Iterated Integrals and
Area in the Plane
By Dr. Julia Arnold
Courtesy 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 1
Evaluate an Iterated Integral
Definition of an Iterated Integral
Just 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 = b
then the double integral (or iterated integral) of f(x,y)
over R is defined by
The first integration gives us a function in x
second gives us a numerical value.
Let’s look at an example
while the
Example 1: Evaluate the iterated integral
3 2
2
x
  y dydx
The order the dx
dy is in determines
which you do first.
0 1
3
2
0
1
2
[
x
  y dy] dx
 x 2 y 2 2
 x 2 22 x 212   3 x 2
0  2 1 dx and since  2  2    2


3
 3x 2
0  2
3

3 x 3 x 3 3 27
27
dx




0


6
2 0 2
2




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
x

2 ye x dydx
This is the solution to the first integral:
1 1
 x

x
1  1 2 ye dy dx


4
4
  xe
x
x
 2 ye
x
dy  y e
1

 e x dx using parts
0
 xe
x


2 x
4
1 4
4
1
 4e  e   4
1
e e
2
x
 x e  x  12 e  x  xe  x  e  x
1
Hint: start with u = e-x and dv=(x – 1)
Objective 2
Use an iterated integral to find
the area of a plane region
Let’s begin by finding the area of a rectangular
y
region.
d
R
c
a
b
x
y
If we integrate with
respect to y first we
would go from c to d.
d
Then integrate with
respect to x and we
would go from a to b.
R
c
a
b
x
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.
y
Y goes from 0 to

x2  y 2  4
4  x2
x goes from 0 to 2

2
4 x 2
 

0
dydx
Using a table of integrals
0
2
 4 x 2 
1
x2
2
2


dy
dx

4

x
dx

x
4

x

4
arcsin


0  0  0
2
2 0


1
2 1
0
2
2
4

4

4
arcsin

0
4

0

4
arcsin




2
2 2
2
2

x






2 arcsin1  2 arcsin 0  2

2
 20  
Since we know this is ¼ of a circle we can verify
by using the traditional formula.
Exercise: Use an iterated integral to find the area of the
region bounded by the graphs of the equations.
xy  9, y  x, y  0, x  9
First let’s sketch
the bounded area.
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.




x











Exercise: Use an iterated integral to find the area of the
region bounded by the graphs of the equations.
xy  9, y  x, y  0, x  9
Triangle area
1
9
3 3  and

2
2
3 x
x2 3 9
0 0 dydx  2 0  2
area on right
 9x 
9
9
9
9


dy
dx

dx

ln
x

9
ln
9

9
ln
3

9
ln
 9 ln 3
3  0  3 x
3
3




9
Total Area +9ln3
2
9
y










x











You may be wondering if you
integration. The answer is
be easier than the other.
Exercise: Sketch the region
order of integration.
0
can switch the order of
yes. However, one way may


R of integration and switch the


4 2

y

f ( x, y )dydx
y
y  x  2 and 0  y  4
y  x2  4





x











Exercise: Sketch the region
order of integration.
R of integration and switch the
y

Switched
4 2

0
f ( x, y )dxdy
y
y  x  2 and 0  y  4
y  x2  4


2
x2
0
0
  f ( x, y)dydx







x











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.
0  x  4  y 2 and  2  y  2
2 4 y 2
 
2
2
dxdy 

2
0
 2

y3 2
4  y dy  4 y 

3 2
2
3


2
 
4  2 
  4  2  

3
3

3

  8  8  8  8  16  16  32

3
3
3
3

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.
0  x  4  y 2 and  2  y  2
x  4  y2
y2  4  x
y   4 x
We will need
two double
integrals in
this order.
4
4 x
 
0
4

0
4
dydx  
0

dydx 
0  4 x
4
4
0
0
1
2
4  xdx   4  xdx  2   4  x  (1)dx
0
4
1
2
2   4  x  (1)dx 
4  4  x 
3
0
4  4  4 
3
3
2

4  4  0 
3
3
2
0
3
2
4
0
32 32

3
3
For comments on this
presentation you may email the
author
Dr. Julia Arnold at
jarnold@tcc.edu.