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Homework Help: Integration - areas

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the area enclosed by the y axis, the line y = 3 and the curve x = y^2

    2. Relevant equations

    [/b]3. The attempt at a solution[/b]

    Area = [tex]\int[/tex] 3 to 0 y^2.dy
    = (1/3 y^3) 3 to 0
    = (1/3 x 27)
    = 9 sq units

    Im not really confident on the y axis questions, can you please confirm my understanding. If this same area is then rotated 360 degrees about the y -axis, can I seek guidence on how to do this. Cheers P
     
  2. jcsd
  3. Jun 5, 2010 #2

    Mentallic

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    What you did is correct. If you aren't comfortable with questions involving the y-axis, you can always change the question so it can be equivalently integrated with the x-axis. For example, x=y2 between y=3 and the y-axis has the same area as y=x2, x=3 and the x-axis :smile:

    Do you know the formulae (or even better, derive the formulae) for volumes of revolution?

    If y=f(x) is rotated about the x-axis between a and b, the volume is [tex]\pi\int_a^b\left(f(x)\right)^2dx[/tex]
     
  4. Jun 5, 2010 #3
    Thanks Mentallic

    Yes, I realise I know the volume formulae. I calculate the volume would be 48.6 units cube - do you mind checking?

    Cheers P
     
  5. Jun 6, 2010 #4

    Mentallic

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    Um, no. Show me your working and I'll point to where you went wrong.
     
  6. Jun 7, 2010 #5
  7. Jun 7, 2010 #6

    Mentallic

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    Yeah exactly, so why are you saying it's 48.6 u3? Whatever happened to the [itex]\pi[/itex]? :tongue:
     
  8. Jun 7, 2010 #7

    Sorry, oooopppsss, can I say 48.6pi u3? I got the impression that I had gone horribly wrong. Manythanks for you responses.

    Cheers P
     
  9. Jun 7, 2010 #8

    Mentallic

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    Of course, all you did was convert the fraction to a decimal, so there's no point but yeah, sure I guess :tongue:
     
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