Answer
$2 \pi$
Work Step by Step
We calculate the volume as follows:
Volume $=2 \int_0^{\infty} \pi (e^{-x/2})^2 \ dx\\=2\pi \int_0^{\infty} (e^{-x/2})^2 \ dx \\=2\pi \lim\limits_{R \to \infty}[-e^{-x}]_0^R\\=-2 \pi \lim\limits_{R \to \infty} (e^{-R}-1)\\=2 \pi (1-0)\\=2 \pi$
