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While going through the first few chapters of my multivariable calculus book, I came across the following:

The graph of a function of two variables is a surface in $\mathbb{R}^3$ and is a level set of a function of three variables. However, not all level sets of functions of three variables are graphs of functions of two variables.

I am finding trouble grasping the notion of this intuitively. Is it actually impossible to find an arbitrary graph that corresponds to a given level set (for the cases given above)?

Could I possibly ask for a concrete example that demonstrates the statement?

2 Answers2

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The statement is a bit vague but I think this is probably what the author means.

The graph of a function of two variables... is a level set of a function of three variables. That is, if $f(x,y)$ is a function of two variables, then the graph $z=f(x,y)$ can be written as $$z-f(x,y)=0\ .$$ If you now define a function of three variables $$g(x,y,z)=z-f(x,y)\ ,$$ then the above is simply $$g(x,y,z)=0\ .$$ This is an example of a level set of $g$, that is, it is $g(x,y,z)={}$ constant.

However, not all level sets of functions of three variables are graphs of functions of two variables. For example, consider $$g(x,y,z)=x^2+y^2+z^2\ .$$ One of its level sets is $x^2+y^2+z^2=1$, a sphere. However, this cannot be written in a form where $z$ is a function of $(x,y)$, because many values of $(x,y)$ give two possibilities for $z$.

David
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Consider a three-dimensional sphere $$x^2+y^2+z^2 = 1.$$ It is the level set of the function $f: \mathbb{R}^3 \to \mathbb{R}$ given by $$f(x,y,z) = x^2 +y^2 +z^2$$ for $f(x,y,z) = 1$. You cannot write any of the $x,y,z$ in terms of the other two. For example, if you isolate $z$ in terms of $x,y$ you will arrive at $$z^2 = 1-x^2-y^2.$$ This will force you to make a choice. Either $z \geq 0 $ or $z \leq 0$. You cannot have both simultaneously. Thus $$z = \sqrt{1-x^2-y^2}$$ or $$z= - \sqrt{1-x^2-y^2}.$$ The former takes the upper part of the sphere, while the latter takes the lower part. In both cases there are points of the sphere not covered by the function.

Mark Fantini
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