LECTURE-29-DOUBLE INTEGRAL FOR ENGINEERING MATHEMATICS.ppt

Engineering Maths Double Integral

1. DEPARTMENT OF
MATHEMATICS
[YEAR OF ESTABLISHMENT – 1997]
DEPARTMENT OF MATHEMATICS, CGU

2. MATHEMATICS - I
FOR BTECH 1
st
SEMESTER COURSE [COMMON TO ALL BRANCHES OF
ENGINEERING]
DEPARTMENT OF MATHEMATICS, CVRCE
TEXT BOOK:
ADVANCED
ENGINEERING
MATHEMATICS
BY ERWIN
KREYSZIG
Double Integrals
[Chapter – 9.3]
LECTURES –29

3. 10/28/25 LINE INTEGRAL CONTI..........
Content:

►
Introduction to Double Integral

►
Definition and Properties

►
Computation of Double Integral

►
Applications of Double Integrals

►
Examples

►
Change of Variables

►
Examples

►
Practice Problems

4. DOUBLE INTEGRAL
Let f(x, y) be any function. Let R be a region in a plane. Let us
subdivide the region in to a large number of small
rectangular subregions. Let the k
th
region has the
dimensions say x
k
and y
k .
Let (x
k,y
k ) be any arbitrary point on
the k
th
subregion and  A
k = x
k y
k
be the area of the k
th
rectangular
subregion. Next consider the
sequence of real numbers (J
1
,
J
2
, . . . , ), where
1
( , )
n
n k k k
k
J f x y A

 
x
k
y
k  A
k
= x
k
y
k
x
y
(0,0)

5. Assuming that f(x, y) is continuous in R and R is
bounded by finitely many smooth curves. It can
be shown that the sequence (J
1, J
2, . . . , )
converges to a limit that is independent of the
choice of subdivisions and corresponding points
(x
k,y
k). This limit is called the double integral of
f(x, y) over the region R and is denoted by
(, ) or (, )
R R
f xydxdy f xydA 
DOUBLE INTEGRAL

6. 10/28/25 LINE INTEGRAL CONTI..........

•area of the region: finding area using double integral,
volume of an elliptic paraboloid.
• average value of a function: calculating avarage
strom rainfall, of a wire;
• Surface integral;
• double and volume integral.
Applications of Double Integrals

7. 1. [ ( , ) ( , )] ( , ) ( , )
A A A
f x y g x y dA f x ydA g x ydA    
2. (,) (,)
A A
cfxydA c fxydA 


AA
dAyxgdAyxf ),(),(
1.Properties Linearity
If f(x,y) ≥ g(x,y) for all (x,y) in R, then2.Comparison
S
ome Properties of Double Integral

8. 10/28/25 LINE INTEGRAL CONTI..........



2121
),(),(),(
AAAA
dAyxfdAyxfdAyxf
3.Additively
If A
1
and A
2
are non-overlapping regions then
A
1
A
2
10/28/25 LINE INTEGRAL CONTI..........
S
ome Properties of Double Integral

9. •If f (x,y) is continuous on rectangle R=[a,b]×[c,d]
then double integral is equal to iterated integral
a b
x
y
c
d
x
y


b
a
d
c
d
c
b
aR
dydxyxfdxdyyxfdAyxf ),(),(),(
fixed fixed
10/28/25 LINE INTEGRAL CONTI..........
c
omputation of Double Integral

10. •If f (x,y) is continuous on
A={(x,y) | x in [a,b] and h (x) ≤ y ≤ g (x)} then
double integral is equal to iterated integral
a b
x
y
h(x)
g(x)
x


b
a
xg
xhA
dydxyxfdAyxf
)(
)(
),(),(
A
10/28/25 LINE INTEGRAL CONTI..........
c
omputation of Double Integral

11. •If f (x,y) is continuous on
A={(x,y) | y in [c,d] and h (y) ≤ x ≤ g (y)} then
double integral is equal to iterated integral
d
x
y


d
c
yg
yhR
dxdyyxfdAyxf
)(
)(
),(),(
c
h(y) g(y)
y
A
10/28/25 LINE INTEGRAL CONTI..........
c
omputation of Double Integral

12. If f (x, y) = φ (x) ψ(y) is continuous on rectangle
R=[a,b]×[c,d] then double integral is equal to
iterated integral












 
d
c
b
a
d
c
b
aR
dyydxxdxdyyxdAyxf )()()()(),( 
10/28/25 LINE INTEGRAL CONTI..........
c
omputation of Double Integral

13. S
ome applications of double integral
2. Volume V beneath the surface z = f(x,y) (>0) and above
the region R in the xy-plane is given by
,
R
V f xy dxdy
1.Area A of a plane region R is given by
R
A dA

14. 3. Total Mass M of a mass distribution of the density f(x,y)
(= mass per unit area) enclosing a region R in the xy-plane
is given
S
ome applications of double integral
,
R
M fxydxdy
4. Let be the coordinates of the center of gravity of a plane
region R enclosed in xy-plane with mass M of a mass distribution of
the density f(x,y) (= mass per unit area). Then we have.
 
