Using the disk method, how do you find the volume of the solid generated by revolving about the x-axis the area bounded by the curves x=0x=0, y=0y=0 and y=-2x+2y=2x+2?

1 Answer
Sep 11, 2015

V = (4pi)/3V=4π3

Explanation:

Plot of y = 2 - 2xPlot of y = 2 - 2x

Looking at the graph of y = 2 - 2xy=22x, imagine the area between the line, the x-axis, and the y-axis, being revolved around the x-axis. You'll end up with a cone, with the point/tip at x = 1x=1, and the center of the circular base (which has radius 2) at the origin.

Next, imagine looking at a cross-section parallel to the x-z axis - parallel to the circular base. Every cross-section will be a circle. At x = 0x=0, the circle has radius 2, and at x = 1x=1, the circle has radius 0. In fact, for any xx, the cross-sectional circle has radius 2 - 2x22x.

To find the volume of the solid of revolution, we can imagine that our solid is composed of infinitely many disks, of infinitesimal width, and radius equal to 2 - 2x22x. To find the volume of the solid, we sum up (integrate) each disk. This process is commonly called the method of disks.

The general formula for the method of disks is:

V = int_a^b pi*(f(x))^2 dxV=baπ(f(x))2dx

where f(x)f(x) is the curve we're revolving about the x-axis, and [a,b][a,b] is the interval we're concerned with. Important to note is the pi*(f(x))^2π(f(x))2 - this just means "area of a circle with radius f(x)f(x)." Multiplying this by dxdx gives the volume of a cylinder (disk) of width dxdx and radius dxdx. And integration allows us to sum all these disks up, exactly what we want to accomplish.

So, in our case, we'll note that f(x) = 2 - 2xf(x)=22x, and [a,b] = [0,1][a,b]=[0,1], and substitute:

V = int_0^1 pi*(2 - 2x)^2 dxV=10π(22x)2dx

Well, that was the difficult part; setting up the integral. From here, evaluating the integral should be fairly easy. I'll leave it to you as an exercise, but the answer should come out to

V = (4pi)/3V=4π3