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Integration of circle?

  1. Jul 28, 2012 #1
    1. The problem statement, all variables and given/known data
    -8[itex]\int^{3}_{0}\sqrt{9-x^2}dx[/itex]


    2. Relevant equations



    3. The attempt at a solution

    am i right in thinking this the area of a circle in the first quadrant so my answer is-8([itex]\frac{9\pi}{4})[/itex] = -18[itex]\pi[/itex]

    Thanks for reading?
     
  2. jcsd
  3. Jul 28, 2012 #2

    Dick

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    Sure it is.
     
  4. Jul 28, 2012 #3
    Hi Dick.
    thanks for reply but...
    are you saying "it sure is" or are you asking "are you sure it is?"

    "sure it is" is the way my friends would sarcastically say "your wrong":confused:
     
  5. Jul 28, 2012 #4

    Dick

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    It's the not sarcastic 'it sure is'. The integral is -18*pi and you can deduce that from its being a quarter circle. You could also do it with a trig substitution and get the same thing.
     
  6. Jul 28, 2012 #5
    thanks a million Dick. :smile:
     
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