An area preserving diffeomorphism between a disk and an ellipse

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This is a self-answered question. I post it here since (embarrassingly) it took me some time to realize that the solution is obvious.

Let $D \subseteq \mathbb R^2$ be the closed unit disk and let $E$ be an ellipse with the same area, i.e. with minor and major axes of lengths $a<b$ and $ab=1$. $$ E=\{(x,y) \, | \, \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \} $$

Question: Can we construct explicitly an area preserving diffeomorphism $f:D \to E$?

(i.e. $Jf=1$ identically on $D$).

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We have $A(E)=\pi ab$, so $A(E)=A(D)=\pi$ if and only if $ab=1$.

The linear map $f:(x,y) \to (\tilde x ,\tilde y)=(ax,by)$ is an area preserving diffeomorphism $D \to E$.

Clearly $Jf=ab=1$. We just need to make sure that $f$ maps $D$ onto $E$.

Indeed, $$ (\tilde x ,\tilde y) \in E \iff (\frac{\tilde x}{a})^2 + (\frac{\tilde y}{b})^2 \le 1 \iff x^2+y^2 \le 1 \iff (x.y) \in D. $$