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