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Volume of revolution

  • #1
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Homework Statement


A container with height [tex]4.5[/tex] is created by rotating the curve [tex]y = 0.5x^2[/tex] [tex]0 \leq x \leq 3 [/tex] around [tex] x = -3 [/tex] and putting a plane bottom in the box. Find the volume [tex]V[/tex] of the box.

Homework Equations




The Attempt at a Solution


I want to solve this by using the shell method. I have put up the following integral, which will be the volume of the revolution. It is however not correct, and I haven't really used the information about the height given in the text.. could anyone help me?

[tex]2\pi\int_0^3 (x+3)(0.5x^2) \, dx[/tex]
 
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Answers and Replies

  • #2
mjc123
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You are omitting the central cylinder (between x = 0 and -6).
Personally I think it would be easier to take circular slices in the y direction.
 
  • #3
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How do you go about calculating the volume of the central cylinder?
 
  • #4
Charles Link
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I think you just edited it with the ## 2 \pi ##, and it looks correct to me.
 
  • #5
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I think you just edited it with the ## 2 \pi ##, and it looks correct to me.
No, the post haven't been edited. I think it´s true what mjc123 said about omitting the central cylinder. I am however not exactly sure how to visualize it.
 
  • #6
Charles Link
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No, the post haven't been edited. I think it´s true what mjc123 said about omitting the central cylinder. I am however not exactly sure how to visualize it.
I am a little puzzled by this one also, because it doesn't look like a box. ## \\ ## Edit: I think I see what they are wanting, but I'll let you work on it for a few minutes.
 
  • #7
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I am a little puzzled by this one also, because it doesn't look like a box.
I´ve translated it from my own language, so box is probably the wrong description. Let's say container instead. (edited first post)
 
  • #8
mjc123
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Oh, and you're also integrating under the curve, i.e. outside the box. Replace 0.5x2 by (4.5 - 0.5x2).
Try drawing a diagram of this to see what the "box" looks like (it's like a bowl).
 
  • #9
mjc123
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This is what I think you're being asked to find the volume of. What is the volume of the bowl?
bowl.png
 

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  • #10
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Oh, and you're also integrating under the curve, i.e. outside the box. Replace 0.5x2 by (4.5 - 0.5x2).
Try drawing a diagram of this to see what the "box" looks like (it's like a bowl).
Is it like this? We want to find the volume which results when rotating the gray colored area around x = -3, but we also need to include the "inner" volume which I´ve colored orange. The volume of this orange part should be pi*3^2*4.5..?

42982625_431839237221407_8407967947338809344_n.jpg
 

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  • #11
mjc123
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Looks like you've got it.
 
  • #12
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Looks like you've got it.
I still get wrong. My answer is (135/2)*pi. Is it the same as you?
 
  • #13
mjc123
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No, I get 114.75*pi, by both methods.
 
  • #14
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No, I get 114.75*pi, by both methods.
Excellent, had forgotten that the graph was 0.5x^2 and not x^2... got it now!
 

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