3D graphing on a 3-D plane?

Question

How would this function look hand drawn using the x,y,z axis: $$||x^2+y^2||$$

Answer 1

Consider the range of the function, then draw level curves:

f(x,y) = |x+y| = 0

f(x,y) = |x+y| = 1

f(x,y) = |x+y| = 2

Etc.

Answer 2

We can verify this with 2-D intersections.

Imagine (yet unknown) 3-D object intersected by a plane $z=w$. We know the object equation is $z=f(x,y)\equiv |x+y|$ and the plane is parallel to the $xy$ plane for every $w$.

Now it's all about computing and imagining those intersections for different values of $w$ i.e. for different $z$ coordinates.

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For $w<0$ we have:

$$w=|x+y|<0$$

with no solutions; none of the $z=w$ planes intersects with the object in this case.

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For $w=0$ the equation is:

$$0=|x+y|$$ $$0=x+y$$ $$y=-x$$

This is the "crease" mentioned in vadim123's comment. At this point we already know your drawing is wrong.

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For $w=1$:

$$1=|x+y|$$ $$(x+y=-1) \vee (x+y=1)$$ $$(y=-x-1) \vee (y=-x+1)$$

So there are two parallel lines at $z=1$.

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For $w=2$ we have…

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The two lines are more distant than those previous ones.

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For $w=3$…

And so on.