m.e._liberti ha scritto:Come si ci arriva a calcolare semplicemente [...]?
Dato che: \[
\text{area}(A) := \iint\limits_A 1\,\text{d}x\,\text{d}y,
\quad \quad
x_G:=\frac{1}{\text{area}(A)}\iint\limits_A x\,\text{d}x\,\text{d}y,
\quad \quad
y_G:=\frac{1}{\text{area}(A)}\iint\limits_A y\,\text{d}x\,\text{d}y
\] allora: \[
\text{volume}(V) = \alpha\;\text{distanza}(G,\text{asse})\,\text{area}(A)
\] e quando \(\alpha=2\pi\) e l'asse di rotazione coincide con l'asse x, si ha: \[
\text{volume}(V)=2\pi\,|y_G|\,\text{area}(A) = 2\pi\left|\iint\limits_A y\,\text{d}x\,\text{d}y\right| = \pi\int_0^1 \left(e^{-x}\sqrt{x}\right)^2\text{d}x=\frac{\pi}{4}\left(1-\frac{3}{e^2}\right).
\]