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Finding Area of Region

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    x + y^2 = 42, x + y = 0

    Find the area of the region enclosed.

    2. Relevant equations

    Integral of the top to the bottom.

    3. The attempt at a solution

    At first I tried to use S sqrt(42-x)+x dx and i get -2/3 (42 - x) ^ (3/2) - 1/2 x^2

    However when I set them equal to one another 42-x=x^2 and get the root values -6 and 7. Plugging it in doesn't get me the right answer...what am I doing wrong?
  2. jcsd
  3. Jan 29, 2009 #2


    Staff: Mentor

    Your first antiderivative looks right, but the second (and easier) one is off by a sign. It should be +1/2 x^2.

    You should be working with limits of integration (a definite integral) if you want to come out with a number. To get the limits of integration you absolutely need a graph of the region. Have you done this?

    You have two choices on how you can set up the integral: using vertical area elements or using vertical area elements. If you use vertical area elements, the top of each area element will always be y = +sqrt(42 -x), but the bottom of the elements are different depending on whether -7 <= x <= 6 or 6 <= x <= 42. In the first interval I listed, y = -x. In the second interval, y = -sqrt(42 -x). This means you need two definite integrals, set up like so:
    [tex]\int_{-7}^6 f(x)dx + \int_{6}^{42} g(x)dx [/tex]

    An easier way is to use horizontal area elements. The area of a typical area element is [itex][-y^2 + 42 - (-y)]\Delta y[/itex], and you need only one integral, and it's much easier to integrate.
  4. Jan 29, 2009 #3
    That makes sense, I can't get a picture of what it looks like though, however, what intervals would i use for the y then?

    I set the x's equal to one another....
    then I make it like -y^2+42+y and the values would be 6 and -7, but that isn't the case...I Plug them in and I end up with 119.6 and that's not the right answer.
    Last edited: Jan 29, 2009
  5. Jan 29, 2009 #4


    Staff: Mentor

    You need a graph of the region to be able to get the limits of integration.
    x + y^2 = 42 ==> y^2 = -x + 42 ==> y = +/-sqrt(-(x - 42))
    You can graph this relation, can't you?

    Concerning the work you show, you have
    x = -y^2 + 42
    x = -y
    ==> -y^2 -y + 42 = 0 ==> y^2 + y - 42 = 0
    Solving this quadratic gives you y = -7 or y = 6.
    These are the y values at the points of intersections of the two graphs. You can use either equation to find the corresponding x values.

    In your first post you attempted to find an antiderivative of something, and then you apparently substituted y = -7 and y = 6 into the antiderivative. That makes no sense, which is why you didn't get the right answer.
  6. Jan 29, 2009 #5
    So I should find the x values to get the intervals used in the FTC for -y^2+42+y. i recall its integral right minus leftt and isn't that how I should do it. I'm not sure what I am doing wrong still...
  7. Jan 29, 2009 #6


    Staff: Mentor

    No, if you go with the easier tactic I described in post #2, you need the y-values at the two intersection points.

    For my typical area element, the length (horizontal) is from the x value on the line to the x value on the parabola. The width (vertical) is [itex]\Delta y[/itex]. What is the interval along which the [itex]\Delta y[/itex]'s run?

    Have you graphed the area yet?
  8. Jan 29, 2009 #7
    I know it willl havea parabola at the top and a line underneath.
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