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Find the volume of th eresulting solid by and method

  1. Feb 14, 2007 #1

    tony873004

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    The region bounded by the curves is rotated about the specified axis. Find the volume of th eresulting solid by and method.

    [tex]x^2 + \left( {y - 1} \right)^2 = 1[/tex]

    First question want it rotated about the x-axis, the second question wants it rotated about the y-axis.

    So I re-write it in terms of x. Seems easier than in terms of y since there's only 1 x term.
    [tex]x = \sqrt { - y^2 + 2y}
    [/tex]


    The problem is when I graph it,
    [​IMG]

    So nothing is bounded. There is no solid to rotate. Should I have solved in terms of y instead? How would I do that?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Feb 14, 2007
  2. jcsd
  3. Feb 15, 2007 #2

    cepheid

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    Don't you need to take both the positive and negative square roots?
     
  4. Feb 15, 2007 #3

    tony873004

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    Thanks for your reply.

    It still seems like there is no bounded region. All the other problems in this section give a second curve, usually a line like y=2, just to bound the curve. What am I doing wrong? Did I properly solve for x?
    [​IMG]
     
  5. Feb 15, 2007 #4

    HallsofIvy

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    You have lost a "sign". x2+ (y-1)2= 1 is obviously a circle of radius 1 with center at (0, 1), not the hyperbola you show.

    [tex]x= \pm \sqrt{1-(y-1)^2}= \pm \sqrt{2y- y^2}[/tex]
    and
    [tex]y= 1\pm \sqrt{1- x^2}[/tex]
     
  6. Feb 15, 2007 #5

    tony873004

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    lol... and I should have recognized that the answer in the back of the book, 4pi/3 is the volume of a unit sphere. Without the r^3 in the formula I didn't recognize it.

    Thanks, Halls
     
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