Which is the image and the area of this set?

Question

Lets have this application defined by: $$ \begin{cases} x = u^2-v^2 \\ y = 2uv \end{cases} $$

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Which are the image and the area of $\Omega$ by the transformation of the rectangle $R$, which vertexes are $(1,1), (2,1), (2,3)$ and $(1,3)$ ?

Answer 1

We will follow the path $(1,1) \rightarrow (2,1) \rightarrow (2,3) \rightarrow (1,3) \rightarrow (1,1)$ and explicitly give the curves in the form $y = f(x)$.

$(1,1) \rightarrow (2,1)$: Here $v = 1$ and $u \in [1,2]$ $$ \begin{cases} x = u^2 -1 \\ y = 2u \end{cases} \Longrightarrow \begin{cases} u = \sqrt{x+1}\\ y = 2u \end{cases} \Longrightarrow y = 2\sqrt{x+1} $$ for $x \in [0,3]$

$(2,1) \rightarrow (2,3)$: Here $u = 2$ and $v \in [1,3]$ $$ \begin{cases} x = 4 -v^2 \\ y = 4v \end{cases} \Longrightarrow \begin{cases} v = \sqrt{4-x}\\ y = 4v \end{cases} \Longrightarrow y = 4\sqrt{4-x} $$ for $x \in [-5,3]$

$(2,3) \rightarrow (1,3)$: Here $v = 3$ and $u \in [1,2]$ $$ \begin{cases} x = u^2 -9 \\ y = 6u \end{cases} \Longrightarrow \begin{cases} u = \sqrt{x+9}\\ y = 6u \end{cases} \Longrightarrow y = 6\sqrt{x+9} $$ for $x \in [-8,-5]$

$(1,3) \rightarrow (1,1)$: Here $u = 1$ and $v \in [1,3]$ $$ \begin{cases} x = 1 -v^2 \\ y = 2v \end{cases} \Longrightarrow \begin{cases} u = \sqrt{1-x}\\ y = 2v \end{cases} \Longrightarrow y = 2\sqrt{1-x} $$ for $x \in [-8,0]$

See the plot on WA.