Using Cavalieri to derive a formula for $n$-dim $L^p$ ball let $p>0$ and $B_p^n:=\{x\in \mathbb{R}^n : |x_1|^p+\cdots + |x_n|^p\leq 1\}$. Show that $B_p^n$ is Borel-measurable with $$\lambda^n(B_p^n)=\frac{(2\Gamma(1+\frac{1}{p}))^n}{\Gamma(1+\frac{n}{p})}$$
My idea:
I have a hint saying that I have to look at the integral $\int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x)$. I know that I could just use power laws and fubini to achieve a gaussian integral that I could compute. But I have to use Cavalieri's principle.
I started like this: We have:$$\int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x)=\int \limits_{0}^{\infty}\lambda^n(\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\})\, \mathrm{d}t$$
Now I have to apply Cavalieri's principle to $\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\}$.
we have $\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\}=\{x\in \mathbb{R}^n : ||x||_p\leq (\log(t))^{1/p}\}\subset \mathbb{R}\times  \mathbb{R}^{n-1}$.
For all $y\in \mathbb{R}^{n-1}$ we cut $A_y=\{x\in \mathbb{R} : ||x||_p\leq \log(t)^{1/p}\}$. Now, we have:$$\lambda^n(\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\})=\int \limits_{\mathbb{R}^{n-1}}\lambda(A_y)\, \mathrm{d}y$$ How can I compute this integral?
 A: By Cavalieri, we can write
$$\int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x)=\int \limits_{0}^{\infty}\lambda^n(\{x\in \mathbb{R}^n : ||x||_p^p \leq \log (1/t)\})\, \mathrm{d}t. $$
Notice that the set we are measuring on the right is empty for $t > 1$ and is just the scaled up unit $L^p$-ball for $0 \leq t \leq 1$,
$$ \{x\in \mathbb{R}^n : ||x||_p^p \leq \log (1/t)\} = \log(1/t)^\frac1p \cdot B_p^n. $$
Therefore, by the scaling property of the Lebesgue measure ($\lambda^n (a B) = a^n \lambda^n (B)$), and recalling a well-known integral formula for the Gamma function, we write
$$ \int \limits_{0}^{\infty}\lambda^n(\{x\in \mathbb{R}^n : ||x||_p^p \leq \log (1/t)\})\, \mathrm{d}t = \lambda^n(B_p^n) \cdot \int_0^1 \log(1/t)^\frac{n}{p} \mathrm d t = \lambda^n (B_p^n) \cdot  \Gamma(1 + n/p). $$
At this point it is reasonable to expect
$$ \int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x) = (2 \Gamma (1 + 1/p))^n $$
which completes the derivation. This essentially is a generalization of the argument used to find the volume of the standard $L^2$-ball in $n$-dimensions, though rather than partitioning the domain of integration explicitly by $n-1$-dimensional spheres and radii we do it instead writing the spheres as level sets.
