## Calculus (3rd Edition)

$\dfrac{776 \pi}{15}$
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, $R_{outside}=2+x^2+2=x^2+4$ and $R_{inside}=4$ Now, $V=\pi \int_0^2 [(x^2+4)^2-(2)^2] \ dx \\ = \pi \int_0^2 [x^4 +16+8x^2-4] \ dx \\=\pi \int_0^2 [x^4+8x^2+12] \ dx \\=\pi (\dfrac{x^5}{5}+\dfrac{8x^3}{3}+12x )_0^2 \ dx \\=\dfrac{776 \pi}{15}$