University Calculus: Early Transcendentals (3rd Edition)

$$\dfrac{2 \pi}{3}$$
We need to integrate the integral to compute the volume. $$Volume= (\pi) \times \int_{0}^{2} (1-x+\dfrac{x^2}{4}) dx \\ = (\pi x-\dfrac{\pi x^2}{2}+\dfrac{\pi x^3}{12})]_{0}^{2} \\=\pi (2-2+\dfrac{2}{3}) \\=\dfrac{2 \pi}{3}$$