1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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


    User Avatar
    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

    "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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted