Answer
$\dfrac{\pi}{2}$
Work Step by Step
Area of a cross-sectional region: $A=\pi \times (e^{-x})^2=\pi e^{-2x}$
Now, $Volume =\int_{0}^{\infty} \pi e^{-2x} dx= \pi \lim\limits_{a \to \infty} \int_0^a e^{-2x} dx$
or, $= \pi \lim\limits_{a \to \infty} [-\dfrac{1}{2} e^{-2a} -(-\dfrac{1}{2} e^{0})]$
or, $= \pi (0+\dfrac{1}{2})$
or, $=\dfrac{\pi}{2}$
