Calculate area of a non-compact submanifold of dimension 2

Question

Let $M \subset \mathbb{R}^3$ given by \begin{align} M:= \{ (x,y,z) \in \mathbb{R}^3 : z=xy, x^2 + y^2 <1 \} \end{align} Calculate the area of the submanifold $M$.

The solution says:

$M$ is a submanifold, since it is the graph of the continuously differentiable function $f(x,y):= xy$ defined on the open set $U:= B(0,1)$. From this, we get immediately a local parametrization $\phi(x,y):= (x, y, xy)$ which extends to a map $\phi: \mathbb{R}^2 \rightarrow \mathbb{R}^3$, defining a 2-dimensional submanifold $N \subset \mathbb{R}^3$. The 2-Jacobian of $\phi$ is given by \begin{align} J_2(d\phi) = \sqrt{1 + |\nabla f|^2} = \sqrt{1 + y^2 + x^2} \end{align} The area of $M$ is defined as \begin{align} Area(M) := \text{sup} \Bigg\{ \int_M f \, \, d\text{vol}_2 : f \in \mathscr{C}^0_{c}(M, [0, 1]) \Bigg\} \end{align} where $\mathscr{C}^0_{c}(M, [0, 1])$ denote the set of all continuos functions $ f: M \rightarrow [0,1]$ with compact support in $M$.

Let $\epsilon >0$.

We have some trouble to understand the following part of the solution:

\begin{align} \int_{B(0, 1- \epsilon)} J_2(d \phi) d\mu \leq Area(M) \leq \int_{B(0, 1+ \epsilon)} J_2(d \phi) d\mu \end{align}

(It says that this inequality is true because the Jacobian is alway positive but I don't see how to get it)

Any suggestions? Thanks in advance!

Answer 1

The remark that the Jacobian is always positive is merely meant to point out that it is nonzero. In fact the surface in question is the graph of a smooth function and as such the jacobian will always be $>0$. In your case, the Jacobian as $\sqrt{1+x^2+y^2}$ as you pointed out, which is always $\geq1$. In the case when the surface is the graph of a smooth function (in your case, $z=xy$) the integral over the unit ball involving the Jacobian can be taken to be the definition of the area of the surface.

Answer 2

If i correctly understand the question we want to show only the stated double inequality.

Let $\phi:M\to\Bbb R$ be a continuous function with compact support on the given open surface (boundary removed from a surface with boundary), $0\le\phi\le 1$.

Let $S$ be the compact support of $\phi$. The distance between $S$ and the unit sphere, the boundary of the (compact full ball) $\bar B(0,1)$, is $>0$. (Else we consider sequences of points in $S$, respectively on the sphere that realize in distance in limit this zero, pass to convergent subsequences, get a contradiction.)

Choose an $\epsilon>0$ smaller as this distance. Consider $\phi_\epsilon$ to be the continuous function which

• is "radial", $\phi_\epsilon(x)=f(|x|)$ for a suitable real valued function $f$,
• is one on $\bar B(0,1-\epsilon)$, i.e. $f(r)$ is $1$ on $[0,1-\epsilon]$,
• is zero outside $\bar B(0,1-\epsilon/2)$, i.e. $f(r)$ is $0$ on $[1-\epsilon/2,\ \infty)$,
• is "radially linear between the balls", i.e. $f(r)$ is linear on $[1-\epsilon,\ 1-\epsilon/2]$.

Consider a similar function $\phi'_\epsilon$ which is one on $\bar B(0,1)$ and has support in $B(0,1+\epsilon)$. Then we have $$ \phi\le \phi_\epsilon\le \phi'_\epsilon\ .$$ The first inequality tells us that we can restrict to the functions in the "filtrating family" $\phi_\epsilon$, when passing to the supremum over all $\phi$. We obtain thus the first required inequality, the left one. Using $$\phi_\epsilon\le \phi'_\epsilon\le 1_{B(0,1+\epsilon)}$$ we get also the right inequality.