Answer
$1.7 \pi$
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, $V=2\pi \int_{0}^{1} (3-x) (\sqrt x-x^2) \ dx\\= 2 \pi \int_0^1 (3x^{1/2}-3x^2-x^{3/2}+x^3) \ dx \\=2 \pi [ \dfrac{3x^{3/2}}{3/2}-x^3-\dfrac{x^{5/2}}{5/2}+\dfrac{x^4}{4}]_0^1 \\=2 \pi [2-1- \dfrac{2}{5}-\dfrac{1}{4}] \\=1.7 \pi$
