Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits in an area integral

  1. Apr 19, 2012 #1
    Let R be bounded by y=0, x=2 and y=x^2.
    Then ∫∫6xydA= ? (Note the integral is to be evaluated over R)

    Now what will be the lower limit of x. I took it to be 0 and the answer was 32. which turned out to be correct.
    Is their any way in such questions by which we can determine the lower limit of x or do we just take it to be 0?
  2. jcsd
  3. Apr 19, 2012 #2

    What I understand from what you wrote is that the area of integration is the one under the graph of [itex]y=x^2\,\,[/itex] on the interval [itex][0,2][/itex] , and thus

    we must take [itex]0\leq x\leq 2\,\,,\,\,0\leq y\leq x^2[/itex] .

    The lower limit of x is...the lower limit of x: since we're on [0,2] for x, what would you think we could have taken something else? If
  4. Apr 19, 2012 #3
    Look, how did you deduce that the interval was [0,2]. I did'nt write that. And it was y which was 0 not x. So how did you find out or infer that the interval was [ 0,2 ]?
  5. Apr 19, 2012 #4

    Ok, I think I see now what your problem was...Well, very simple: draw down a sketch of the integration area. You shall get the

    graph of the parabola [itex]y=x^2[/itex] in the first quadrant, so I deduced R MUST be the area between the y-axis, the x-axis and the graph

    of the parabola, as they talk of an area BOUNDED by...etc. This gives us x between 0 and 2...

  6. Apr 19, 2012 #5


    User Avatar
    Science Advisor

    An integral of the form [itex]\int_{x=a}^b\int_{y= c}^d f(x,y)dydx[/itex], that is, with numbers as all four limits of itegration is necessarily over a rectangle in the xy- plane, bounded by the four lines x= a, x= b, y= c, and y= d.

    This region of integration has three boundaries- the line y= 0, the line x= 2, and the curve y= x^2. The limits of integration will NOT all be constants since this is not a rectangle. It would be a good idea to graph or at least visualize the graphs of those boundaries.

    The first thing you have to do is decide in which order to to integrate. Suppose we decide to integrate in the order "dydx". Since the result must be a number and not depend on x or y, the limits on the outside integral must be numbers. Since we are differentiating with respect to x, and want to cover the entire region, the limits of integration must be the smallest and largest values of x in that region. Here, we see that y= 0 and [itex]y= x^2[/itex] itersect at (0, 0) while the right boundary is the line x= 2. The lower limit is x= 0 and the right limit is 2.

    But now, the limits of integration for y will depend on x! If you have actually drawn the graphs, now draw a vertical line representing a specific value of x (if you visualized the graph visualize the vertical line). You should see that the lower boundary is the line y= 0 and the upper boundary is [itex]y= x^2[/itex]. Your integral should be
    [tex]\int_{x=0}^2\int_{y= 0}^{x^2} 6xydydx[/tex]

    If, instead, you choose to use the order "dxdy"- that is, integrate with respect to x first, then y, note that the highest point in the region is where [itex]y= x^2[/itex] intersects x= 2 at the point (2, 4) while the lowest points are on line y= 0. So the "outer integral", with respect to y now, has limits y= 0 and y= 4. To find the x-limits draw, or imagine, a horizontal line at each y-value. x will go from the curve [itex]y= x^2[/itex], that is, [itex]x= \sqrt{y}[/itex] (we are clearly in the first quadrant so this is the positive square root) on the left to the line x= 2 on the right. The integral will be
    [tex]\int_{y=0}^4\int_{x=\sqrt{y}}^2 6xy dxdy[/tex]

    It would be a good exercise to do both of those itegrals to so that you really do get the same result.
  7. Apr 19, 2012 #6
    Thanks guys.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook