Integration


  • From: Jen Wilcox
  • Date: 9 May 1999
  • Subject: Integrating squared trig functions

Find the volume generated when the part of the curve y = 3sin2x between the limits x=0 and x=pi/6 is rotated around the x-axis.

Please help me with this problem - every time I try it I get a different answer!


Maths Help suggests:

This is a question about Volumes of Revolution. You need to use the formula
V=pi*INT(y^2)dx
Using the function and the values in your question, you need to find
V=pi*INT((3sin2x)^2)dx

First of all, this simplifies to
V=9pi*INT((sinx)^2)dx

Where most people get stuck is that they do not know how to integrate a squared trig function. At A-level, there is a standard procedure for this:

To integrate sin2x or cos2x use the double angle identities:
cos2A = 2cos2A - 1 from which cos2A = (1 + cos2A)/2
cos2A = 1 - 2sin2A form which sin2A = (1 - cos2A)/2

In your case the basic angle (A) is 2x so the double angle (2A) will be 4x.

You equation becomes:
4.5pi*INT(1-cos4x)dx
and the solution proceeds as follows (remember to use the standard surd form for sin(2pi/3), or make sure your calculator is in radian mode)
V=1.382 (3dp)

The answer given above is in "exact form", which is expected at A-level.
The decimal equivalent is V = 1.382 cubic units (3 d.p.)


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