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Limits of integration

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Given y=x^2 , bounded by the line x=1 and y=1, first quadrant. Fairly simple problem.

    2. Relevant equations

    Solving for integration by y first...say, [tex]\int[/tex][[tex]\int[/tex]dy]dx

    3. The attempt at a solution

    I have solved this problem, integrating by x first and y first. I'm having trouble with a more difficult, yet similar, problem but I feel I should get this fundamental doubt out of the way first.

    Integrating by y first, we get the limits of integration from y=x^2 to y=1 . This is rather clear, looking at the graph. The x integration, however, is where I get confused. I would've thought that the limits of x would be from x=0 to x=[tex]\sqrt{y}[/tex] because in reality, if you follow the graph, much like we did for the y limits, those are the boundaries x is confined to. The correct x limits, however, are from x=0 to x=1. Any help would be appreciated. Thanks.
    Last edited: Mar 11, 2008
  2. jcsd
  3. Mar 11, 2008 #2


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    Homework Helper

    Starting with y first, we describe the region as being bounded between y=x^2 and y=1, but only the part between x=0 and x=1.

    If we instead start with x first, we describe the region as being bounded between [tex]x=0[/tex] and [tex]x=\sqrt{y}[/tex], but only the part between y=0 and y=1.
  4. Mar 12, 2008 #3
    So you mean..starting with y, we think of y as the function part, the dominating part, and x only as the boundaries. When starting with x, x is the boss, the function. y is just the limits, from here to here.
  5. Mar 12, 2008 #4


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    On the contrary, it is not a problem at all! I might assume that you forgot to say "find the area" but even so, there is no region bounded by y= x2, x= 1, and y= 1. There is a region bounded by y= x2 and y= 1. There is another region bounded by y= x2, y= 0, and x= 1. Which do you mean?

    If the problem is to find the area of the region bounded by y= x2 and y= 1, by double integration, using "dy" first, so that the "outer integral" is with respect to x, remember that the limits on the outer integral must be constants. Looking at a graph of y= x2 and y= 1, x must range from -1 to 1. Now, for each x, y must range from the graph of y= x2 up to the line y= 1. Those are the limits of integration and the integral is
    [tex]\int_{x^2}^1 dydx[/tex]

    If, however, the region is bounded by y= x2, y= 0, and x= 1, then x ranges from 0 to 1 and, for each x, y ranges from 0 to x2. In that case, the integral is
    [tex]\int_0^1\int_{0}^{x^2} dydx[/tex]

    Now, what is the region really?
  6. Mar 12, 2008 #5
    You're right, the problem was to find the area bounded by y=x^2 and the line y=1, in the first quadrant only.

    I think this just about sums up the problem. The limits on the outer integral must be constants. I suppose the problem here is that I'm not, or wasn't, convinced that that is a fact, I've never seen it stated as a rule or law. But if that's the case, that just makes it easier!

    Thanks for the help.
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