Answer
$\dfrac{13 \pi }{6} $
Work Step by Step
The disk method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$V= \pi \int_{m}^{n} (R^2) \ dx$
Now, $V=2\pi \int_{0}^{1/2} (x^2+2) \ dx +\pi \int_{1/2}^{2} (x-2)^2 \ dx \\=\pi \int_{0}^{1/2} (x^2+2) \ dx +\pi \int_{1/2}^{2} (x^2 +4-4x) \ dx \\=\pi (\dfrac{x^3}{3}+2x)_0^{1/2} +\pi (\dfrac{x^3}{3}+4x-2x^2)_{1/2}^2 \\= \pi (\dfrac{1}{24}+1)+\pi (\dfrac{2^3}{3}+8-8-\dfrac{1}{24}-2+\dfrac{1}{2}) \\=\dfrac{13 \pi }{6} $
