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Homework Help: Find the Region enclosed

  1. Oct 4, 2004 #1
    Find the area of the region enclosed between [tex]y=3sin(x)[/tex] and [tex]y=4cos(x)[/tex] from [tex]x=0[/tex] to [tex]x=0.9\pi[/tex] .


    does that mean the integral would like this: [tex] \int 4cos(x) - 3sin(x) [/tex] (a=0, b = 0.9*pi) is that correct?
     
  2. jcsd
  3. Oct 4, 2004 #2

    Galileo

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    The cosine is larger than the sine at x=0 and the sine is larger than the cosine at [itex]0.9\pi[/itex] so they cross each other somewhere in that interval.
    Since area is always positive you should split up your integral, and find at which point they intersect.
    A picture may help.
     
  4. Oct 4, 2004 #3

    ah, 0.9pi is equal to 162 degrees right? i graphed it using my calculator. dont know why, but i thought of 0.9pi as 2.872433.

    so would my equation look like this...

    [tex] \int 4cos(x) -3sin(x) (a=0,b=.295\pi) [/tex] + [tex] \int 3sin(x) - 4cos(x) (a=.295\pi,b=9\pi)[/tex] is that correct?
     
  5. Oct 4, 2004 #4

    Galileo

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    The notation is a bit weird, but I think you got the right idea.
    [tex]\int_0^{0.295\pi}(4\cos x-3\sin x)dx+\int_{0.295\pi}^{0.9\pi}(3\sin x-4\cos x)dx[/tex]
     
  6. Oct 4, 2004 #5
    lol yeah, i didnt know how to insert a and b into the integral sign. thanks for the help
     
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