Answer
$$\dfrac{\pi}{2}$$
Work Step by Step
$$Area =\pi r^2=\pi (e^{-x})^2=\pi e^{-2x}$$
Now, $$Volume =\int_{0}^{\infty} \pi e^{-2x} dx \\= \pi \lim\limits_{u \to \infty} \int_0^u e^{-2x} dx \\= \pi \times \lim\limits_{u \to \infty} [-\dfrac{1}{2} e^{-2u} -(-\dfrac{1}{2} e^{0})] \\= \pi (0+\dfrac{1}{2}) \\=\dfrac{\pi}{2}$$
