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Should my volume be negative in this case

  1. Oct 9, 2007 #1
    I have not run into a volume problem that returned a negative value as of yet...until now.

    I have the following problems:
    Find the volume when the region, in the FIRST quadrant,bounded by y=x^2; y=2-x; and x=0 is:
    revolved around the x-axis and then revolved around the line y=2.


    (a) For the x-axis I got:
    [tex]v=\pi\int_0^1[(2-x)^2-(x^2)^2]dx[/tex]
    [tex]=\pi\int_0^1[4-4x+x^2-x^4]dx[/tex]
    [tex]=\pi[4x-2x^2+\frac{x^3}{3}-\frac{x^5}{5}]^1_0=\frac{32\pi}{15}[/tex]

    (b) and for around y=2 I got
    [tex]v=\pi\int_0^1[(x^2)^2-(2-x)^2]dx[/tex]
    [tex]=\pi[\frac{x^5}{5}-4x+2x^2-\frac{x^3}{3}]_0^1=-\frac{22\pi}{15}[/tex]


    But I have my doubts about the latter.

    Any help is appreciated,
    Casey
     
    Last edited: Oct 9, 2007
  2. jcsd
  3. Oct 9, 2007 #2
    How about this question: for part (b), am I using the right integral? To revolve around y=2, that is?
     
  4. Oct 9, 2007 #3
    I give up. Usually Dick is yelling at me by now:frown:
     
    Last edited: Oct 10, 2007
  5. Oct 10, 2007 #4
    Are my bounds reversed on part b? Since the line being rotated about is located "ABOVE" the region being rotated? I know it needs to be positive.
     
    Last edited: Oct 10, 2007
  6. Oct 10, 2007 #5
    This one must be too tough or something.
     
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