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finding a normal domain that is not convex.

I thought of this: $\{(x,y) | 4 0 \}$. I can see easily that this is not convex. However I have troubles showing that this is a normal domain, since I do not find a parametrization.

Does anybody see how to parametrize? Since I believe that if it is parametrizable then it is also normal. Please, do show me the right path.

Definition:

A domain $A\subset \mathbb{R}^{2}$ is a normal domain, if there are piecewise continuously differentiable functions $\alpha_{1},\alpha_{2} : [c,d] \rightarrow \mathbb{R}$ and $\beta_{1}, \beta_{2} : [a,b] \rightarrow \mathbb{R}$ such that : $ A=\{(x,y) | a < x < b, \beta_{1}(x) < y < \beta_{2}(x) \} = \{(x,y) | \alpha_{1}(y) < x < \alpha_{2} (y ) , c< y < d\}$

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    A slightly easier example: x^2+y^2>1, 02011-12-22

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Let $\beta_1,\beta_2:]0,3[\mapsto \mathbb{R}$ be defined by: $\beta_1(x):=\begin{cases} \sqrt{4-x^2} &\text{, if } 0 and: $\beta_2(x):= \sqrt{9-x^2}\; ;$ then it is clear that your set: $\Omega:=\{(x,y)\in \mathbb{R}^2:\ x,y>0 \text{ and } 4 can be represented as: $\Omega=\{(x,y)\in \mathbb{R}^2:\ 0 thus $\Omega$ is a normal domain w.r.t. $x$.