Multiple Integration

Recall our definition of the definite integral of a function of a single variable:

Let f(x) be defined on [a,b] and let x0,x1,,xn be a partition of [a,b]. For each [xi-1,xi], let xi* [xi-1,xi]. Then
b

a 
f(x) dx =
lim
maxΔxi 0 
n
Σ
i = 1 
f(xi*)Δxi
Take a quick look at the Riemann Sum Tutorial

We can extend this definition to define the integral of a function of two or more variables.

Double Integral of a Function of Two Variables

Let f be defined on a closed and bounded region R of the xy-plane. Set up a grid of vertical and horizontal lines in the xy-plane to form an inner partition of R into n rectangular subregions Rk of area ΔAk, each of which lies entirely in R. (Ignore the rectangles that are not entirely contained in R) Choose a point (xk*, yk*) in each subregion Rk. The sum

n
Σ
k = 1 
f(xk*, yk*) ΔAk  
is called a Riemann Sum. In the limit as we make our grid more and more dense, we define the double integral of f(x,y) over R as
  

 
∫∫
R
f(x,y) dA =
lim
maxΔAk 0 
n
Σ
k = 1 
f(xk*, yk*) ΔAk

Notes

  • If this limit exists, we say that f is integrable over the region of integration R.

  • If f is continuous on R, then f is integrable over R.

Geometric Interpretation of the Double Integral

Notice that as we increase the density of our grid, the sum ΣkAk of the individual rectangles better and better approximates the area of region R. In the limit as ΔAk 0, we have

Area of R =  
∫∫
R
dA
Suppose now that f(x,y) 0 on R. Then f(xk*,yk*)ΔAk is the volume of a rectangular parallelopiped of height f(xk*,yk*) and base area ΔAk. Adding up these volumes, we get an appoximation for the volume of the solid above R and below the suface z = f(x,y). Thus, in the limit as ΔAk 0,
Volume of solid
above R and
below the surface
z = f(x,y)
=  
∫∫
R
f(x,y) dA     (for f(x,y) 0 on R)

Note

The interpretaion of the double integral as a volume still holds if f(x,y) takes on both positive and negative values. In this case, we obtain the difference between the volume above the xy-plane between z = f(x,y) and R and the volume below the xy-plane between z = f(x,y) and R.

We next turn to the actual evaluation of double integrals.

Iterated Integrals

In the double integral  
∫∫
R
f(x,y) dA, dA may be viewed informally as an infinitesimal
area of a rectangle inside R with dimensions dy and dx. For the kinds of "ordinary" functions and regions we'll be concerned with,

 
∫∫
R
f(x,y) dA
=
b

a 





g2(x)

g1(x) 
f(x,y) dy




dx = b

a 
g2(x)

g1(x) 
f(x,y) dy dx
=
d

c 





h2(y)

h1(y) 
f(x,y) dx




dy = d

c 
h2(y)

h1(y) 
f(x,y) dx dy

where the limits of integration are determined by the region R over which we are integrating.

Notes

These integrals are called iterated integrals, since we integrate more than once.

We integrate "from the inside out". That is, in ∫ b
a
g2(x)
g1(x)
f(x,y) dy dx, we first integrate
 f(x,y) with respect to y and evaluate it at g2(x) and g1(x).
We then integrate the result with respect to x and evaluate the outcome at a and b.

Iterated triple integrals  
∫∫∫
G
f(x,y,z)dV can be defined in a similar way.

An example will make these ideas more concrete.

Example

Let's evaluate the double integral  
∫∫
R
6xy dA, where R is the region bounded by
y = 0, x = 2, and y = x2. We will verify here that the order of integration is unimportant:

 
∫∫
R
6xy dA
=
2

0 
x2

0 
6xy dy dx
=
2

0 
[ 3xy2 |y = 0x2 ] dx
=
2

0 
3x5 dx
=
1
2
x6 |
|
|
2

x = 0 
=
1
2
(64)- 1
2
(0)
=
32
 
∫∫
R
6xy dA
=
4

0 
2

y 
6xy dx dy
=
4

0 
[ 3x2y |x = y2 ] dy
=
4

0 
( 12y-3y2 ) dy
=
( 6y2 -y3 ) |4y = 0
=
( 6(4)2-43 ) - ( 6(0)2- 03 )
=
32

so  
∫∫
R
6xy dA = 32 here, regardless of the order in which we carry out
the integration, as long as we are careful to set up the limits of integration correctly.

Now for a triple integral...

Example

We will evaluate the triple integral ∫ 2
0
y2
-1
z
1
yz dx dz dy.

2

0 
y2

-1 
z

1 
yz dx dz dy
=
2

0 
y2

-1 
[ (xyz) |x = 1z ] dz dy
Integrate with respect to x first.
=
2

0 
y2

-1 
( yz2-yz ) dz dy
Next integrate with respect to z.
=
2

0 









yz3
3
- yz2
2




|
|
|
y2

z = -1 





dy
=
2

0 





y7
3
- y5
2
+ 5y
6





dy
Finally, integrate with respect to y.
=





y8
24
- y6
12
+ 5y2
12





|
|
|
2

0 
=
264
24
- 64
12
+ 20
12
=
84
12
=
7


Key Concept [index]

Let f(x,y) be defined on a closed and bouned region R of the xy-plane. Then

 
∫∫
R
f(x,y) dA =
lim
maxΔAk 0 
n
Σ
k = 1 
f(xk*, yk*) ΔAk
where each ΔAk gives the area of a rectangle in an inner partition of R.

We evaluate  
∫∫
R
f(x,y) dA as an iterated integral:

 
∫∫
R
f(x,y) dA
=
b

a 
g2(x)

g1(x) 
f(x,y )dy dx
=
d

c 
h2(y)

h1(y) 
f(x,y) dx dy
for "ordinary" regions R and functions f(x,y).