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 x_{0},x_{1},…,x_{n} be
a partition of [a,b]. For each [x_{i1},x_{i}], let x_{i}^{*} ∈ [x_{i1},x_{i}]. Then

∫ 
b
a

f(x) dx = 
lim
maxΔx_{i} → 0


^{n} Σ
i = 1

f(x_{i}^{*})Δx_{i} 



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
(sia R una regione piana contenuta in una regione rettangolare W; se W è suddivisa in rettangolini
mediante rette orizzontali e verticali, l'insieme di tutti i rettangolini contenuti completamente in R
viene chiamato partizione interna di R)
Let f be defined on a closed and bounded region R of the xyplane. Set up a grid of vertical and horizontal lines in the
xyplane to form an inner partition of R into n rectangular subregions R_{k} of area ΔA_{k}, each of which
lies entirely in R (ignore the rectangles that are not entirely
contained in R). Choose a point (x_{k}^{*}, y_{k}^{*}) in each
subregion R_{k}. The sum

^{n} Σ
k = 1

f(x_{k}^{*}, y_{k}^{*}) ΔA_{k }  
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
maxΔA_{k} → 0


^{n} Σ
k = 1

f(x_{k}^{*}, y_{k}^{*}) ΔA_{k} 

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 Σ_{k}A_{k} of the individual rectangles better and better
approximates the area of region R. In the limit as ΔA_{k} → 0, we have
Suppose now that f(x,y) ≥ 0 on R. Then f(x_{k}^{*},y_{k}^{*})ΔA_{k} is the volume of a rectangular parallelopiped
of height f(x_{k}^{*},y_{k}^{*}) and base area ΔA_{k}. 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
ΔA_{k} → 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 xyplane
between z = f(x,y) and R and the volume below the xyplane
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,




∫ 
b
a




∫ 
g_{2}(x)
g_{1}(x)

f(x,y) dy 

dx = 
∫ 
b
a


∫ 
g_{2}(x)
g_{1}(x)

f(x,y) dy dx 
 


∫ 
d
c




∫ 
h_{2}(y)
h_{1}(y)

f(x,y) dx 

dy = 
∫ 
d
c


∫ 
h_{2}(y)
h_{1}(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 
∫ 
g_{2}(x) g_{1}(x) 
f(x,y) dy dx, we first integrate 
 f(x,y) with respect to y and evaluate it at
g_{2}(x) and g_{1}(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 = x^{2}. We will verify here that the order of integration is unimportant:




∫ 
2
0


∫ 
x^{2}
0

6xy dy dx 
 


∫ 
2
0

[ 3xy^{2} _{y = 0}^{x2} ] dx 
 

 

 

 


 





 


∫ 
4
0

[ 3x^{2y} _{x = √y}^{2} ] dy 
 

 

( 6y^{2} y^{3} ) ^{4}_{y = 0} 
 

( 6(4)^{2}4^{3} )  ( 6(0)^{2} 0^{3} ) 
 


 


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


∫ 
y^{2}
1


∫ 
z
1

yz dx dz dy 



∫ 
2
0


∫ 
y^{2}
1

[ (xyz) _{x = 1}^{z} ] dz dy 
 Integrate with respect to x first. 




∫ 
2
0


∫ 
y^{2}
1

( yz^{2}yz ) dz dy 
 Next integrate with respect to z. 




∫ 
2
0






yz^{3} 3

 
yz^{2} 2


   
y^{2}
z = 1



dy 




∫ 
2
0




y^{7} 3

 
y^{5} 2

+ 
5y 6


dy 
 Finally, integrate with respect to y. 





y^{8} 24

 
y^{6} 12

+ 
5y^{2} 12


   
2
0












 

Key Concept
[index]
Let f(x,y) be defined on a closed and bouned region R of the
xyplane. Then
∫∫
R

f(x,y) dA = 
lim
maxΔA_{k}→ 0


^{n} Σ
k = 1

f(x_{k}^{*}, y_{k}^{*}) ΔA_{k} 

where each ΔA_{k} gives the area of a rectangle in an inner
partition of R.
We evaluate 
∫∫
_{R} 
f(x,y) dA as an iterated integral: 




∫ 
b
a


∫ 
g_{2}(x)
g_{1}(x)

f(x,y )dy dx 
 


∫ 
d
c


∫ 
h_{2}(y)
h_{1}(y)

f(x,y) dx dy 

 

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