Answer
$\dfrac{3 \pi }{10} $
Work Step by Step
The 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)(\sqrt x-x^2) \ dx\\= 2\pi \int_{0}^{1} (x^{3/2}-x^3) dx \\=2 \pi [\dfrac{2x^{5/2}}{5}-\dfrac{x^4}{4}]_{0}^{1} \\=2 \pi [\dfrac{2}{5}-\dfrac{1}{4}] \\=\dfrac{3 \pi }{10} $
