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This problem is driving me crazy.

"Let $S$ be a parametrized surface by $\varphi:[a,b]\times[c,d]\rightarrow\mathbb{R^3}$, $\varphi$ is of class $C^1$. Show that exists an open, connected set $D\subset\mathbb{R^2}$ and a bijection $f:D\rightarrow[a,b]\times[c,d]$ of class $C^1$ such that $\varphi\circ f$ preserves areas."

Here is what I did:

I started supposing that $D$ has no holes, doing this it's possible to calculates the area of any region $D\ '\subset D$ by $$\int_{\alpha_1}^{\beta_1}g_1(x)-h_1(x) \ dx +\ldots+\int_{\alpha_n}^{\beta_n}g_n(x)-h_n(x) \ dx$$

where each $g_i(x)$ is the upper function and $h_i(x)$ is the lower function. To make it simpler, but not losing generality, I may consider only one region like this, so we have that $$\int_{\alpha}^{\beta}g(x)-h(x) \ dx$$

is the area of $D\ '$.

*Note that $(\varphi\circ f)(x,y)=\Big(\varphi_1(f(x,y)),\varphi_2(f(x,y)),\varphi_3(f(x,y)\Big)$.
The corresponding area in S is the area of $(\varphi\circ f)(D\ ')$, it's is given by $$\int_{D\ '} \bigg| \bigg( \frac{\partial\varphi_1(f)}{\partial x},\frac{\partial\varphi_2(f)}{\partial x},\frac{\partial\varphi_3(f)}{\partial x}, \bigg)\times\bigg( \frac{\partial\varphi_1(f)}{\partial y},\frac{\partial\varphi_2(f)}{\partial y},\frac{\partial\varphi_3(f)}{\partial y}, \bigg) \bigg|\cdot |J_f| dA=$$ $$=\int_\alpha^\beta\int_{h(x)}^{g(x)} \bigg| \bigg( \frac{\partial\varphi_1(f)}{\partial x},\frac{\partial\varphi_2(f)}{\partial x},\frac{\partial\varphi_3(f)}{\partial x}, \bigg)\times\bigg( \frac{\partial\varphi_1(f)}{\partial y},\frac{\partial\varphi_2(f)}{\partial y},\frac{\partial\varphi_3(f)}{\partial y}, \bigg) \bigg|\cdot |J_f|\ dy\ dx.$$

This takes me to think that $\bigg| \bigg( \frac{\partial\varphi_1(f)}{\partial x},\frac{\partial\varphi_2(f)}{\partial x},\frac{\partial\varphi_3(f)}{\partial x}, \bigg)\times\bigg( \frac{\partial\varphi_1(f)}{\partial y},\frac{\partial\varphi_2(f)}{\partial y},\frac{\partial\varphi_3(f)}{\partial y}, \bigg) \bigg|\cdot |J_f|=1$, that way I have the equality I'm looking for. This is probably wrong because all I have to do to solve this is equation is to find some $f$, it doesn't depend on $D$.

Please, someone could point out what I'm doing wrong and show me the correct way.

Thank you very much for your help.

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1 Answer

up vote 2 down vote accepted
+50

I assume that your representation $\phi$ is regular, i.e., that $$w(x,y):=|\phi_x(x,y)\times \phi_y(x,y)|\ne 0$$ for all $(x,y)\in R:=[a,b]\times[c,d]$. Instead of $f:\ D\to R\ $ we shall construct its inverse $g:\ R\to D$.

Note that $w(x,y)$ represents the local area dilatation of the map $\phi$. We now take care that $g$ has the same local area dilatation $w(x,y)$. To this end let $$u(x,y):=\int_a^x w(t,y)\ dt$$ and put $$g(x,y):=\bigl(u(x,y),y\bigr)\qquad \bigl((x,y)\in R\bigr)\ .$$ Then $g$ maps horizontal lines $y=\eta$ to the same lines $v=\eta$ in the $(u,v)$-plane, and as $u_x(x,y)>0$ for all $(x,y)$ each such horizontal is mapped in a strictly increasing fashion onto its parallel in the $(u,v)$-plane. It follows that the map $g$ is injective onto some image domain $D$. Furthermore the Jacobian of $g$ computes to $$J_g(x,y)=u_x(x,y)=w(x,y)>0\ .$$ It follows that $g$ is regular and thus has a $C^1$-inverse $f$. In addition the area dilatation of $g$ is $w(x,y)$, and therefore the area dilatation of $f:=g^{-1}$ at the point $g(x,y)$ is ${1\over w(x,y)}$.

From this we conclude that $\phi\circ f$ has area dilatation $1$ at all points, in other words: that $\phi\circ f$ preserves areas.

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sorry for taking so long to comment. I have a question. How do guarantee that $g$ has a $C^1$ inverse from the fact that $g$ is regular. I thought you were using the inverse function theorem, but this theorem only gives a bijection in some neighborhood inside $D$. Could you explain me this part? Thanks. –  Integral Apr 8 '13 at 4:06
    
@Integral: As $g$ maps $R$ bijectively onto $D$ by construction the inverse $f:\ D\to R$ is well defined to start with. The inverse function theorem guarantees its differentiability. –  Christian Blatter Apr 8 '13 at 7:46
    
I did understand the $C^1$ part. What I didn't understand is why $g$ is bijective. –  Integral Apr 8 '13 at 15:30
    
@Integral: It's explained in the paragraph beginning with "Then $g$ maps horizontal $\ldots$". –  Christian Blatter Apr 8 '13 at 15:45
    
thank you for helping me Blatter, I finally understand your solution. –  Integral Apr 8 '13 at 18:16
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