There are many various ways to derive the formula for the volume of a $p$-norm ball in $\mathbb{R}^n$
$$V_{n,p,R}=\frac{(2\Gamma(1+\frac{1}{p}))^n}{\Gamma(1+\frac{n}{p})}R^n$$
And I found many articles who use this to find the surface area (of $n-1$ dimensions) of the boundary sphere for $p=2$, using $\int_0^R S_{n,\color{red}2,r}\text{d}r=V_{n,\color{red}2,R}$ and concluding that $S_{n,\color{red}2,r}=\frac{\text{d}V_{n,\color{red}2,R}}{\text{d}r}$. This seems very intuative and does in-fact work.
My question is, how come this can't be generalized for arbitrary $p$? You would think that $p$-norm balls can also be divided into shells and by the same process achieve
$$S_{n,p,R}=\frac{(2\Gamma(1+\frac{1}{p}))^n}{\Gamma(\frac{n}{p})}pR^{n-1}$$
However, I found no articles doing this, and did find some sources which state there probably isn't any closed form for the surface area.
The point is that $$ \frac{d}{dR} V(R) = A(R) $$ works for the Euclidean norm because if $A$ and $B$ are balls with respect to such norm, $A+B$ still is a ball with respect to such norm. Such fact allows to derive a short proof of the isoperimetric inequality (in arbitrary dimension!) through the Brunn-Minkowski inequality, for instance. The same does not hold for different norms. Let us consider $n=2$ and the $4$-norm. The length of the curve $\{(x,y):x^4+y^4=1\}$ can be computed as $$ 4\int_{0}^{1}\sqrt{1+\frac{x^6}{(1-x^4)^{3/2}}}=\int_{0}^{1}x^{-3/4}\sqrt{1+\frac{x^{3/2}}{(1-x)^{3/2}}}\,dx=\int_{0}^{+\infty}\frac{\sqrt{1+x^{3/2}}\,dx}{x^{3/4}(1+x)^{5/4}} $$ which is a highly non-elementary integral, $\approx 7.017697943564$. The enclosed area is a bit better: $$ 4\int_{0}^{1}(1-x^4)^{1/4}\,dx = \int_{0}^{1}x^{-3/4}(1-x)^{1/4}\,dx = \frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{\pi}} $$ thanks to Euler's Beta function. In higher dimensions it only gets worse.