Answer
$$2\pi$$
Work Step by Step
$$Volume =\int_{0}^{\infty} 2 \pi x \times e^{-x} dx \\=2 \pi \lim\limits_{a \to \infty} \int_0^a xe^{-x} dx \\ =2 \pi \times \lim\limits_{a \to \infty} \int_0^a xe^{-x} dx \\=2 \pi \times \lim\limits_{a \to \infty} [ (-x e^{-x})_0^a +\int_0^a e^{-x} dx \\ = 2 \pi \lim\limits_{a \to \infty} [-a e^{-a} +(-e^{-a} +e^{-0})] \\= 2 \pi \times \lim\limits_{a \to \infty} [-a e^{-a}]-2\pi \times (0)+2 \pi $$
We need to apply L'Hopital's rule:
$$Volume=2 \pi \times \lim\limits_{a \to \infty} \dfrac{-a}{e^{a}}+2 \pi \\=2 \pi \times \lim\limits_{a \to \infty} \dfrac{-1}{e^{a}}+2 \pi \\= 2\pi$$
