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

Answer:

$V = \frac{\pi^2}{8}$

$V = 1.23245$

Step-by-step explanation:

Given

$y = \cos 2x$

$y = 0; x = 0; x = \frac{\pi}{4}$

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

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

$a = \frac{\pi}{4}; b =0$

So:

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

Where

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

$a = \frac{\pi}{4}; b =0$

So:

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

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

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

Apply the following half angle trigonometry identity;

$\cos^2(x) = \frac{1}{2}[1 + \cos(2x)]$

So, we have:

$\cos^2(2x) = \frac{1}{2}[1 + \cos(2*2x)]$

$\cos^2(2x) = \frac{1}{2}[1 + \cos(4x)]$

Open bracket

$\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\cos(4x)$

So, we have:

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

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

Integrate

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

Expand

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [\frac{0}{2} + \frac{1}{8}\sin(4*0)])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [0 + 0])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})])$

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}\sin(\pi)])$

$\sin \pi = 0$

So:

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}*0])$

$V = \pi *[{\frac{\pi}{8}]$

$V = \frac{\pi^2}{8}$

or

$V = \frac{3.14^2}{8}$

$V = 1.23245$