# Volume and symmetry

## Homework Statement

Your supposed to set up the integral and do it from the calculator.
y = , y = (cosx)^2, -pi/2 <= x <= pi/2
a. x axis
b. y axis

## Homework Equations

I = ∏∫(cosx)^4 dx = 3.70110 and for part b
I = ∏∫((1)-(cosx)^2)^2 dx = 3.70110

## The Attempt at a Solution

Did I set this up right? It would be the same volume in both cases right?

vela
Staff Emeritus
Homework Helper
It's not clear to me what exactly you've been asked to do. Could you state explicitly what you're trying to do?

Yes, given y =0 , y = (cosx)^2, -pi/2 <= x <= pi/2
set up the integral to revolve it around a.) The x axis and b.) the y axis
It is just a volume of revolution problem the book just says do it on the calculator. That's it I just want to know if my integral are right.Thanks

vela
Staff Emeritus
Homework Helper
Your answer for part (a) looks fine. I don't understand what you're doing for part (b).

Part b wants you to revolve it around the line y = 1.
Does that help?
So I did 1 - cosx^2

vela
Staff Emeritus
Homework Helper
Yes, it helps when you tell us what the actual problem is. In the original post, you said part (b) was about revolving the area about the y-axis, and then you repeated that in post 3. And somehow we were supposed to read your mind and realize the problem was actually about revolving around the horizontal line y=1?

In any case, your integral for part (b) isn't correct. The volume you calculated is if you took the area bounded below by ##\cos^2 x## and above by y=1 and rotated it about y=1. You want the area bounded above by ##\cos^2 x## and below by y=0 and rotate that area about y=1.

I'm sorry. I don't understand your above and below. I think just the second part. I want the area bounded above by ## cos(x)^2## and below y =0 ? I don't understand this. Do you mean that I need the area below y = 1 and above ## cos(x)^2##? So could I just multiply my original answer by two? Sorry for the confusion from the earlier post.

I don't understand why the integral would not be the same. Can anyone clarify? I just looked at this thing I can't figure out what you are saying about the areas.

vela
Staff Emeritus
Homework Helper
You were given an area bounded by y=0 and y=cos2 x, right? You're supposed to revolve this area around the line y=1. You're looking at a completely different area if you look at the region between y=1 and y=cos2 x.

You were given an area bounded by y=0 and y=cos2 x, right? You're supposed to revolve this area around the line y=1. You're looking at a completely different area if you look at the region between y=1 and y=cos2 x.

Yes, you are correct. The first part (a) you rotate about (cosx)^2 around the x axis. Next you rotate about y = 1. Look at the graph. I wish I had a way to draw it. When you do around y = 1 you have (1-cos(x)^2).
So you have ∏∫(1-(cos(x)^2)^2 dx = ∏∫(sin(x))^4 dx
When you do this between (-∏/2) and (∏/2) you get 3.7011
Which is equivalent to the answer you get when you integrate cos(x)^2 around the x axis. So that integral is ∏∫(cos(x))^4 dx between (-∏/2) and (∏/2) you get 3.7011.
I don't see how I'm wrong here.

vela
Staff Emeritus