Expert: Paul Klarreich - 3/20/2007

Question
QUESTION: I've been trying to work problem to try and find the area for this problem for like 3 days and I can't seem to solve the problem.  Here's what I got so far
find the area of surface of revolution for the problem
r = 4cos(2theta)   revolving about the polar axis
from 0 to pi/4

         pi/4
S = 2pi Integral 0      4cos2theta*(sintheta)*sq((4cos2theta)^2 + (-8sin2theta)^2)) dtheta.
Then I worked it down until I get to this point and I can't seem to simplify it any further.
         pi/4
S = 32pi Integral 0      cos2theta*(sintheta)*sq((cos^2(2theta)) + (4sin^2(2theta)))dt

I can't find a rule to simplify the part under the sqroot sign for the double angles when they are with cos^2 and sign^2.  I'm stuck.

Thanks
Jeff
ANSWER: Questioner:   Jeff
Category:  Calculus
Private:  No
 
Subject:  area of surface of revolution
Question:  I've been trying to work problem to try and find the area for this problem for like 3 days and I can't seem to solve the problem.  Here's what I got so far
find the area of surface of revolution for the problem
r = 4cos(2t)   revolving about the polar axis
from 0 to pi/4

         pi/4
S = 2pi Integral 0      4cos2t (sin t) sqrt((4cos2 t)^2 + (-8sin 2t)^2)) dt.

Then I worked it down until I get to this point and I can't seem to simplify it any further.
         pi/4
S = 32pi Integral 0      cos2theta*(sintheta)*sq((cos^2(2theta)) + (4sin^2(2theta)))dt

I can't find a rule to simplify the part under the sqroot sign for the double angles when they are with cos^2 and sign^2.  I'm stuck.

Thanks
Jeff
.............................................
Hi, Jeff,

I don't have a complete solution for you right now, but I'll keep looking at it.

I think you have something extraneous in the integral, but it might not matter.

[Notation:  Theta is 't', and my usual notation: C = cos t, S = sin t


An arc length element is:  ds = sqrt(r^2 + (dr/dt)^2 )

And you rotate that around the polar axis, which is really the x-axis.  So the area element is:  circumference * arc length element =

2 pi x ds = 2 pi cos t sqrt(r^2 + (dr/dt)^2 )

and the total area will be:

{pi/4
|     2 pi cos t sqrt(r^2 + (dr/dt)^2 ) dt
}0

And:  r = 4 cos (2t),  and  dr/dt = - 8 sin(2t)

The radical is:

sqrt( 16 cos^2(2t) + 64 sin^2(2t))

Let's fiddle with that a bit:

4 sqrt( cos^2(2t) + 4 sin^2(2t))

4 sqrt( 1 + 3 sin^2(2t))

At this point I am not sure how to proceed.  I have a couple of ideas, but they'll take some time.


---------- FOLLOW-UP ----------

QUESTION: I appreciate your help on this one, my book shows the Area formula as
         B
S = 2pi Integral f(t)*sin(t)sqroot([f(t)]^2 + [f'(t)]^2)d(t)
         A
for revolving around the polar axis , my previous email was probably a little sloppy.  I was curious how you reduced the value of  cos^2(2t) + 4 sin^2(2t))
down to ( 1 + 3 sin^2(2t)) .  Thanks again for your help

Answer
Hi, Jeff,

Unfortunately, I didn't make progress beyond that point, BUT:

QUESTION: I appreciate your help on this one, my book shows the Area formula as
         B
S = 2pi Integral f(t)*sin(t)sqroot([f(t)]^2 + [f'(t)]^2)d(t)
         A
>> AHA!  You are indeed correct.  I am afraid I blew it there and you do NOT have something extraneous in the integral.  I had simply forgotten that  x = r cos t, and, of course,  r = f(t).  I will go back and work on it some more.
...........................
I was curious how you reduced the value of  cos^2(2t) + 4 sin^2(2t))
down to ( 1 + 3 sin^2(2t)) .  Thanks again for your help

>> About this part, it's nothing exciting:

cos^2(2t)     + 4 sin^2(2t) =
1 - sin^2(2t) + 4 sin^2(2t) =
1          + 3 sin^2(2t) =

I'll let you know when I have more.