Find the volume when revolving about the y-axis the region enclosed by the graphs y=9-x^2 and y=9-3x, [0, 2]?

Answer 1

For starters, figure out which one of the graphs is on top. You can doa quick check and see that the quadratic is on top of the line for the given domain.

As for the rest of the setup, we should decide if we're going to solve using disks/washers or shells. Since we're revolving around the y-axis and since the equations are already in y= format, we can go with the shell method.

The shell method uses a cylinder with a height y2 - y1, a radius of x (and thus a circumference of 2pix), and a thickness of dx. If we uncurl the cylinder, we have a flat rectangular prism that we can find the volume of. So we can set up and solve as follows (I'm using S as the integration symbol):

V = S (y2 - y1)(2pix)dx
V = 2pi S ((9 - x^2) - (9 - 3x))x dx
V = 2pi S (-x^2 + 3x)x dx
V = 2pi S (-x^3 + 3x^2)dx
V = 2pi((-x^4)/4) + x^3), from 0 to 2
V = 2pi((-2^4)/4 + 2^3) - ((0^4)/4 + 0^3)
V = 2pi(4 - 0)
V = 8pi