Answer
$\dfrac{\pi}{3} \ cubic \ units $
Work Step by Step
The volume of a solid generated by revolving region $R$ about x-axis can be calculated by using Washer Method as:
$V=\pi \int_p^q (R^2_{outer}-R^2_{inner} ) \ dx=\pi \int_p^q [f(x)^2- g(x)^2] dx\\=\pi \int_{0}^{1} [(\sqrt[4] x)^2-x^2] \ dx \\=\pi \int_0^1 (\sqrt x-x^2) \ dx \\=\pi (\dfrac{2x^{3/2}}{3}-\dfrac{x^3}{3}]_0^1\\=\pi [\dfrac{2}{3}-\dfrac{1}{3}]-0 \\=\dfrac{\pi}{3} \ cubic \ units $