1 1
, and y ,
R R
x xf x y dxdy yf x y dxdy
M M
  
x,y

15. 5. Moments of inertia I
x
and I
y
about x and y axes,
respectively, of a mass with a mass distribution of the density
f(x,y) (= mass per unit area) in a region R in xy-plane are given
as follows.
S
ome applications of double integral
 
2 2
, and ,
x y
R R
I y f x y dxdy I x f x y dxdy  
Polar Moment of inertia I
0
about the origin of a mass with a
mass distribution of the density f(x,y) (= mass per unit area)
in a region R in xy-plane is given as follows.
 
2 2
0
,
x y
R
I I I x y f x ydxdy   

16. 1.Evaluate the following double integral:
2 4
2 2
0 0
( )x y dxdy
2 4
2 2
0 0
( )x y dxdy
4
2 3
2
0 0
3
x
xy dy
 
 
 
 

128 32 160
3 3 3
  
ANS
S
ome problems involving double integral
Solution:
2 4
2 2
0 0
( )x y dxdy
 
  
 

2
3
0
64 4
3 3
y y 
 
 
 
2
2
0
64
4
3
y dy
 
 
 
 


17. 4 2
2 2
0 0
( )x y dxdy
4
2
0
8
2
3
x dx
 
 
 
 

32 128 160
3 3 3
  
ANS
S
ome problems involving double integral
2.Evaluate the following double integral:
Solution:
2 4
2 2
0 0
( ) byorderreversed.x y dxdy
4 2
2 2
0 0
( )x y dy dx
 
  
 

2
4 3
2
0 0
3
y
yx dx
 
 
 
 

4
3
0
8 2
3 3
x x 
 
 
 

18.  
3
2 2
0
y
y
x y dxdy


3
3 3 4
0 0
8 8 8 81
54
3 3 4 3 4
y y
dy

   


ANS
3.Evaluate the following double integral: 
3
2 2
0
y
y
x y dxdy


Solution:  
3
2 2
0
y
y
x y dx dy

 
  
  

33
2
0
3
y
y
x
xy dy

 
3 3 3
3 3
0
3 3
y y
y y dy
     
       
     

S
ome problems involving double integral

19.  
3
2 2
0
4.Findthevalue of byorderreversed.
y
y
x y dxdy


Solution:
-3,3 0,3 3,3
0,0
x=-y
x=y
Y=3A
O
CB
10/28/25 LINE INTEGRAL CONTI..........
S
ome problems involving double integral
Y-axis
X-axis

20. For changing the order of integration, we
divide the region of integration into vertical
strips . The region of integration is divided
into two parts OAB and OBC
For the region OAB, y varies from (-x)
to 3 and x varies from (-3) to 0. for the
region OBC, y varies from x to 3 and x
varies from 0 to 3
S
ome problems involving double integral
Given that the region of integration is bounded by x = - y , x
= y, y = 0 and y = 3, i.e. the region of integration is portion
OAC in the Figure.

21.      
3
2 2 2 2 2 2
0

y
y OAB region OBC region
x y dxdy x y dydx x y dydx
  
       
3 3
0 3 3 3
2 2
3 0
3 3
x
x x x x
y y
x y dx x y dx

  
    
   
0 3 3 3
2 3 2 3
3 0
3 9 3 9
3 3
x
x x
x x
x x dx x x dx

 
         
              
         
 
10/28/25 LINE INTEGRAL CONTI..........
S
ome problems involving double integral
   
30 3 3
2 2 2 2
3 0
yx
x y x x y x
x y dy dx x y dy dx

   
   
      
      
  

22. 3
0
4
3
0
3
3
4
3
99
3
x
xxxx
x


542727)272727()272727(0 
ANS
10/28/25 LINE INTEGRAL CONTI..........
S
ome problems involving double integral

23. C
hange of variables in double
i
ntegrals: jacobian
The formula for a change of variables in double
integrals from x,y to u,v is
That is, the integrand is expressed in terms of u and v,
and dxdy is replaced by dudv times the absolute value
of the Jacobian “J” which is defined as follows.
  , , , ,
R R
f x y dxdy f x uv y uv Jdudv

 

24. C
hange of variables in double
i
ntegrals: jacobian


,
,
x x
xy x y x yu v
J
y yuv u v v u
u v
 
     
    
     
 
Here we assume that the functions x = x(u,v) and y = y(u,v)

