Expert: Steve Holleran - 11/20/2007

Question
Hi! I am trying to find the volume generated by rotating the following graphs aroudn the x-axis: y = sin(2x), x=0, x=pi, and y=0.  We are supposed to use disks.  I have set up the integral: pi times the integral from 0 to pi of (sin 2x)^2.  I just don't know how to solve this integral.  I have tried setting u equal to 2x and to sin 2x but I can't seem to get anywhere.  My textbook suggests that I use a half-angle formula, but I don't see how that would help.  I have heard that it can be solved by integration by parts, but we have not learned that in class yet.  Can you give me any advice?
Thanks!

Answer
Hi Pete,

Okay, there's a lot to go through here, so let's take it a piece at a time.

On the curve itself, y = sin 2x.  This curve has a period of pi.  That means it completes a full cycle at x = pi.  So, since the arches of the sine wave are congruent, we can take the volume generated from x = 0 to x = pi/2, and double it to get the volume from 0 to pi.

Okay, then by disks, we have

V = pi * INT[ y^2 * dx] = pi * INT[ (sin 2x)^2 * dx]

Then, let's go to the double angle formula for cosine:

cos(2 * theta) = 1 - 2 * sin^2 (theta)

If we let theta= 2x, then we have:

cos(2 * 2x) = cos(4x) = 1 - 2 * sin^2 (2x)

so cos(4x) - 1 = -2 * sin^2 (2x)

or 1 - cos(4x) = 2 * sin^2 (2x)

or 1/2[1-cos(4x)] = sin^2 (2x).

Then the integration is :

V = 2 * pi * INT (x = 0 to x = pi/2) [1/2(1 - cos (4x) * dx]

= pi * INT(x=0 to x=pi/2) [1 - cos(4x) * dx]

Splitting these up :

V = pi*([ INT(0 to pi/2) [dx] - INT(0 to pi/2) [cos 4x * dx])

= pi * ([x](0 to pi/2)  - 1/4 * INT(0 to pi/2) [cos 4x * 4dx])

= pi * ([pi/2 - 0] - 1/4 * sin (4x) (0 to pi/2)])

= pi * ( pi/2 - 1/4 * (sin 2pi - sin 0))

= pi * (pi/2 - 1/4 * ( 0 - 0 ) )

= pi^2 / 2.

I'm pretty sure this is what you want.  Using the double angle formula for cosine does the substitution you need, not the half-angle formula.

Good luck!

Steve