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How do l integrate this ?

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]
    18\int_0^4 \sqrt{4- (y-2)^2}dy
    [/tex]

    2. Relevant equations

    According to the textbook l am supposed to use [tex]y=2sin\theta [/tex] for substitution

    3. The attempt at a solution

    [tex]y=2sin\theta [/tex]

    [tex]dy=2cos\theta d\theta [/tex]

    [tex]
    18\int_0^4 \sqrt{4- (2sin\theta-2)^2}2cos\theta d\theta
    [/tex]
    ( I am stuck at converting the limits of my integral using [tex]y=2sin\theta [/tex] )


    [tex]
    18\int_0^4 \sqrt{4- (4sin^2\theta-4sin\theta-4sin\theta+4)}2cos\theta d\theta
    [/tex]

    [tex]
    18\int_0^4 \sqrt{4- (4sin^2\theta-8sin\theta+4)}2cos\theta d\theta
    [/tex]


    ( How do l simplify this integral )
     
  2. jcsd
  3. Feb 26, 2009 #2

    gabbagabbahey

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    Gold Member

    Ermmmm.... how can [itex]y[/itex] ever equal 4 with this substitution?
     
  4. Feb 26, 2009 #3


    That is where l am stuck and also on how to simply this integral
     
  5. Feb 26, 2009 #4

    gabbagabbahey

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    Try a different substitution!:smile:

    ....maybe [itex]y=4\sin^2\theta[/itex] :wink:
     
  6. Feb 26, 2009 #5

    Mark44

    Staff: Mentor

    I would make two substitutions: u = y - 2, so du = dy, and then the trig substitution.

    As far as your limits of integration, you can carry them along all the way through until you finally undo both substitutions, when your antiderivative will be in terms of y.

    You can remind yourself that the limits of integration are y values by doing adding "y = " on the lower limit of integration, like so:
    [tex] 18\int_{y = 0}^4 \sqrt{4- (2sin\theta-2)^2}2cos\theta d\theta [/tex]

    That should help you remember that these are y values.

    Alternatively, you can change the limits of integration for each substitution. If you're careful, both techniques will work.
     
  7. Feb 26, 2009 #6

    Engineeringcalculus.jpg

    Is this correct.
     
  8. Feb 26, 2009 #7

    gabbagabbahey

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    Sure, but you didn't finish evaluating it by substituting in the limits....

    Incidentally, there is also a neat little trick to evaluating the integral [tex]\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta[/tex].... If you sketch a graph of both sin^2 and cos^2 over that interval, you see they both have the same area underneath them

    [tex]\implies \int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2} (\cos^2\theta+\sin^2\theta) d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2} (1)d\theta=\frac{\pi}{2}[/tex]
     
  9. Feb 26, 2009 #8

    Thanks very much.
     
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