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Even number problem in book volume

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Homework Statement


Hey, I got an even number from my book trying to double check it so if you wouldn't mind I would appreciate some comfirmaion or denial.

Find the volume of the solid obtained by rotated te region bounded by the given curves about the specificed line.
y = x, y =0, x =2, x =4 : about x = 1

Homework Equations



I set up this integral

I = ∏∫ dy - ∏(1-y^2)^2 dy

I did this integral between [0,1]

The Attempt at a Solution




I got ∏( y-(y^5/5) + (2/3)y^3 - 1)
I worked it out and got (7/15)∏

Yes no maybe so ?Thx
 

Answers and Replies

  • #2
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Is my integral set up properly?
 
  • #3
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I'm not sure about the above integral, but maybe this ?

[itex]V=2\pi\int (x-1)x\,dx[/itex]
 
  • #4
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I don't know I'm trying to see if mine is correct.
 
  • #5
LCKurtz
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Homework Statement


Hey, I got an even number from my book trying to double check it so if you wouldn't mind I would appreciate some comfirmaion or denial.

Find the volume of the solid obtained by rotated te region bounded by the given curves about the specificed line.
y = x, y =0, x =2, x =4 : about x = 1

Homework Equations



I set up this integral

I = ∏∫ dy - ∏(1-y^2)^2 dy

I did this integral between [0,1]

The Attempt at a Solution




I got ∏( y-(y^5/5) + (2/3)y^3 - 1)
I worked it out and got (7/15)∏

Yes no maybe so ?Thx
Definitely no, for several reasons.

One thing you should be learning about integrals is when to choose ##x## or ##y## as your integration variable, which is the same choice as whether to use disks or shells. In this problem the best choice would be ##x## (shells). Do you see why that would be preferred, just from the figure and before any calculations are done?

If you insist on using ##y## (disks), then your first mistake is the limits. ##y## doesn't go from ##0## to ##1##. And your integrand doesn't make any sense either. Have you drawn a picture of the relevant region and shaded the area that is being rotated? That's where you should start.
 
  • #6
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The only two methods that I see in this section are A = pi(radius)^2 and A = pi(radius outer)^2 - pi(radius inner)^2
 
  • #7
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Definitely no, for several reasons.

One thing you should be learning about integrals is when to choose ##x## or ##y## as your integration variable, which is the same choice as whether to use disks or shells. In this problem the best choice would be ##x## (shells). Do you see why that would be preferred, just from the figure and before any calculations are done?

If you insist on using ##y## (disks), then your first mistake is the limits. ##y## doesn't go from ##0## to ##1##. And your integrand doesn't make any sense either. Have you drawn a picture of the relevant region and shaded the area that is being rotated? That's where you should start.
How does y not go from 0 to 1 I'm revolving around x=1. So the intersection of the line y = x and x =1 is (1,1) so the bounds go from [0,1] for the integral.
 
  • #8
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I have come up with this integral
I = ∏∫(y^2 -2y +1) dy
from [0,1]
 
  • #9
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Wait I keep getting this damn thing confused on which is going which way. I did it the same as my last post but with I = PI (integral) (1-x)^2 dx. between 0 and 1.
I got the same answer
 
  • #10
LCKurtz
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How does y not go from 0 to 1 I'm revolving around x=1. So the intersection of the line y = x and x =1 is (1,1) so the bounds go from [0,1] for the integral.
Instead of contradicting what I just told you, do like I said, draw a picture of the region. Show me the shaded region that is being rotated. Start there. Nowhere else if you want another reply from me.
 
  • #11
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I did that dude. I have it right here on my desk. You have the line y=x then you have the line x = 1. So you have that little triangle that you are going to rotate. I'm sure you know what I'm looking at. OK now explain this method please.
 
  • #12
LCKurtz
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You haven't earned the right to address me as "dude". It wastes both of our time when you are working from an incorrect picture, which you have just described, and we don't know it. No wonder your integrals don't make sense. So go back to square one, figure out the correct region, and post an image of it. You might then even understand why I made the comments I did in post #5.
 
  • #13
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How is that not the correct picture. The problem states
18.) y = x, y = 0, x = 2, x = 4: about x =1
So I draw the line y=x in my graph and I draw the lines x = 2 and = =4

How is my graph incorrect. I have no way to put an image on here.
 
  • #14
LCKurtz
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How is that not the correct picture. The problem states
18.) y = x, y = 0, x = 2, x = 4: about x =1
So I draw the line y=x in my graph and I draw the lines x = 2 and = =4

How is my graph incorrect.
What do ##x=2## and ##x=4## have to do with this problem?
 
  • #15
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The bounds ? So it is the chunk between 2 and 4?
 
  • #16
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I'm confused though do I include the chunk between x = 2 and x = 4? So I will have a hole between x=2 and the line x = 1 that I'm revolving it around?
 
  • #17
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OK I want the chunk between 2 and 4. But what do I do about the line y = x? Because it makes that block between 2 and 4 cut weird.
 
  • #18
LCKurtz
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OK I want the chunk between 2 and 4. But what do I do about the line y = x? Because it makes that block between 2 and 4 cut weird.
So what? y=x is the top. This is where you decide whether disks or shells is easiest. Then you work the problem.
 
  • #19
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It says to rotate it about x = 1. All I have in the chapter is disk or shells. If I'm going to do it around x = 1. So I will do it as a function of y. So then that makes x = 4 the top and x = 2 the bottom. But what do I do about the line y = x that cuts it? Do I have to subtract off the little triangle that would be included if I just integrated between x = 4 and x = 2?
 
  • #20
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OK, I tried. This is what I did.

I said the radius furthest out R = 3 , the difference between the axis of rev. x =1 and the line x = 4
I did the same thing for the x =2 and this radius r = 1
I then figured out the area of the triangle that was included where y =x cuts between x =2 and x = 4
Got (1/2)base x height and got 2. Just subtracting the points
Then I did

I = ∏∫(3)^2 dy - ∏∫dy - 2
between 0 and 4.

I got ∏(3y-y) and when I plugged in valued I got 8∏-2 = 23.13
 
  • #21
LCKurtz
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That extra little triangle is exactly why this problem should be worked with shells. See if you can figure out why the integral given in post #3 is what you should do, assuming you figure out the correct limits.
 
  • #22
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##V=2\pi\int (x-1)x\,dx##

Because you can rotate it about the x axis? My book only gives disk method at this point. Is my integral right or wrong when I subtracted the triangle? Would the limits be [2,4]? Since that is the spot I'm interested in? See the book only has disk at this point.
 
  • #23
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Oh and I did the integral and got 25.13 which is 2 more then when I did it and subtracted the triangle. So...I'm confused. Would I of got the answer if I didn't subtract the triangle?
 
  • #24
LCKurtz
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Oh and I did the integral and got 25.13 which is 2 more then when I did it and subtracted the triangle. So...I'm confused. Would I of got the answer if I didn't subtract the triangle?
If you work out that integral you should get ##\frac{76\pi} 3## which is not ##25.13##. And you can't just "subtract the triangle" in your calculations. You have to subtract the volume you get from revolving the triangle.
 
  • #25
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Actually I got that just now. I did it the annoying way like my book wanted but I got that answer.
Thanks dooode:)
 

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