I'd like to add something to the discussion made above.

Let $f:]0,+\infty[ \to [0,+\infty[$ a measurable function, $N\in \mathbb{N}$, $x\in \mathbb{R}^N$ and $t\in \mathbb{R}$. Consider the set:

$$E_f:=\{ (x,t)\in \mathbb{R}^{N+1}:\ |x|\leq |f(t)|\}$$

which is the *body of revolution generated by $f$* with respect to the $t$ axis in $\mathbb{R}^{N+1}$.
Using cylindrical coordinates, one finds for the Lebesgue measure $\mathcal{L}^{N+1}(E_f)$ the following expression:

$$\mathcal{L}^{N+1} (E_f)=\omega_N \int_0^{+\infty} f^N (t)\ \text{d} t\; ,$$

where $\omega_N$ is the volume of the unit ball in $\mathbb{R}^N$ (i.e. $\omega_N:=\pi^{\frac{N}{2}}/ \Gamma (\frac{N}{2} +1)$), hence $\mathcal{L}^{N+1}(E_f)=\omega_N \| f\|_{L^N}^N$ and $\mathcal{L}^{N+1} (E_f)$ is finite iff $f\in L^N(]0,+\infty[)$.

On the other hand, if $f$ is also *Lipschitz continuous*, it is easy to compute the *surface area* (or *De Giorgi perimeter*) $\mathcal{P} (E_f)$ of $E_f$: using cylindrical coordinates one finds:

$$\mathcal{P} (E_f)=N\omega_N \int_0^{+\infty} f^{N-1}(t)\ \sqrt{1+|f^\prime (t)|^2} \text{d} t\; ,$$

thus:

$$N\omega_N \| f\|_{L^{N-1}}^{N-1} \leq \mathcal{P} (E_f) \leq N\omega_N\sqrt{1+\| f^\prime \|_{L^\infty}^2}\ \| f\|_{L^{N-1}}^{N-1}$$

and $\mathcal{P} (E_f)$ is finite iff $\| f\|_{L^{N-1}}^{N-1}$ does.

Let $\mathcal{S} (E_f)$ be the measure of the sections of $E_f$ obtained cutting the set with any hyperplane $\Pi:=\{ (x,t)|\ \langle a,x\rangle =0\}$ ($|a|=1$) containing the axis of revolution; then:

$$\mathcal{S} (E_f) =\omega_{N-1} \int_{0}^{+\infty} f^{N-1}(t)\ \text{d} t\; ,$$

and also $\mathcal{S} (E_f)$ is finite iff $f\in L^{N-1} (E_f)$.

Since $f$ is defined in the interval $[0,+\infty[$, one in general doesn't have $f\in L^{N}\Rightarrow f\in L^{N-1}$ not even if $f$ is Lipschitz (e.g., $f(x):=\chi_{[0,1[}(x)+x^{1-N}\chi_{[1,+\infty[} (x)$ is $L^N$ but not $L^{N-1}$): thus in general it is always possible to pick a Lipschitz function $f$ such that $\mathcal{L}^{N+1} (E_f)$ is finite and $\mathcal{P} (E_f)$, $\mathcal{S} (E_f)$ are not.

**N.B.**: Instead of the De Giorgi perimeter, one can use the *Hausdorff measure* $\mathcal{H}^{N}$ or the *Minkowski content* $\mathcal{M}$ as well: in fact there is equality among $\mathcal{P} (E_f)$, $\mathcal{H}^{N} (E_f)$ and $\mathcal{M} (E_f)$ because the boundary $\partial E_f$ is sufficiently regular when $f$ is Lipschitz.