Answer
$0.3 \pi$
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:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$
Now, $V=2\pi \int_{0}^{1} x (\sqrt x-x^2) \ dx\\= 2 \pi \int_0^1 (x^{3/2}-x^3) \ dy \\=2 \pi [ \dfrac{x^{5/2}}{5/2}-\dfrac{x^{4}}{4}]_0^1 \\=2 \pi [ \dfrac{2}{5}-\dfrac{1}{4}] \\=0.3 \pi$
