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Given this function:

$$f(x,y)=(x^2-y^2,2xy)$$

Sketch the image under $f$ of the set:

$$S=\left\{(x,y)\:|\: x^2+y^2 \leq a^2 \:\&\&\: x\geq 0 \:\&\&\: y\geq 0 \right\}$$

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closed as unclear what you're asking by george2079, José Antonio Díaz Navas, Michael E2, Coolwater, MarcoB Apr 15 '18 at 14:38

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  • $\begingroup$ Convert to polar coordinates? $\endgroup$ – Michael E2 Apr 12 '18 at 0:28
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    $\begingroup$ is this even a mathematica question? Whats an "imagem"? $\endgroup$ – george2079 Apr 12 '18 at 3:55
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Perhaps this:

Block[{a = 2},
 ParametricPlot[{x^2 - y^2, 2 x y}, {x, y} ∈ Disk[{0, 0}, a, {0, Pi/2}]]
 ]

Update: Perhaps an answer to @Rom38's region interpretation:

Block[{a = 2},
 RegionPlot@
  ParametricRegion[{{x^2 - y^2, 2 x y}, {x, y} ∈ Disk[{0, 0}, a, {0, Pi/2}]}, {x, y}]
 ]
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  • $\begingroup$ To show the output seems a bit of a spoiler to me. This is a nice, easy exercise by hand or in one's head. $\endgroup$ – Michael E2 Apr 12 '18 at 0:32
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    $\begingroup$ You've missed the condition x>=0 and y>=0 $\endgroup$ – Rom38 Apr 12 '18 at 3:45
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    $\begingroup$ @Rom38 Yeah, I guess doing someone else homework completely is better for the world than leaving such details for the reader....Anyway someone else couldn't resist, so why not add the missing restriction? Thanks for pointing it out. $\endgroup$ – Michael E2 Apr 12 '18 at 11:15
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    $\begingroup$ I've pointed it out because on my view the question was more about region definition than about plotting of the functions. But I agree with you - the solution of school tasks is not for this site. $\endgroup$ – Rom38 Apr 13 '18 at 4:43
  • $\begingroup$ @Rom38 I've updated my answer in response to your interpretation to the question being about regions. $\endgroup$ – Michael E2 Apr 13 '18 at 11:05
1
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How about this?

Manipulate[
  ParametricPlot[{x^2 - y^2, 2 x y}, {x, 0, a}, {y, 0, Sqrt[a^2 - x^2]}, 
    PlotPoints -> 60], 
  {{a, 1}, 0.1, 10}]
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