Answer:
[tex]V = \frac{\pi^2}{8}[/tex]
[tex]V = 1.23245[/tex]
Step-by-step explanation:
Given
[tex]y = \cos 2x[/tex]
[tex]y = 0; x = 0; x = \frac{\pi}{4}[/tex]
Required
Determine the volume of the solid generated
Using the disk method approach, we have:
[tex]V = \pi \int\limits^a_b {R(x)^2} \, dx[/tex]
Where
[tex]y = R(x) = \cos 2x[/tex]
[tex]a = \frac{\pi}{4}; b =0[/tex]
So:
[tex]V = \pi \int\limits^a_b {R(x)^2} \, dx[/tex]
Where
[tex]y = R(x) = \cos 2x[/tex]
[tex]a = \frac{\pi}{4}; b =0[/tex]
So:
[tex]V = \pi \int\limits^a_b {R(x)^2} \, dx[/tex]
[tex]V = \pi \int\limits^{\frac{\pi}{4}}_0 {(\cos 2x)^2} \, dx[/tex]
[tex]V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx[/tex]
Apply the following half angle trigonometry identity;
[tex]\cos^2(x) = \frac{1}{2}[1 + \cos(2x)][/tex]
So, we have:
[tex]\cos^2(2x) = \frac{1}{2}[1 + \cos(2*2x)][/tex]
[tex]\cos^2(2x) = \frac{1}{2}[1 + \cos(4x)][/tex]
Open bracket
[tex]\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\cos(4x)[/tex]
So, we have:
[tex]V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx[/tex]
[tex]V = \pi \int\limits^{\frac{\pi}{4}}_0 {[\frac{1}{2} + \frac{1}{2}\cos(4x)]} \, dx[/tex]
Integrate
[tex]V = \pi [\frac{x}{2} + \frac{1}{8}\sin(4x)]\limits^{\frac{\pi}{4}}_0[/tex]
Expand
[tex]V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [\frac{0}{2} + \frac{1}{8}\sin(4*0)])[/tex]
[tex]V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [0 + 0])[/tex]
[tex]V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})])[/tex]
[tex]V = \pi ([{\frac{\pi}{8} + \frac{1}{8}\sin(\pi)])[/tex]
[tex]\sin \pi = 0[/tex]
So:
[tex]V = \pi ([{\frac{\pi}{8} + \frac{1}{8}*0])[/tex]
[tex]V = \pi *[{\frac{\pi}{8}][/tex]
[tex]V = \frac{\pi^2}{8}[/tex]
or
[tex]V = \frac{3.14^2}{8}[/tex]
[tex]V = 1.23245[/tex]
