Question

Let R be the region in the first quadrant enclosed by the graph of `y=2cosx`, the x-axis, and the y-axis, how do you find the volume of the solid obtained by revolving region R about the x-axis?

Answer 1

Please see below.

Explanation:

Each disk has volume `pir^2 * "thickness" = pi(2cosx)^2dx`

`x` goes from `0` to `pi/2`, so

`V = int_0^(pi/2) 4pi cos^2x dx`

Using `cos^2x = 1/2(1+cos(2x))` and integrating by substitution, we get:

`= {: 2pi(x+sinxcosx)]_0^(pi/2) = pi^2`