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Integration along y axis

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Compute the area as an integral along the y-axis:
    f(x) = x^2 - 6, g(x) = 6 - x^3


    2. Relevant equations
    N/A


    3. The attempt at a solution

    I solve for x in terms of y for both equations and end up with:
    f(y) = +/-√(y+6), g(y) = (6-y)^(1/3)

    I then look for interception points of the functions f(y) = g(y)
    and I find y = -2.

    My question is, if there is only one interception point how can I compute the area
    between these two functions? I tried plugging into wolfram, and even it says
    "cannot compute integral".

    Am I reading the problem wrong? Or am I doing something wrong?

    EDIT: When the question states "along the y-axis" does it perhaps mean the line x = 0 as a lower bound?
    And then to just integrate from x = 0 to the point of interception (2)? That's the only way I can think of doing this.
     
    Last edited: Sep 10, 2013
  2. jcsd
  3. Sep 11, 2013 #2
    You are right when you say it means to take x=0 as a lower bound, at least that would make the most sense according to the problem statement.
     
  4. Sep 11, 2013 #3
    Does the problem say something to the effect of the area bounded by the functions, f(x), g(x), and the y axis?
     
  5. Sep 11, 2013 #4

    Zondrina

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    Homework Helper

    I find this question peculiar as the graphs ##f(y)## and ##g(y)## will intersect at ##(-2,2)##. This creates two different regions ( areas to compute ) in the second quadrant bounded by the line ##x=0##.

    Is there more to this problem at all?
     
  6. Sep 11, 2013 #5
    Yeah, it would be helpful to see EXACTLY what the problem says.
     
  7. Sep 11, 2013 #6
    Verbatim:

    "In Exercises 27-44, sketch the region enclosed by the curves and compute its area as an integral along the x- or y-axis."

    ...

    28). y = x^2 - 6, y = 6 - x^3, y-axis
     
  8. Sep 11, 2013 #7
    Ok, yes that says the region bounded by those two curves and the y axis, its not saying that you have to integrate with respect to y, however it would make the problem easier.
     
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