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Simple area between curves question

  1. Jan 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region enclosed by the following curves:
    f(y)=1-y^2
    g(y)=y^2-1

    2. Relevant equations

    3. The attempt at a solution
    I'm confused by the graph because the region enclosed has positive and negative parts, and I can't determine whether f(y)>g(y), g(y)>f(y), or what. I'm not sure what to integrate here.

    Thanks, I appreciate the help.
     
  2. jcsd
  3. Jan 1, 2007 #2

    Hootenanny

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    You could use the modulus function;

    [tex]A = \int^{a}_{b} \left\|f(y)\right\| + \left\|g(y)\right\|\;dy[/tex]

    Also, note that f(y) = -g(y)
     
  4. Jan 1, 2007 #3

    Gib Z

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    Homework Helper

    The Solution to the integral [tex]\int f(y) g(y) dy[/tex] is zero, as f(y)=-g(y).

    You could find the definite integral of each, absolute valued. Or a simpler method would be to use symmetry arguments to realize the solution is [tex]2\int^{1}_{-1} 1-y^2 dy[/tex], which you should be fine with.
     
    Last edited: Jan 1, 2007
  5. Jan 1, 2007 #4
    Are you sure that integral is zero? You might want to recheck your math unless you meant that product to be a sum.
     
  6. Jan 1, 2007 #5

    matt grime

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    The curves meet at +1 and -1. You need the area between the curves as y goes from +1 to -1. I'm not sure why you can't tell that f(y) =>g(y) in that region: f(0)=1 and g(0)=-1, so it is clear which is 'on top'. The area is just the integral of f(y)-g(y); I don't mind telling you this (we aren't supposed to just hand out answers) since it is just what you were told in class/in the book. It doesn't matter whether f or g are positive or negative individually: you only care about one relative to the other, the actual signs of f and g don't matter.

    For any functions f, and g, the area bound between them between a and b is always

    [tex] \int_a^b \max(f(t),g(t)) - \min(f(t),g(t))dt[/tex]

    a=-1, b=1 here, and f(t)=>g(t) for all t in [-1,1].
     
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