## Calculus (3rd Edition)

$\dfrac{2 \pi}{5}$
Shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by: $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$ Now, $V=2\pi \int_0^{1} (x)(x^3) \ dx\\=2\pi \int_0^{1} (x^4) \ dx \\=2 \pi [\dfrac{x^5}{5}]_0^{1} \\=\dfrac{2 \pi}{5}$