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Math Insight

Examples of changing the order of integration in double integrals

Given a double integral

of a function over a region , y ou may be able to write it as two different iterated integrals. Y ou can

integrate with respect to first, or y ou can integrate with respect to first. If y ou integrate with respect to

first, y ou will obtain an integral that looks something like

and if y ou integrate with respect to first, y ou will obtain an integral that looks something like

We often say that the first integral is in order and the second integral is in order.

One difficult part of computing double integrals is determining the limits of integration, i.e., determining

what to put in place of the boxes in the above integrals. In some situations, we know the limits of

integration the order and need to determine the limits of integration for the equivalent integral in

order (or v ice versa). The process of switching between order and order in double

integrals is called changing the order of integration (or reversing 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 . From

the picture, y ou can determine the corners and edges of the region , which is what y ou need to write

down the limits of integration.

We demonstrate this process with examples. The simplest region (other than a rectangle) for reversing the

integration order is a triangle. Y ou can see how to change the order of integration for a triangle by

comparing example 2 with example 2' on the page of double integral examples. In this page, we give some

further examples changing the integration order.

Example 1

Change the order of integration in the following integral

f(x,y) dA∬D

f(x,y) D

x y

x

f(x,y) dA = ( f(x,y) dx)dy,∬D

∫□

□∫

□

□

y

f(x,y) dA = ( f(x,y) dy)dx.∬D

∫□

□∫

□

□

dx dy dy dx

□dx dy

dy dx dx dy dy dx

D

D

f(x,y)dx dy.∫1

0∫

ey

1

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(Since the focus of this example is the limits of integration, we won't specify the function . The

procedure doesn't depend on the identity of .)

Solution: In the original integral, the integration order is . This integration order corresponds to

integrating first with respect to (i.e., summing along rows in the picture below), and afterwards

integrating with respect to (i.e., summing up the values for each row). Our task is to change the

integration to be , which means integrating first with respect to .

We begin by transforming the limits of integration into the domain . The limits of the outer integral

mean that and the limits on the inner integral mean that for each value of the range of is

The region is shown in the following figure.

The maximum range of over the region is from 0 to 1 , as indicated by the gray bar to the left of the figure.

The horizontal hashing within the figure indicates the range of for each value of , beginning at the left

edge (blue line) and ending at the right curve edge (red curve).

We have also labeled all the corners of the region. The upper-right corner is the intersection of the line

with the curve . Therefore, the value of at this corner must be , and the point is

.

To change order of integration, we need to write an integral with order . This means that is the

variable of the outer integral. Its limits must be constant and correspond to the total range of over the

region . The total range of is , as indicated by the gray bar below the region in the following

figure.

Since will be the variable for the inner integration, we need to integrate with respect to first. The

vertical hashing indicates how, for each value of , we will integrate from the lower boundary (red curve) to

the upper boundary (purple line). These two boundaries determine the range of . Since we can rewrite the

equation for the red curve as , the range of is . (The function

indicates the natural logarithm, which sometimes we write as .)

f(x,y)f

dx dy

x

y

dy dx y

D dy

0 ≤ y ≤ 1, dx y x

1 ≤ x ≤ .ey D

y

x y

x = 1 x = ey

y = 1 x = ey x e = = ee1

(e, 1)

dy dx x

x

D x 1 ≤ x ≤ e

y y

x

y

x = ey y = log x y log x ≤ y ≤ 1 log x

lnx

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In summary , the region can be described not only by

as it was for the original integral, but also by

which is the description we need for the new integration order. This latter pair of inequalites

determine the bounds for integral.

We conclude that the integral with integration order reversed is

Example 2

Sometimes y ou need to change the order of integration to get a tractable integral. For example, if y ou tried

to evaluate

directly , y ou would run into trouble. There is no antiderivative of , so y ou get stuck try ing to compute

the integral with respect to . But, if we change the order of integration, then we can integrate with respect

to first, which is doable. And, it turns out that the integral with respect to also becomes possible after

we finish integrating with respect to .

According to the limits of integration of the given integral, the region of integration is

which is shown in the following picture.

