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Find the area enclosed by x + y^2 = 12 and x + y = 0

  1. Dec 8, 2009 #1
    i sketched the graph
    found that i should integrate from -3 to 4

    then set up the integral as the integral from -3 to 4 of (-y) - (12-y^2)

    after evaluating it i got -34.66666667 - 22.5
    which is -57. 1666667

    but the right area is 57.16666667

    what did i do wrong what got me the wrong answer?
  2. jcsd
  3. Dec 8, 2009 #2
    Try making two calculations. One for the boundaries y=-3 to y=0 and another for the boundaries y=0 to y=4. Then add the areas.

    If you are still getting a negative value, take a look at your signs when you integrate.

    A final thing to try might be integrating with the equations in the y(x) form instead of the x(y) form like you are doing
  4. Dec 8, 2009 #3
    Sorry, I just gave you a couple of different ways to do the integration that might help you see what you did wrong.

    Looking at your setup, you have everything right except for the order of the equations in your integration. Switch the equations and your sign will change.

    The reason for this is you are integrating from left to right on the x-axis (negative to positive). If you were doing this same problem with respect to x, would you write sqrt(12-x)-(-x) or -x-sqrt(12-x)? Your answer to this should help you understand why you need to switch the order of your equations.
  5. Dec 8, 2009 #4
    i switched the equations and got the right answer but to find the area this way doesnt f(x) have to be greater than or equal to g(x) so therefore y = -x is greater than x = 12- y^2 when you graph them
  6. Dec 8, 2009 #5
    Looking at it in terms of f(x) and g(x), let [tex]f\left(x\right)=-x[/tex] and [tex]g\left(x\right)=\pm\sqrt{12-x}[/tex].

    From the point where they intersect in quadrant II, (-4, 4) all the way to (12, 0), g(x) > f(x). It is important to realize that g(x) is not a function. For any value in the domain x with the exception of x=12, there are two corresponding values in the codomain. This is why we have the +- in front of the square root. This lets us divide g(x) into two functions, one for the + and one for the -. Grab a graphing calculator and plot y=-x, y=sqrt(12-x) and y=-sqrt(12-x). Then look at the relations at the different boundaries (y=-3, y=0, or y=4). At each of these boundaries and everywhere in between, there is a value in the relation g(x)=-sqrt(12-x) where g(x)>f(x) with the exception of the intersection point at (-4, 4)

    So, you integrate with the greater function minus the lesser function to find the are in between.
  7. Dec 8, 2009 #6


    Staff: Mentor

    Your area elements are horizontal strips that extend, left to right, from the line (x = -y) to the parabola (x = -y^2 + 12), and from y = -4 to y = 3. The x-values on the parabola are greater than those on the line, so the integral below will give you the area as a positive number.

    Also, it is much simpler in this problem to integrate with respect to y (using horizontal strips) than to integrate with respect to x (using vertical strips).
    [tex]\int_{y = -4}^3 (-y^2 + 12) -(-y)~dy[/tex]

    Are you sure that the correct answer is 57.16666? I get 50.16666.
  8. Dec 8, 2009 #7
    [tex]\int_{y = -3}^4 (-y^2 + 12) -(-y)~dy[/tex]

    [tex]\left[12y-\frac{1}{3}y^{3}+\frac{1}{2}y^{2} \right]^{4}_{-3}[/tex]

    Evaluating the above gives


    Your boundaries are wrong. Look at the graph and find where x+y2=12 and x+y=0 intersect. There will be two points. Take the y values from these intersection points as your limits of integration.

    If you are going to think about integration in terms of the rectangular strips with width dy, you don't want left to right, which is for integrating with respect to dx (tiny intervals measured on the x-axis). You have to think about top to bottom because we are integrating with respect to the dy (tiny intervals measured on the y-axis).
  9. Dec 8, 2009 #8


    Staff: Mentor

    Yes. I made a sign error in the middle term of my quadratic and got y = -4 and y = 3, instead of y = 4 and y = -3.
    I think we're both saying the same thing here in different words. The long dimension in each of my horizontal strips runs from the x-value on the line to the x-value on the parabola, and the short dimension is dy.
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