Volume and symmetry

  • Thread starter Jbreezy
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  • #1
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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?
 

Answers and Replies

  • #2
vela
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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
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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
vela
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Your answer for part (a) looks fine. I don't understand what you're doing for part (b).
 
  • #5
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Part b wants you to revolve it around the line y = 1.
Does that help?
So I did 1 - cosx^2
 
  • #6
vela
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Yes, it helps when you tell us what the actual problem is. :wink: 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
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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
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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
vela
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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
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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.
 
  • #11
vela
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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?
 

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