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Question regading integration (Volume generated)

  1. Feb 2, 2005 #1
    I have my question,solution,doubt in the attachment that followed.I hope that anyone will help me figure out this problem.Thanks for anybody that spend some time on this question.

    Attached Files:

  2. jcsd
  3. Feb 2, 2005 #2
    Help, I really need somebody to look out this question for me whether I have done wrong or not.I have my solution to the question in the attachment.
  4. Feb 2, 2005 #3
    I might be slightly paranoid, but why would I open a Word document from a complete stranger? Besides, there's a perfectly good way to write (mostly) anything you need here on the forum, using LaTeX.
  5. Feb 2, 2005 #4
    I don't know how to write equation in LateX form , so I post up the question with attachment.Trust me ,the ataachment don't have virus and won't affected your computer.
  6. Feb 2, 2005 #5
    You crashed Word!


    (PS. Major accomplishment, but can't help ya' since I can't see it, sorry.)
  7. Feb 2, 2005 #6


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    can you ask your question in plain old words?
  8. Feb 3, 2005 #7


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    The question asks to evaluate the volume generated by revolving the region S in the first quadrant, bounded by the coordinate axes , the line x=3 and the curve [itex]y=\sqrt{1+x2}[/itex] around the y-axis.

    Check out this thread for LateX:
    You can also click the latex equations from other posters to see the exact code used. It's fast and easy to learn this way. I advise making access to the problems easier if you want people to help you.

    Anyway, you seem to have a made a slight mistake in calculating the 'total volume'
    The volume generated by rotating the rectangle bounded by x=0, y=0, x=3 and y=[itex]\sqrt{10}[/itex] is [itex]3^2\pi \sqrt{10}[/itex], instead of [itex]3 \pi \sqrt{10}[/itex].

    The volume use have to subtract is:
    [tex]\pi \int_1^{\sqrt{10}}(y^2-1)dy=\pi\left[\frac{10\sqrt{10}}{3}-\sqrt{10}+2/3\right][/tex]
    You got that part right.

    If you subtract the above from [itex]\pi 9\sqrt{10}[/itex] you get the right answer.
    Last edited: Feb 3, 2005
  9. Feb 3, 2005 #8
    [itex]\pi 9\sqrt{10}[/itex]
  10. Feb 3, 2005 #9
    Soory,I was just trying with the LateX equation and thanks for Galileo for seeking out the question for me and correct my errors.
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