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Darn Volume of Revolution...

  1. Jul 24, 2015 #1
    1. The problem statement, all variables and given/known data
    The area under y = (x^2/9) + 1 from x = 0 to x = 3 , and the area enclosed by the y= 0 , y=2 , x=3 , and x=4, are rotated about the y-axis , and the solid generated represents a metal ash tray , the units being cm. Calculate the volume of a metal.

    2. Relevant equations


    3. The attempt at a solution
    for the volume between x=3 and x=4 i wanted to rotate it about the y-axis but I got confused... so I just did about the x-axis. Also I only sketched the part of the curve I would need.
    I can't see why I am not getting the answer, shouldn't I just find those two volumes and ADD them together to get the volume of the ash tray? This is so frustrating...

    Is there something I am missing? All that I see done in the examples in this text book is draw a rectangle perpendicular to the axis of rotation and then boom you have your formula,

    either ∫πy2 dx or ∫πx2 dy


    2w5lvkj.jpg
     
  2. jcsd
  3. Jul 24, 2015 #2

    SammyS

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    It seems that you only know the "Disk" method. An extension of this is the "Washer" method, sometimes referred to as the "Donut" method.

    A useful method for this problem as well as you recent problem, https://www.physicsforums.com/threads/volume-of-revolution-question.824553/#post-5177463, is the "Shell" method.

    I suggest you study these methods and then come back to this problem.

    Rotating about the x-axis rather than the y-axis gives a very different result.
     
  4. Jul 24, 2015 #3
    Okay, I find that so weird, because the textbook I am using only has that method. So it should mean that method can be used. I will look up those methods though..
    Also how would you even rotate an area under y=2 around the y axis. Since the integral would be something like ∫πx2 dy. There is no x to square , is there?
     
  5. Jul 24, 2015 #4

    SammyS

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    You would need to use subtraction. Find the volume of the large cylinder, (radius 4cm, symmetry axis is the y-axis). Then subtract the volume of the piece inside the parabolic portion.

    In fact, for this problem, that's not too bad a technique.
     
  6. Jul 24, 2015 #5
    OKAY so what I did now was instead of really using calculus, I rotated the area bounded by y=2 from x=4 to x=0. That's a cylinder with radius 4 and height 2
    so the volume would be 32Pi. Then from my previous work the volume bounded by the parabola x=3 and x=0 about the y-axis was 9pi/2.

    Subtracting the volume of the parabolic portion like you said from the larger cylinder I get 27(Pi/2). WHich is the answer, but I don't think i fully understand. Okay the way I was visualizing the shape made by the parabolic portion was like maybe a bowl of some sort? and then the volume that I would get from y=2 ,x=3 and x=4 would be the outside of the ash tray, so that is why I thought I would just add them...

    Subtracting the volume formed form the parabolic portion wouldn't that give the larger cylinder a huge hole? Sorry I am just trying to visualize this ash tray haha...
     
  7. Jul 25, 2015 #6
    Like previous posters said, the washer or shell method would make this a trivial problem. Do you want the proof for the shell method?
     
  8. Jul 25, 2015 #7

    HallsofIvy

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    Personally, I would do this as two separate parts, first the section under the parabola from x= 0 to x= 3.

    The remaining portion can also be done in two parts. The entire rectangle, from x= 0 to x= 4 rotates into a disk with radius 4 and thickness 2 so volume [itex]\pi (4)^2(2)= 32\pi[/itex]. The part that we don't want, from x= 0 to x= 3 is a disk with radius 3 and thickness 2 so has volume [itex]\pi (3)^2(2)= 18\pi[/itex]. Subtract that inner portion, that we don't want from the entire disk to get the outer part.
     
  9. Jul 25, 2015 #8
    I had got the answer from subtracting the section under the parabola from x =0 to x=3 from the cylinder formed by rotating the area under y=2. Thank you for all your responses
     
  10. Jul 25, 2015 #9

    SammyS

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    You should have subtracted the volume above the paraboloid from the cylinder volume.

    It turns out that the volume above the paraboloid is the same as the volume above y=1 and below the paraboloid.
     
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