Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify your results using the integration capabilities of a graphing utility.y = cos 2xy = 0x = 0x = pi/4

Answer 1

`V = (\pi^2)/(8)`

`V = 1.23245`

Explanation:

Given

`y = \cos 2x`

`y = 0; x = 0; x = (\pi)/(4)`

Required

Determine the volume of the solid generated

Using the disk method approach, we have:

`V = \pi \int\limits^a_b {R(x)^2} \, dx`

Where

`y = R(x) = \cos 2x`

`a = (\pi)/(4); b =0`

So:

`V = \pi \int\limits^a_b {R(x)^2} \, dx`

Where

`y = R(x) = \cos 2x`

`a = (\pi)/(4); b =0`

So:

`V = \pi \int\limits^a_b {R(x)^2} \, dx`

`V = \pi \int\limits^{(\pi)/(4)}_0 {(\cos 2x)^2} \, dx`

`V = \pi \int\limits^{(\pi)/(4)}_0 {\cos^2 (2x)} \, dx`

Apply the following half angle trigonometry identity;

`\cos^2(x) = (1)/(2)[1 + \cos(2x)]`

So, we have:

`\cos^2(2x) = (1)/(2)[1 + \cos(2*2x)]`

`\cos^2(2x) = (1)/(2)[1 + \cos(4x)]`

Open bracket

`\cos^2(2x) = (1)/(2) + (1)/(2)\cos(4x)`

So, we have:

`V = \pi \int\limits^{(\pi)/(4)}_0 {\cos^2 (2x)} \, dx`

`V = \pi \int\limits^{(\pi)/(4)}_0 {[(1)/(2) + (1)/(2)\cos(4x)]} \, dx`

Integrate

`V = \pi [(x)/(2) + (1)/(8)\sin(4x)]\limits^{(\pi)/(4)}_0`

Expand

`V = \pi ([((\pi)/(4))/(2) + (1)/(8)\sin(4*(\pi)/(4))] - [(0)/(2) + (1)/(8)\sin(4*0)])`

`V = \pi ([((\pi)/(4))/(2) + (1)/(8)\sin(4*(\pi)/(4))] - [0 + 0])`

`V = \pi ([((\pi)/(4))/(2) + (1)/(8)\sin(4*(\pi)/(4))])`

`V = \pi ([{(\pi)/(8) + (1)/(8)\sin(\pi)])`

`\sin \pi = 0`

So:

`V = \pi ([{(\pi)/(8) + (1)/(8)*0])`

`V = \pi *[{(\pi)/(8)]`

`V = (\pi^2)/(8)`

or

`V = (3.14^2)/(8)`

`V = 1.23245`