Answer
$\dfrac{56 \pi}{15}$
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}=-x^2+4x-3-(-1)=-x^2+4x-2 $ and $ R_{inside}=0-(-1)=1$
Now, $V=\pi \int_{1}^{3} [(-x^2+4x-2)^2 -1] \ dx \\ = \pi \int_1^3 [x^4-8x^3+20x^2-16x+3 ] \ dx \\=\pi [\dfrac{x^5}{5}-2x^4+\dfrac{20x^3}{3}-8x^2+3x]_{1}^{3} \ dx \\=\pi [\dfrac{18} {5}+\dfrac{2}{15}] \\=\dfrac{56 \pi}{15}$
