# Volume and symmetry

1. Sep 1, 2013

### Jbreezy

1. The problem statement, all variables and given/known data

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
2. Relevant equations

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

3. The attempt at a solution

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

2. Sep 1, 2013

### vela

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

3. Sep 1, 2013

### Jbreezy

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

4. Sep 1, 2013

### vela

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

5. Sep 1, 2013

### Jbreezy

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

6. Sep 2, 2013

### vela

Staff Emeritus
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.

7. Sep 2, 2013

### Jbreezy

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.

8. Sep 2, 2013

### Jbreezy

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.

9. Sep 2, 2013

### vela

Staff Emeritus
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.

10. Sep 2, 2013

### Jbreezy

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.

11. Sep 2, 2013

### vela

Staff Emeritus
Depending on how the problem was worded, you could be right, but I think you're not. Could you please post the problem here exactly as it was written?