Answer
$2 \pi$
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:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$
Now, $V=2\pi \int_{0}^{2} (x)[1-|x-1|) \ dx\\= 2\pi \int_{0}^{1} x [1+(x-1) ] \ dx +2\pi \int_{1}^{2} x [1-(x-1) ] \ dx \\= 2\pi \int_{0}^{1} x^2 \ dx +2\pi \int_{1}^{2} (2x-x^2) \ dx\\= 2 \pi [\dfrac{x^3}{3}]_0^1 +2 \pi [x^2-\dfrac{x^3}{3}]_1^2 \\=2 \pi(3-\dfrac{6}{3})\\=2 \pi$
