Apply Gauss's theorem when the metric is unknown

[semigroups]

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 $$4F_1=\{(u^1,u^2)\in \mathbb{R}^2:|u^1|\leq1, |u^2|\leq1\}$$ 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 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?

 

[jvicens]

Thank you for bringing up this interesting problem. As you correctly pointed out, your computation assumes that the metric $g$ is also the Euclidean metric on $F_1$. To modify your computation to be fully correct, we need to take into account the fact that the metric $g$ is unknown on $F_1$.

To do this, we can use the Gauss theorem as stated in the problem, but instead of using the vector field $X=u^1\frac{\partial}{\partial u^1}+u^2\frac{\partial}{\partial u^2}$, we can use a new vector field $Y=u^1\frac{\partial}{\partial u^1}+u^2\frac{\partial}{\partial u^2}$ that is defined as follows:

- On $F_1$, we define $Y$ to be the zero vector field, as the metric $g$ is unknown on $F_1$ and we cannot use the Gauss theorem in this region.

- Outside of $F_1$, we define $Y$ to be equal to $X$, i.e. $Y=u^1\frac{\partial}{\partial u^1}+u^2\frac{\partial}{\partial u^2}$.

With this new vector field $Y$, we can still use the Gauss theorem to compute the surface area of $f(F_2)$, but now we have to take into account the fact that $Y$ is not a constant vector field on $F_1$. This means that we have to modify the integral in the Gauss theorem to take into account the variation of $Y$ on $F_1$.

To do this, we can write $Y$ as $Y=u^1\frac{\partial}{\partial u^1}+u^2\frac{\partial}{\partial u^2}=u^1\frac{\partial}{\partial u^1}+u^2\frac{\partial}{\partial u^2}+(u^1-u^1)\frac{\partial}{\partial u^1}+(u^2-u^2)\frac{\partial}{\partial u^2}$. This allows us to write $Y$ as the sum of two vector fields $Y_1=u^1\frac{\partial}{\partial u^1}+u^