Iterated Integrals and Iterated Integrals and Area in the PlaneArea in the Plane

Iterated Integrals and Iterated Integrals and Area in the PlaneArea in the Plane

By Dr. Julia ArnoldBy Dr. Julia ArnoldCourtesy of a CDPD grantCourtesy of a CDPD grant

Objectives of this section:

1. Evaluate an Iterated Integral

2. Use an iterated integral to find the area of a plane region

Objective 1Objective 1Objective 1Objective 1

Evaluate an Iterated Integral

Definition of an Iterated IntegralJust as we can take partial derivative by considering only one of the variables a true variable and holding the rest of the variables constant, we can take a "partial integral".  We indicate which is the true variable by writing "dx", "dy", etc.  Also as with partial derivatives, we can take two "partial integrals" taking one variable at a time.  In practice, we will either take x first then y or y first then x.  We call this an iterated integral or a double integral. 

Notation:Let f(x,y) be a function of two variables defined on a region R bounded below and above by                 y  =  g1(x)          and          y  =  g2(x)and to the left and right by                x  =  a          and          x  =  bthen the double integral (or iterated integral) of f(x,y) over R is defined by

The first integration gives us a function in x while the second gives us a numerical value.

Let’s look at an example

Example 1: Evaluate the iterated integral

3 22

0 1

3 22

0 1

3 2 2 2 2 2 2 2

0

3 2 3 3

0

[ ]

2 2 1 3

12 2 2 2

33 3 27 270

02 6 2 2 2

x y dydx

x y dy dx

x y x x xdx and

x x xdx

since

The order the dx dy is in determines which you do first.

We integrate with respect to y holding x term like a constant. Evaluate it at its limits.

Then we integrate with respect to x and evaluate it at its limits.

Example 2: Evaluate the iterated integral

4

1 1

422 2

1 1 1

4

0

4 14

2

2 2 11

4 1 44

1

xx

x xx x x x x x x

x x

x

ye dydx

xye dy dx ye dy y e x e e xe e

xe e dx parts

xe e ee e

using

This is the solution to the first integral:

Hint: start with u = e-x and dv=(x – 1)

Objective 2Objective 2Objective 2Objective 2

Use an iterated integral to find the area of a plane region

Let’s begin by finding the area of a rectangular region.

x

y

a b

c

d

R

x

y

a b

c

dR

If we integrate with respect to y first we would go from c to d.

Then integrate with respect to x and we would go from a to b.

( )( )b d b

a c a

dydx d c dx d c b a

Which is the same as length times width.

Example 2: Use an integral to find the area of the region.

x

y

2 2 4x y Y goes from 0 to

x goes from 0 to 2

2

2

2 4

0 0

2 4 22 2

0 0 0

2

214 4 4arcsin

02 2

1 2 1 02 4 4 4arcsin 0 4 0 4arcsin

2 2 2 2

2arcsin1 2arcsin 0 2 2 02

x

x

dydx

xdy dx x dx x x

Using a table of integrals

Since we know this is ¼ of a circle we can verify by using the traditional formula.

24 x

Exercise: Use an iterated integral to find the area of the region bounded by the graphs of the equations.

9, , 0, 9xy y x y x

First let’s sketch the bounded area.

x

y

It looks like we might need to divide this into two problems.

Since the left area is of a right triangle we could save time and use the formula.

Exercise: Use an iterated integral to find the area of the region bounded by the graphs of the equations.

9, , 0, 9xy y x y x

x

y

3 2

0 0

99 9

3 0 3

31 9 93 3

02 2 2 2

99 9ln 9ln9 9ln3 9ln 9ln3

3 3

x

x

xand dydx

dy dx dx xx

area on right

9Total Area +9ln3

2

Triangle area

You may be wondering if you can switch the order of integration. The answer is yes. However, one way may be easier than the other.

Exercise: Sketch the region R of integration and switch the order of integration.

4 2

0

2

( , )

2 0 4

4

y

f x y dydx

y x and y

y x

x

y

Exercise: Sketch the region R of integration and switch the order of integration.

4 2

0

2

( , )

2 0 4

4

y

f x y dxdy

y x and y

y x

x

y

Switched22

0 0

( , )x

f x y dydx

Last Exercise

Sketch the region R whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area.

20 4 2 2x y and y

242 2 32

2 0 2

3 3

24 4

23

2 2 8 8 16 324 2 4 2 8 8 16

3 3 3 3 3 3

y ydxdy y dy y

Last Exercise

Sketch the region R whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area.

2

2

2

0 4 2 2

4

4

4

x y and y

x y

y x

y x

4 4 4 0

0 0 0 4

4 4 4 1

2

0 0 0

34 1 2

2

0

3 3

2 2

4 4 2 4 ( 1)

4 4 42 4 ( 1)

03

4 4 4 4 4 0 32 320

3 3 3 3

x

x

dydx dydx

xdx xdx x dx

xx dx

We will need two double integrals in this order.

For comments on this presentation you may email the author Dr. Julia Arnold at jarnold@tcc.edu.