#### Answer

$\dfrac{136 \pi}{15}$

#### Work Step by Step

The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ 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:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
where, $R_{outside}=4-(x^2-1)=5-x^2 $ and $ R_{inside}=4-(2x-1)=5-2x$
Now, $V=\pi \int_0^2 [(5-x)^2 -(5-2x)^2] \ dx \\ = \pi \int_0^2 [-14x^2+x^4+20x ] \ dx \\=\pi [ -\dfrac{14 x^3}{3}+\dfrac{x^5}{5}+10x^2 ]_0^2 \ dx \\=\pi [ -\dfrac{14 (2^3)}{3}+\dfrac{(2^5)}{5}+10(2^2)] \\=\dfrac{136 \pi}{15}$