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Homework Help: Rotating solid's volume

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    the region bounded by parabola y=x2, the line x=1, and x axis is rotated about y-axis. Find volume of the resulting solid.

    2. Relevant equations



    3. The attempt at a solution
    Please check my answer

    Area: Pi*x4
    V=integral 1->0 Pi*x4 dx=.....=1/5Pi
     
  2. jcsd
  3. Aug 4, 2010 #2
    [STRIKE]yep, that's right.[/STRIKE]
     
    Last edited: Aug 4, 2010
  4. Aug 4, 2010 #3

    Dick

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    No, I don't think it is right. You are thinking about the wrong region. Read the problem again.
     
  5. Aug 4, 2010 #4
    I think he is talking about the region beneath the parabola, between x=0 and x=1 bounding x-axis.
     
  6. Aug 4, 2010 #5
    [STRIKE]height of function is x^2....distance from axis of revolution is x...so isn't it
    [/STRIKE]


    EDIT: nvm...oops I just realized it's not pi*r^2....it's 2*pi*r.

    OP, listen to Dick :).
     
  7. Aug 4, 2010 #6
    Should I use the formula 2 Pi R, which is
    2 Pi x2

    v=integral 1-->0 2 Pi x2 dx=2 Pi x3/3=2/3 Pi.

    Is this the answer, Dick?
     
  8. Aug 4, 2010 #7
    r = x = distance from y axis (axis of rotation). Also, you need to multiply that by the height of the shell, which is y(x) = x^2.
     
  9. Aug 4, 2010 #8
    I am sorry, I don't understand what you are saying...
     
  10. Aug 4, 2010 #9
  11. Aug 4, 2010 #10
    Is this the correct beginning?
     
  12. Aug 5, 2010 #11
    No, r is x, not x^2. Look/learn the explanation on wikipedia.
     
  13. Aug 5, 2010 #12
    Please check if I got the problem right.

    In this problem, y=x2
    x=sqrt(y)
    A=Pi r2
    r2=x2=y
    V=integral 1->0 Pi y dy= pi y2/2 =Pi 1/2=

    1/2 Pi

    Please confirm if I got it right or no?
     
  14. Aug 5, 2010 #13

    Dick

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    The answer is right but the region is wrong. But you are getting there. What you have is the volume of the region bounded by y=1, x=0 and y=x^2, rotated around the y-axis. Are you drawing a sketch of these regions? It would really help. Compare that with the region you were supposed to work with. You could get the answer by subtracting what you got from a cylinder of radius 1 and height 1. If you've drawn the regions, you'll know why that's true.
     
  15. Aug 5, 2010 #14
    That's not right, the volume if the region is bounded by the line x=1 and x axis.
     
  16. Aug 5, 2010 #15
    No, you misunderstood Dick. He said "What you have is the volume of the region bounded by y=1, x=0 and y=x^2, rotated around the y-axis."

    He's knows this is the incorrect region. But this is what you have when you showed your solution in the previous post. Your answer, [tex] \frac{\pi}{2} [/tex], is the volume obtained from this incorrect region. Fix your region and you will fix your answer.
     
  17. Aug 5, 2010 #16
    The mistake you are making is you are not realizing that the distance [tex] r [/tex] from the axis of revolution is INDEPENDENT of the function [tex] y(x). [/tex] You need to be able to visualize how this volume is found.

    1. Break your region up into infinitely many vertical lines. (from x = 0 to x = 1)
    2. Rotate each line about your axis of revolution (the y-axis). Each vertical line creates a hollow shell when rotated.
    3. Find the formula for the surface area of each one of these shells. What's the equation for this?
    4. You get your volume by adding up all the infinitely many surface areas, i.e, take the integral of your formula from step 3.
     
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