Since we can also describe the region by

D

0 ≤ y ≤ 11 ≤ x ≤ ey

dx dy

1 ≤ x ≤ e

log x ≤ y ≤ 1,

dy dx

f(x,y)dx dy∫ 10 ∫ ey

1

f(x,y)dy dx.∫e

1∫

1

log x

dy dx∫1

0∫

1

x

ey 2

ey 2

y

x y

x

0 ≤ x ≤ 1x ≤ y ≤ 1,

0 ≤ y ≤ 10 ≤ x ≤ y,

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the integral with the order changed is

With this new order, we integrate first with respect to

Since the integration with respect to gave us an extra factor of , we can compute the integral with

respect to by using a -substitution, , so . With this substitution, rannges from 0 to

1 , and we calculate the integral as

Example 3

Here's an example that's a bit more tricky . Reverse the order of integration in the following integral.

Solution: The region described by this integral is

as shown in the following image, where the total range on is shown by the gray bar below the region, and

the variable boundaries for are shown by the blue and cy an curves.

One trick for changing variables with this region is correctly dealing with the lower boundary .

When we solve this boundary equation for as a function of , we may be tempted to write it as

and may be even think that in the region.

Looking closely at the picture, we see this cannot be the case. In fact, the lower boundary for as a function

of (the blue curve) has to be both the upper and lower boundaries for as a function of , as shown by the

red and purple curves in the below figure.

dy dx = dx dy∫1

0∫

1

x

ey 2 ∫1

0∫

y

0ey2

dx dy x

dx dy = x dy = y dy.∫1

0∫

y

0ey 2 ∫

1

0ey2 ∣∣

x=y

x=0∫

1

0ey 2

x y

y u u = y 2 du = 2y dy u

dx dy∫1

0∫

y

0ey 2

= y dy∫1

0ey 2

= du = = (e − 1).∫1

0

12

eu 12

eu∣∣∣1

0

12

f(x,y)dy dx∫5π/2

π/2∫

1

sin x

D

π/2 ≤ x ≤ 5π/2sinx ≤ y ≤ 1.

x

y

y = sin(x)x y

x = arcsin(y) x ≤ arcsin(y)

y

x x y

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To get the formula for these boundaries, we have to remember how the inverse of the sinusoid, , is

defined. In order to define the inverse of , we need to restrict the function to an interval where it

takes on each value only one time. The standard way to define is to restrict to values of

in the interval as ranges from to 1 in that interval. This means that ranges

from as goes from to 1 .

For the upper boundary of (in purple), ranges from to . If we let , then

when and when , as required. For the lower boundary of (in red), we

need to be a decreasing function of , starting at when and decreasing to

when . These conditions are satisfied if we choose . If y ou are an expert at y our

trignometric identifies, y ou can verify that the equations for both of these curves are just different inverses

of , as taking the sinusoid of these equations reduces them to .

Since in the region, ranges over the interval (gray bar to the left of the region), we can describe the

region with the inequalities

This description of is what we need to change the order of integration, and we find that

More examples

If y ou'd like more double integral examples, y ou can study some introductory double integral examples..

Y ou can also take a look at double integral examples from the special cases of interpreting double integrals

as area and double integrals as volume.

See alsoIntroduction to double integrals

Double integrals as iterated integrals

Double integral examples

Double integrals where one integration order is easier

Cite this as

arcsin(y)sin(x)

arcsin(y) sin(x) x

[−π/2,π/2] sin(x) −1 arcsin(y)[−π/2,π/2] y −1

x x 3π/2 5π/2 x = arcsin(y) + 2π

x = 3π/2 y = −1 x = 5π/2 y = 1 x

x y x = 3π/2 y = −1 x = π/2y = 1 x = π − arcsin(y)

sin(x) y = sin(x)

y [−1,1]D

−1 ≤ y ≤ 1π − arcsiny ≤ x ≤ arcsiny + 2π.

D

f(x,y)dy dx = f(x,y)dx dy.∫5π/2

π/2∫

1

sin x

∫1

−1∫

arcsin y+2π

π−arcsin y

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Ny kamp DQ, “Examples of changing the order of integration in double integrals.” From Math Insight.

http://mathinsight.org/double_integral_change_order_integration_examples

Key words: change integration order, double integral, integral

Examples of changing the order of integration in double integrals by Duane Q. Ny kamp is licensed under a

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