If we have $f(r) = r^n$, where $r^n = (x^2+y^2+z^2)^{n/2}$, and one wishes to examine the level curves of $f(r)$, namely $f(r) = k$ for some real number $k$, then one might proceed as follows: there would be two cases - (i) take $n>0$ and (ii) $n<0$.
Now, for case (i), taking $n=1$, we'll have the upper part of a sphere in 3-space of radius $\sqrt{k}$. Taking $n=2$, we'll obtain a whole sphere of radius $\sqrt{k}$. But what about taking $n > 2$? That's where I'm getting lost. What kind of a surface are we going to get?
For case (ii), it becomes even more complicated. For example, for $n=-2$, we'll have something like a "reversed sphere".
The question is that we can actually view these surfaces in two ways:
(1) For example, for $n=5$, we can arrive at $x^2+y^2+z^2=k^{2/5}$. For $n=-2$, we can arrive at $x^2+y^2+z^2=\frac{1}{k}$, and both are just spheres in 3-space.
(2) We can expand (for $n=5$) the expression $(x^2 + y^2 + z^2)^{5/2}= k$ and get a very long expression with powers of 4 and less.
Which approach is "more correct" - (1) or (2)?
$\newcommand{\Reals}{\mathbf{R}}$If $n$ is a positive real number and $f_{n}:\Reals^{3} \to \Reals$ is defined by $$ f_{n}(x, y, z) = (x^{2} + y^{2} + z^{2})^{n/2}, \tag{1} $$ then the level sets of $f_{n}$ are spheres centered at the origin, independently of $n$. (The level set at "height" $0$ consists of the origin alone, but may be viewed as a sphere of radius $0$.)
Specifically, if $k \geq 0$, the set \begin{align*} f_{n}^{-1}(\{k\}) &= \{(x, y, z) \in \Reals^{3} : f(x, y, z) = k\} \\ &= \{(x, y, z) \in \Reals^{3} : x^{2} + y^{2} + z^{2} = k^{2/n}\} \end{align*} is the sphere of radius $k^{1/n}$ centered at the origin.
If $n < 0$, the formula (1) defines a function on "punctured" $\Reals^{3}$, from which the origin has been removed, but the level set analysis is the same: If $k > 0$, the set $f_{n}^{-1}(\{k\})$ is the sphere of radius $k^{1/n}$.
The underlying point of confusion in comparing Questions 1 and 2 may stem from a misconception. In fact, the expression $(x^{2} + y^{2} + z^{2})^{n/2}$ is not a polynomial in $(x, y, z)$ unless $n$ is a non-negative even integer.
As a general matter, if $u$ is a real-valued function on $\Reals^{3}$, such as $u(x, y, z) = r^{2} = x^{2} + y^{2} + z^{2}$, and if $f$ is a real-valued function of one variable, such as $f(u) = u^{n/2}$, then:
Every level set of $f \circ u:\Reals^{3} \to \Reals$ is a union of level sets of $u$.
If $f$ is monotone, then each level set of $f \circ u$ is a level set of $u$.