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Finding the volume using cylindrical shells

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data
    x=1+(y-2)^2, x=2. Rotating about the x-axis


    2. Relevant equations
    Volume=(2∏y)(1+(y-2)2(Δy)

    Limits of integration would be from 1 to 3
    2∏∫(y)(1+(y-2)2dy
    2∏∫y3-4y2+5y dy


    3. The attempt at a solution
    2∏[y4/4-4y3/3+5y2/2]

    Plug in the limits and I get 32∏/3. The answer is 16∏/3. It works if for the circumference you use 2∏ instead of 2∏y.
     
  2. jcsd
  3. Oct 27, 2013 #2

    eumyang

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    Here's where you've gone wrong, I think. The representative rectangles need to be inside the parabola, not outside, as you have set the integral up.
     
  4. Oct 27, 2013 #3
    How would you fix this?
     
  5. Oct 27, 2013 #4

    eumyang

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    (To help visualize what you did earlier, I attached two pics. The red region is what is being rotated around the x-axis. The "Wrong.bmp" file shows what you did, and the "Right.bmp" file shows what the problem is asking.)

    The way you had set it up, your heights of the representative rectangle (or the distance from the y-axis to the parabola) is x (which equals 1 + (y - 2)2). Given that the distance from the y-axis to the line x=2 is 2, can you tell me the height of a rectangle from the parabola to the x=2 line?
     

    Attached Files:

  6. Oct 27, 2013 #5
    So it would be 3+(y-2)2?

    edit: I tried this and it didn't work
     
  7. Oct 27, 2013 #6

    eumyang

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    No, you don't want to add 2 and 1 + (y - 2)2.
     
  8. Oct 27, 2013 #7
    I subtracted 2 from it. So my integral is 2∏∫(y)(-1+(y-1)2)dy. I got a negative number when I did this.
     
  9. Oct 27, 2013 #8

    eumyang

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    Almost. Switch the order of subtraction.
     
  10. Oct 27, 2013 #9
    Ok thank you so much for your help. :)
     
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