Q: Consider a surface $M$ with regular parametrization $X:U_{open}\subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ and define the parallel surface $M_t$ by
$$Y(u,v)=X(u,v) + tN(u,v)$$
where $N(u,v)$ is the unit normal on $X$ at point $(u,v)$ and $t$ is a constant. And Show that $$\text{Area}(M_t)=\text{Area}(M)-2t \int_ M H \text{ d}A+t^2\int_M K\text{ d}A$$
where $H$ and $K$ are the mean curvature and Gaussian curvature, respectively.
First, I want to find out what $Y_u \times Y_v$ equals to so that I can apply
$$\text{Area}(M_t)=\int_{M_t}\| Y_u \times Y_v\| \text{ d}u\text{d}v$$ $Y_u=X_u+tN_u=X_u(1-tk_1)$
$Y_v=X_v+tN_v=X_v(1-tk_2)$
where $k_1$ and $k_2$ are the principal curvature of $X$
Therefore,
$$Y_u \times Y_v=(1-2tH+t^2K)X_u \times X_v$$
And,
\begin{equation} \begin{split} \text{Area}(M_t) & =\int_{M}\| (1-2tH+t^2K)X_u \times X_v\| \text{ d}u\text{d}v \\ & = \int_{M} (1-2tH+t^2K)\|X_u \times X_v\| \text{ d}u\text{d}v \\ & = \text{Area}(M) -2t \int_{M}H \| X_u \times X_v \|\text{ d}u\text{d}v + t^2\int_{M}K \|X_u \times X_v \| \text{ d}u\text{d}v \end{split} \end{equation}
I don't know how to simplify this to
$$ \text{Area}(M_t) = \text{Area}(M) -2t \int_{M}H \text{ d}A + t^2\int_{M}K \text{ d}A$$
