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Find the volume of the solid of revolution.

  1. Nov 9, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Problem:
    Find the volume of the solid of revolution obtained by rotating the area bounded by the curves y = x^2 – 2 and y = 0 about the line y = -1. Consider only that part above y = -1.

    Solution:
    The solution is attached as TheSolution.jpeg.

    2. Relevant equations
    Integration.

    3. The attempt at a solution
    I get everything the solution did except for the part with the PQ = 1 - |y| = 1 - (-y) = 1 + y = y + 1.

    Could someone please explain that part for me?

    Any input would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Nov 9, 2013 #2

    LCKurtz

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    It is just the radius of revolution of the dydx element, : ##y_{upper}-y_{lower}=y -(-1)##.
     
  4. Nov 9, 2013 #3

    s3a

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    Sorry, I double-posted.
     
  5. Nov 9, 2013 #4

    s3a

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    Y_lower = the line about which we are rotating, right?

    As for the upper part of the radius, I don't see why y_upper = y.
     
  6. Nov 9, 2013 #5

    LCKurtz

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    You have a little dydx square located at the variable point (x,y) in the interior of your area that you are integrating over the area. It is the distance from that variable (x,y) point to the line y = -1 that is the radius of revolution.
     
  7. Nov 10, 2013 #6

    s3a

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    Why is ##y_upper## = y (a variable) if the parabola is bounded above by a constant function (y = 0)?

    Why isn't it ##y_upper – y_lower## = 0 - (-y) = y (such that the third dimension of the volume is 2π × y instead of 2π × (y + 1))?

    Also, why does the solution say 1 - |y| = 1 - (-y) = 1 + y = y + 1 instead of y - (-1) = y + 1? (I know the final answer is the same; I'm asking about the difference in getting to the final answer.)
     
  8. Nov 10, 2013 #7

    LCKurtz

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    You want the radius of rotation of the little dydx square. You want its distance from ##y=-1##.

    We are talking about the upper and lower ends of the radius of rotation. The upper end is at the dydx square and the lower end is ##y=-1##.

    I have no idea why he wrote it that way.
     
  9. Nov 10, 2013 #8

    vela

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    I'd guess that the way the person who wrote the solution was thinking about it is that the distance from the line y=-1 to the point (x,y) is 1, the distance between the x-axis and the line y=-1, minus |y|, the distance from (x,y) to the x-axis.
     
  10. Nov 11, 2013 #9

    s3a

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    Thanks guys, I get it now! :)
     
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