Find the volume of revolution by rotating the region y=16-x, y=3x+12, and x=-1, about the x-axis using the disk or washer method.
Let's take the I don't know what method (say hollow disk).
Your hollow disks will have an area of $\displaystyle \pi(f_2(x))^2 - \pi(f_1(x))^2 = \pi(16-x)^2 -\pi(3x - 12)^2 =112\pi+40\pi x-8\pi x^2 $*and an heigth of [tex]*dx [tex]
You must find the intersection to know your bound of integration.
[tex] 16-x = 3x+12 \implies 1 \geq x \geq -1[tex]
Finally
$\displaystyle V = \int_{-1}^1 (112\pi+40\pi x-8\pi x^2) dx = -\frac{1}{3}\pi (16-x)^3-\frac{1}{9}(1/9)\pi(3x-12)^3 = \frac{656\pi}{3}$