Thread: change order of integration of triple integral

Express the integral in the order .

As I am having difficulty getting my head around the region (I cant graph it accurately by hand), are there any ways of graphing it on computer?

I have been told that the final result will be a sum of six integrals in the order dxdzdy. Why is that? What is the reason as to why it cannot be done in just only integral? (For example they have said to consider and separetely.)

Notice that by Fubini's theorem you can write where is the region, on the plane, which encloses the area defined by the integration's limits, and .

Since the problem asks you to change the order from to , you can work the problem over the plane.
The reason why you should split the intervals is because in the middle of the process you will have to figure it out how to express the same region in other order.

Is this clear?

Originally Posted by whiteboard

Express the integral in the order .

So the "outside" integral is with respect to y in both instances? That simplifies a lot- it is just a matter of swapping x and z.

As I am having difficulty getting my head around the region (I cant graph it accurately by hand), are there any ways of graphing it on computer?

I have been told that the final result will be a sum of six integrals in the order dxdzdy. Why is that? What is the reason as to why it cannot be done in just only integral? (For example they have said to consider and separetely.)

Draw a two dimensional graph- an x-z graph- that should be easy. First, the dx integral is from x= y to x= 1 so on the x-axis, mark 1 and y (you don't know y but it is fixed and you know 0< y< 1 so mark some point between 0 and 1 and label it "y". Now, for each x, z is between z= x and z= 2x. Draw those two lines. The area you want is the "wedge" bounded by z= x, z= 2x, x= y (fixed remember) and x= 1.

You want to reverse the order of integration so you want to be able to write z values that do not depend upon x and then value for x that do depend upon z. You can see that the vertical line x= 1 crosses the two lines z= x and z= 2x and z= 1 and z= 2. You can also see that the vertical line x= y crosses the two lines at z= y and z= 2y. That is, the lowest value that z takes on is y (less than 1 remember) and the largest is 2. We would like to be able to write the z-integral as going from z= y to z= 2.

To determine the limits of integration for the x-integral, we look at horizontal lines, indicating x value for fixed z. Here is where we run into a problem. For z from y to 2y, the left end of such a line will be x= y and for z< 1, the right end will be at the line x= z. That is, the integral would be " ". But, then, once z> 2y, which can happen since y< 1 and z goes up to 2, the left end of a horizontal line is on the line z= 2x or x= a/2. Further, for z> 1, the right end will be on the vertical line x= 1, not the line z= x. That is why you will need to break the integral into several (possibly 6) different parts- to allow for those different possible endpoints.

Ok, I think I have done this correctly, however, I can only see how to make it into three separate regions (D1,D2,D3). As both the hint in the question and you have stated the result will be a sum of six different regions, I am lost.

Please tell me if I am on the right track here.