frac{1}{4\pi t}\int_{\partial \bar{B}(\bar{x},t)}\bar{g}\,\mathrm{d}\bar{S}\bigg)+\frac{1}{4\pi t}\int_{\partial \bar{B}(\bar{x},t)}\bar{h}\,\mathrm{d}\bar{S},$$ where $\bar{B}(\bar{x},t)$ denotes the ball in $\mathbb{R}^3$ with center $\bar{x}$ and radius $t>0$ and $\mathrm{d}\bar{S}$ denotes the two-dimensional surface measure on $\partial\bar{B}(\bar{x},t)$.
My question is the text in the following: How can we deduce that $$\int_{\partial\bar{B}(\bar{x},t)}\bar{g}\,\mathrm{d}\bar{S}=2\int_{B(x,t)}g(y)\sqrt{1+|D\gamma(y)|^2}\,\mathrm{d}y?$$ in which $\gamma(y)=\sqrt{t^2-|y-x|^2}$. I think it may be more helpful to explain by using some figures.