Understanding how to reverse the integration order

1. Oct 13, 2011

Taturana

Hello,

I am talking about double integrals. I have seen many examples of how to reverse the integration order when current order is too difficult to calculate. But I still don't fully understand how to do it. I would like you to help me understanding this.

Suppose the integral:

∫∫f(x,y) dydx

and a region defined by:

x = 0 to 1
y = 1 - x^2

I want to change the order of integration from dydx to dxdy. I have some understanding of this, but I don't know if it is correct. I will ask some questions in order to understand.

1 - In the more external integral the limits have numbers, so we can get a number out of the integration. Reversing the order we will first integrate in x and then in y (y goes in external integral). To figure out what are the new y limits (in the external integral) I just need to plug x = 0 and x = 1 in (1 - x^2) equation to find the down limit and the upper limit, respectively, is that right? So we will integrate from y = 1 to y = 0?

2 - We know that y depends on x accordingly with y = 1 - x^2. But we need to figure out how does x depends on y, and then we plug this relation in the internal integral, is that right?

3 - The most important question. How do find how does x depends on y? Is there a rule to that? Could I just find the inverse function? I know that we can plot the region in the xy plane and try to find how x depends on y looking at the graph and figuring out, but isn't there a rule for that? A method for doing this? A method that will always works (or at least in the major cases).

2. Oct 13, 2011

HallsofIvy

Staff Emeritus
I assume that you are integrating x from 0 to 1 and y from 0 up to $1- x^2$.

Almost. You integrated with respect to y from 0 to $1- x^2$ so your integral will be from 0 to the maximum value of $1- x^2$ which is 1. So your "outer integral" with be from 0 to 1, not from 1 to 0 (which would change the sign).

I'm not clear what you are saying here. When you say "plug this relation in the internal intregral" are you talking about the integrand? If so, no, you make this one of limits of integration, then "plug in" that value after the integration.

If $y= 1- x^2$, then $x^2= 1- y$ and so $x= \pm\sqrt{1- y}$. Because your region of integration is was originally from x= 0 to x= 1, it is in the first quadrant so you take x from 0 to $\sqrt{1- y}$
$$\int_{x=0}^1 \int_{y=0}^{1- x^2} f(x,y)dy dx= \int_{y= 0}^1\int_{0}^{\sqrt{1- y}} f(x,y) dx dy$$

Take a simple case, such as f(x,y)= 1 and actually integrate
$$\int_{x=0}^1\int_{y= 0}^{1- x^2} dydx= \int_{y=0}^1\int_{x=0}^{\sqrt{1- y}}dxdy$$.
You should get the area under the parabola, 2/3, either way.

The way I think about problems like this is:
First find the lowest and highest values of the "outside" integral. For the problem above those would be x= 0 to 1 if I were integrating "dydx" or y= 0 to 1 for "dxdy".

Then I would imagine (or actually draw) a horizontal (for dxdy) or vertical (for dydx) line across the figure. The endpoints of those lines give the limits of integration.

To take another example, suppose we wish to integrate f(x,y) above the parabola $y= x^2$ and below the horizontal line y= 1. If I wish to integrate with the order "dxdy", is would note that the parabola crosses the line at (-1, 1) and (1, 1) and that the entire region of integration lies between x= -1 and x= 1. For each x in that interval a vertical line would go from $y= x^2$ to $y= 0$. The integral would be
$$\int_{x=-1}^1\int_{y= x^2}^1 f(x,y)dydx$$

If I wish to integrate in the order dydx, I note that the entire region lies between y= 0 and y= 1. For every y in that region, a horizontal line would go between two points on the parabola, having same y of course. Since $y= x^2$, solving for x, $x= -\sqrt{y}$ and $x= \sqrt{y}$. The integral would be $$\int_{y= 0}^1\int_{-\sqrt{y}}^\sqrt{y} f(x,y)dxdy$$.

(Note my addition of "x= " and "y= " on the lower limits. That's a very good way of keeping exactly what you are doing straight- especially when you start working with integrals in three dimensions!)

Last edited: Oct 13, 2011
3. Oct 13, 2011

Staff: Mentor

For your example, which is this:
$$\int \int_R f(x, y)~ dy~dx$$
R is the region bounded by y = 1 - x2, 0 <= x <= 1.

When you do this integration you are taking small rectangles whose area is dx * dy, and stacking them up vertically between y = 0 and y = 1 - x2. The inner integral does this. The outer integral sums these stacks between x = 0 and x = 1 to get the area of this region.

To reverse the order of integration, you are describing the region in a different way. In essesnce, the inner integral stacks the dx * dy rectangles horizontally between x = 0 and the x value on the curve, and then the outer integral sums these horizontal slices between y = 0 and y = 1.

Each horizontal slice has a left endpoint of x = 0. The right endpoint is given by $x = +\sqrt{1 - y}$.

As a result, the integral with the order of integration reversed is
$$\int_{y=0}^1 \int_{x=0}^{\sqrt{1-y}} f(x, y)~ dx~dy$$

4. Oct 13, 2011

Taturana

Thank you for the replies. It gave me a great advance. But I still have some questions.

I notice that everyone when teaching this reverse order of integration thing tells: look at the figure, analyze it etc.

1 - Isn't there a RULE for finding the new OUTER bounds? By rule I mean an algebraically way of doing it without looking the graph. Suppose you have a very complicated region, where drawing and analyzing is not an option.

2 - Isn't there a RULE for finding the new INNER bounds? I found myself that the rule is finding the inverse function. For example: my function was y = 1 - x, the inverse function is x = 1 - y; my function was y = 1 - x^2, the in verse function is x = sqrt(1-y). But I am not sure that this rule I realized is always valid, is really a rule. Does finding the inverse function always works?

Thank you for the help!

5. Oct 13, 2011

Staff: Mentor

As far as I know, there is no rule for getting the new integration limits.

6. Oct 13, 2011

Taturana

Not even the inverse function thing?

Thanks!

7. Oct 13, 2011

Staff: Mentor

Nope.

One of the boundaries of the region was the graph of x = y2 + 1. Here x is a function of y, but the inverse isn't even a function. If you solve for y, you get
$y = \pm\sqrt{x - 1}$, and you have to choose the correct root, which in this case is the positive square root. And this is a fairly simple boundary curve.