Change order of integration examples
Given a double integral
Df(x,y)dA |
Df(x,y)dA = ∫
⎕⎕ dy, | (1) |
Df(x,y)dA = ∫
⎕⎕ dx. | (2) |
The difficult part often is how to determine what the limit of integration are, i.e., to determine what to put where the boxes ⎕ are. In particular, sometimes we start with an iterated integral of form (1) and want to change it to an iterated integral of form (2) (or vice versa). This process is called changing the order of integration.
Changing the order of integration is slightly tricky because its hard to write down a specific algorithm for the procedure. The easiest way to accomplish the task is through drawing a picture of the region D. From the picture, you can determine the corners and edges of the region D, which is what you need to write down the limits of integration.
Example 1
This process is shown best by an example. Let’s say you are given the integral
| ∫ 01 ∫ 1ey f(x,y)dxdy. |
| 0 ≤ y ≤ 1 | ||
| 1 ≤ x ≤ ey. |

We’ve simply drawn that y ranges from 0 to 1, and for a given value of y, the variable x ranges from 1 to ey. The curvey edge of the region is given by x = ey. The left edge is x = 1 and the top edge is y = 1. The upper-right corner is the point where both y = 1 and x = ey. This implies that x = e1 = e, so the point is (e, 1).
If we move x to be the outer integral, its limits must be constant. The limits will be the largest range of x. From picture, we see that 1 ≤ x ≤ e.
For a given value of x, how far does y range? It must be less than the top edge y = 1 and greater than the curvey edge x = ey. We can write the curvey edge as y = log x. Hence, the range of y is log x ≤ y ≤ 1.
Integral with integration order reversed is
| ∫ 1e ∫ log x1f(x,y)dydx. |
Example 2
Sometimes you need to change the order of integration to get a tractable integral. For example, if you tried to evaluate
| ∫ 01 ∫ x1ey2 dydx |
In this case, the region of integration is
| 0 ≤ x ≤ 1 | ||
| x ≤ y ≤ 1, |

We can also describe the region by
| 0 ≤ y ≤ 1 | ||
| 0 ≤ x ≤ y. |
| ∫ 01 ∫ x1ey2 dydx = ∫ 01 ∫ 0yey2 dxdy |
∫
01 ∫
0yey2
dxdy = ∫
01x dy = ∫
01yey2
dy |
∫
01 ∫
0yey2
dxdy = ∫
01yey2
dy = ∫
01 eudu = = (e - 1) |






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