I'm studying multiple variable functions, and a doubt appeared: When I have the function$f(x,y) = (x, y, x^2 + y^2)$, how can I see the image of the function? Because, the first two coordinates are valid in all the $\mathbb{R}^3$, but the third only in $\mathbb{R}^3 > 0$, so, how can I know the image of the function $f$?
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$\begingroup$Here, $ f:\Re^2 \rightarrow \Re^3 $ , $~ f(x,y) = (x,y,x^2+y^2) $.
Therefore the image set is $ z= \{x,y\in\Re~|~x^2+y^2 \} $
For visualization purpose, as in $2D~$, function $f(x) = y^2$ is a parabola.
On similar grounds, in $3D~$, function $f(x,y) = x^2+y^2 $ is a paraboloid.
Like other people have said in their comments (@saulspatz, @drhab), the image of $f$ is the graph of the cone given by $z=x^2 + y^2$. But there is more to that...
In general, we mathematicians (and at some point you might be one) do what we can to represents objects graphically. Our vision is severely constrained by the three dimensions we can observe, and all too often we have to resort to sketchy drawings on the board or on paper. Of course one can (and it is often a good idea) ask a computer program to plot a function for us.
In this case, do note that the first two entries of $(x,y,x^2 + y^2$ are precisely the two arguments of the function $f$. So it should be enough to 'draw' $x,y,x^2 + y^2$ in order to see the 'image of the function'. This time round, we know that this corresponds to the cone above, so you would be done with the drawing part.
Replying to your comment asking if the image was the set $\{(x,y,z)|x,y,\in\mathbb{R}, z>0\}$, the answer is no. Mind that not every positive $z$ will be in the image for every positive pair $(x,y)$. For each pair $(x,y)$, only the $z\in\mathbb{R}$ such that $z = x^2 + y^2$ will be there. So, in notation, this means that the image is precisely the set$$\{(x,y,x^2 + y^2)|x,y,\in\mathbb{R}\}.$$
If you want to see it visually, cut a slice parallel to the $XY$ plane at any height $h\geq 0$. This slice will intersect the cone at the points that are at a distance $\sqrt{h}$ from the $OZ$ axis. The image will be formed by the collection of all the points in the intersections of the many slices with the cone.
Please reply if you have further questions, I will be happy to help.
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