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Let $f:U \to \mathbb{R}^3$ be a surface, where $U=\{(u^1,u^2)\in \mathbb{R}^2:|u^1|<3, |u^2|<3\}.$ Consider the two closed square regions $F_1=\{(u^1,u^2)\in \mathbb{R}^2:|u^1|\leq1, |u^2|\leq1\}$and $F_2=\{(u^1,u^2)\in \mathbb{R}^2:|u^1|\leq2, |u^2|\leq2\}.$

While the first fundamental form $g$ is unknown in the inner region $F_1$, it is the Euclidean metric outside of $F_1$:$g=(\mathrm{d}u^1)^2+(\mathrm{d}u^2)^2$ at all $u\in U-F_1$ Also, it is known that the vector field $X=u^1\frac{\partial}{\partial u^1}+u^2\frac{\partial}{\partial u^2}$ satiesfies $\mathrm{div}X=2$ for $\textbf {all~points}$ of $U$. Show that the total surface area of $f(F_2)$ is 16.

Remark: Recall that the Gauss theorem states: Let $M$ be an oriented surface with a Riemannian metric. Let $X$ be a vector field on $M$. Then for each polygon on $M$ given by $P:F \to M$ it has $\int_{P(F)}(\mathrm{div}X)\mathrm{d}M=\int_{\partial P(F)}l_{X}\mathrm{d}M$ where $l_{X}\mathrm{d}M=-X^2\sqrt|g|\mathrm{d}u^1+X^1\sqrt|g|\mathrm{d}u^2$. Setting $f_1:=-X^2\sqrt|g|$ and $f_2:=X^1\sqrt|g|$ to get
$\int_{P(F)}(\mathrm{div}X)\mathrm{d}M=\iint_F(\frac{\partial f_2}{\partial u^1}-\frac{\partial f_1}{\partial u^2})\mathrm{d}u^1\mathrm{d}u^2. $

In this problem $|g|=1$ outside $F_1$ but it's unknown on $F_1$.

We need to compute $\int_{f(F_2)}\mathrm{d}M=\frac{1}{2}\int_{f(F_2)}(\mathrm{div}X)\mathrm{d}M$. By the convention of notations it has $X^i=u^i$ and thus $\frac{1}{2}\int_{f(F_2)}(\mathrm{div}X)\mathrm{d}M=\frac{1}{2}\int_{-2}^{2}\int_{-2}^{2}(1-(-1))\mathrm{d}u^1\mathrm{d}u^2=16.$

BUT my computation assumes $g$ is also the Euclidean metric on $F_1$, which may not be the case. How to modify my computation for it to be fully correct?

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