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Iterated Integrals - setting up limits of integration

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the region under the graph of f(x,y) = x+y and above the region y2≤x, 0≤x≤9

    3. The attempt at a solution

    From these equations, x will be integrated from 0-9, but i'm not sure about y.

    My thinking is that y will be intgrated from 0-3 because y2≤x and the smallest value of x is 0, and the square root of 0 is 0, so that is the smallest y, and the largest x value possible is 9 and the positive quare root of 9 is 3, so this is the largest value of y,
    So I would integrate:

    ∫0-9∫0-3 (x+y)dydx.

    Is this correct or am I missing something, where the limits of integration also involve using f(x,y)?
  2. jcsd
  3. Sep 30, 2008 #2


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    No, that is not correct. The region [itex]a\le x\le b[/itex], [itex]c\le y\le d[/itex], with a, b, c, d numbers is always a rectangle and the figure here is not a rectangle. But the bounds do NOT involve f(x,y)- that is a "z" value and goes inside the integral as you have it.

    Always draw a picture for problems like this. y^2= x is a parabola, of course, "on its side". The line x= 9 is a vertical line crossing the parabola at (9,3) and at (9, -3). Yes, you can integrate with x going from 0 to 9. On your picture, mark an arbitrary "x" by marking a point on the x-axis between 0 and 9. Now draw a vertical line from one boundary to the other. The y bounds, for that x, are y values of those endpoints: [itex](x, -\sqrt{x})[/itex], and [itex](x, -\sqrt{x})[/itex]. Your integral is
    [tex]\int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}} f(x)dy dx= \int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}}(x+ y) dy dx[/tex]

    Of course, like any double integral, you can reverse the order of integration. If you look at your picture you will see that y ranges, overall, from -3 to 3. Draw a horizontal line across the parabola, representing an arbitrary value of y in that range. It will have left endpoint on the parabola: x= y^2, and right endpoint on the vertical line x= 9. Those will now be the limits of integration for this order:
    [tex]\int_{y= -3}^3\int_{x= y^2}^9 f(x,y)dx dy= \int_{y= -3}^3\int_{x= y^2}^9(x+ y)dx dy[/tex]

    Try it both ways. You should get the same answer.
  4. Sep 30, 2008 #3
    In a similar situation where i have to switch the order of integration from
    ∫0-3∫y2-9 f(x,y) dxdy to dydx,

    Is this also a parabola on it's side, with intercepts through (3,9) and (0,9), meaning that I would be setting up the new limits of integration, fro the portion of the parabola above y=0?
  5. Sep 30, 2008 #4
    So, ∫0-9∫Square root of x -3 f(x,y) dydx
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