# Help in understanding the order of integration

I am having a major headache trying to solve problems based upon changing the order of integration. I have understood the concept, and have also seen and studied various threads in this forum based on similar questions.
Suppose I am given a problem where I have to change the order of a function f(x,y) with some limits. Then I can successfully draw the all the limits, as well as find the region of integration. The problem, however, comes when I have to 'split' the region of integration. I am unable to understand when am I supposed to divide the region and how. I have lots of problems in which the region has to be divided to get the correct answer.
Assuming I get past dividing the region, I dont understand how do I change the limits of integration. If the original function is f(x,y)dxdy, then when the order is changed it will be f(x,y)dydx and the outer limits ( x in this case) will remain constant. But I am unable to get the limits for the 'inner' part(i.e y). I tried reading through the various other threads all over the Internet, but I end up getting even more confused.

Help will be highly appreciated.

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vela
Staff Emeritus
Homework Helper
It would help if you posted a specific example and your attempt at solving it, so we can see where specifically you're running into trouble.

Generally, you have to split the region when a border can't be described by a single function. For example, suppose you have a trapezoidal region of the xy-plane with its vertices at (0,0), (1,1), (2,1), and (3,0). The bottom of the trapezoid lies on the x-axis; the top, on the line y=1; the left side, along the line y=x; and the right side, along the line y=3-x.

Suppose the outer integral is with respect to x. While the lower limit of integration for the y integral is always 0, there's no single formula which describes the upper limit as x runs from 0 to 3, so you have to break the region up into three pieces. When 0<x<1, the limits of y would be 0 to x; when 1<x<2, the limits would be from 0 to 1; and when 2<x<3, the limits would be from 0 to 3-x. In contrast, if the outer integral were with respect to y, the lower limit for the x integral would always be y and the upper limit would always be 3-y, so there'd be no need to split the region into pieces.

OK here is an example...
Change the order of $$\int^{a}_{0}$$ $$\int^{x+3a}_{\sqrt{a^{2}-x^{2}}}$$ f(x,y)dydx.

I drew the limits and got the region. However, I do not know how to further proceed with the problem. I know the region has to be divided in 3 parts for the answer says so. However, I cant seem to determine how should it be divided. All I know is that y varies between 0 to 4a. I dont know how these limits are then divided over the regions.

Help would be appreciated.

vela
Staff Emeritus
Homework Helper
After switching the order of integration, when you integrate with respect to x, you're holding y constant and varying x. In other words, you're moving along a horizontal line in the xy plane. So let's consider a couple of these lines. Take y=3.5a, for instance. For what values of x does this line lie within the region of integration? You can see this from your sketch of the region. At the lower end, it's bounded by the line y=x+3a, and at the upper end its bounded by the line x=a. You can ask the same question for the line y=2a. This time, the lower bound is x=0 and the upper bound x=a. Finally, when the line y=0.5a, the lower bound is the curve y=sqrt(a^2-x^2), and the upper bound is again x=a. Just by looking at your sketch, you should be able to see there are three different cases, you're going to have to deal with. Can you see that?

@vela,

Just one last doubt though. Almost every problem that I have encountered screams the need to draw the limits. But that's not always possible, is it? So how do we solve problems where we don't prefer to draw the limits? I read on other threads that by solving the inequalities, we can algebraically prove that the region under consideration is one and the same with respect to both x and y(or something like that).

vela
Staff Emeritus