# coordinate transformation from $S = \{(u,v)\}\big|0\leq u,v \leq 1\}$ to $D = \{(x,y)\big| 0\leq y\leq2;|x|\leq \frac{y^2}{4}-1 \}$

$S = \{(u,v)\}\big|0\leq u,v \leq 1\}$

$D = \{(x,y)\big| 0\leq y\leq2;\big|x\big|\leq \big|\frac{y^2}{4}-1 \big|\}$

My teacher said S and D has figure like this：

Q1:How to prove this formula is the right figure????

apply transform $$\Phi: \begin{cases} x = u^2+v^2\\ y = 2uv \end{cases}$$ can transform from S to D.

Q2:How to make use of $\Phi$ and $D = \{(x,y)\big| 0\leq y\leq2;\big|x\big|\leq \big|\frac{y^2}{4}-1 \big|\}$ to obtain this? $$S = \{(u,v)\}\big|0\leq u,v \leq 1\}$$ Update: I change the formula of D to $$|x|\leq \big| \frac{y^2}{4}-1 \big|$$ I plot a figure in matlab, and now it's much more lick the PPT's figure:

ezplot('abs(y^2/4-1)=abs(x)')


• I'm not sure I understand your question. Are you wanting to ask how to prove that $\Phi[S]=D$? Jun 10, 2016 at 13:47
• You are correct in that your equation of $D$ doesn't describe the picture of $D$. Maybe the teacher meant $|x| \leq \left| \frac{y^2}{4} - 1 \right|$ and not $|x| \leq \frac{y^2}{4} - 1$? Also, $\Phi$ surely doesn't transform $S$ to $D$ as depicted in the picture. Jun 10, 2016 at 13:53
• @GitGud Yes , I mean how to prove that?
– Long
Jun 10, 2016 at 16:26
• @levap You are right, after add abs value at the right side:$$|x|\leq \big| \frac{y^2}{4}-1 \big|$$ I plot a figure in matlab much more like the one of D in my question, I've updated my question.
– Long
Jun 10, 2016 at 16:30
• @Long Well, it's not true. Take for instance $\Phi(1,1)$. It's not in $D$ even though $(1,1)\in S$. Jun 10, 2016 at 17:11