25. C
hange of variables in double
i
ntegrals: jacobian
and

26. P
roblems involving Change of
v
ariables in double integrals:
j
acobian

2
5. Evaluate , where is region
bounded by the parallelogram 0, 2,
3 2 0,3 2 3.
R
x y dxdy R
x y x y
x y x y

   
   

By changing the variables x,y to the new variables u, v, by the
substitution (transformation) x + y = u, 3x - 2y = v, the given
parallelogram R reduces to a rectangle R* as shown in Figure
given below.
Solution: 
2
Let , =f x y x y

27. x
y
0(0,0)
R
R*
u
v
u=2
v=3
P
roblems involving Change of variables in
d
ouble integrals: jacobian
Transformed region
R* : 0 ≤ u ≤ 2, 0 ≤ v ≤ 3
Given region
R : 0 ≤ x+y ≤ 2, 0 ≤ 3x-2y ≤ 3
0(0,0)
u=0
v=0
(2,0)
(2,3)(0,3)

28. P
roblems involving Change of variables in
d
ouble integrals: jacobian
   
   
1 2 1 1
2 , 2 ,
5 5 5 5
1 3 1 1
3 , 3 .
5 5 5 5
x x
u v u v
u u v v
y y
u v u v
u u v v
      
      
   
      
      
     
   
      
   
Solving and 3 -2 for and we get
1 1
2 and 3 .
5 5
.x y u x y v x y
x u v y u v
  
   

29. 

2 1
, 5 5
3 1,
5 5
2 3 5 1
25 25 25 5
x x
xy u v
J
y yuv
u v
 
  
   
 

 
   
1 1
5 5
J  
P
roblems involving Change of variables in
d
ouble integrals: jacobian

30. 10/28/25 LINE INTEGRAL CONTI..........
P
roblems involving Change of variables in
d
ouble integrals: jacobian
Since we have u = x + y = 0 and u = x + y =2, u
varies from 0 to 2. Since we have v = 3x - 2y = 0 and
v = 3x - 2y =3, v varies from 0 to 3. Thus the given
integral in terms of the new variables u,v is given by

 
2
Also we have , , , =f x uv y uv u
 
2
,
R R
x y dxdy f x y dxdy  

31.  , , ,
R
f x uv y uv Jdudv


P
roblems involving Change of variables in
d
ouble integrals: jacobian
3 2
2
0 0
1
5
u dudv
 

 
 

3 2
2
0 0
1
5
ududv
3 2
2
0 0
1
5
udu dv
 
 
 

2
3 3
0 0
1
5 3
u
dv
 

 
 

3 3
0
1 2
0
5 3
dv
 
 
 
 

3
0
8
15
dv

3
0
8 24 8
.
15 15 5
v  

32. P
roblems involving Change of
v
ariables in double integrals:
j
acobian
 
2 2
6. Evaluate , where is region
0 , 0 using ploar substitution.
x y
R
e dxdy R
x y
 
   

Let us change the variables x,y to the new variables u, v, by
the polar substitution (transformation) x = rcos, y = rsin.
Solution: 
 
2 2
Let , =
x y
f xy e
 

33. P
roblems involving Change of variables
i
n double integrals: jacobian
The region of double integration R is R: 0 < x < , 0 < y < ,
i.e. the first quadrant as shown in the figure given below. By
changing the variables x,y to the new variables r, , by the
polar substitution (transformation) x = rcos, y = rsin, the
transformed region of integration R* is R * : 0 < r < , 0 < 
< /2.
X
(0,0)
Y

34. P
roblems involving Change of variables
i
n double integrals: jacobian
   
   
cos cos , cos sin ,
sin sin , sin cos .
x x
r r r
r r
y y
r r r
r r
   
 
   
 
   
    
   
   
   
   


2 2
Therefore, the Jacobian of transformation is given by
cos sin,
sin cos,
cos sin
x x
rx y r
J
y y rr
r
r r r
 
 

 
 
  
  
 
 
  

35. 10/28/25 LINE INTEGRAL CONTI..........
J r r  
P
roblems involving Change of variables
i
n double integrals: jacobian
 
 
2 2 2 2
2cos sin
Also we have
, , , = .
r r
r
f x r y r e e
 
 
 


Thus the given integral in terms of the polar variables r, is
given by.
 

2 2
= ,
x y
R R
e dxdy f x ydxdy
 
 

36.  
*
= , , ,
R
f x r y r Jdrd  
P
roblems involving Change of variables
i
n double integrals: jacobian
2
2
0 0
r
e rdrd





2
2
0 0
r
e rdrd




 
 
 

2
2
00
1
2
r
e d



 
 
 
 


37. 10/28/25 LINE INTEGRAL CONTI..........

2 2
2
0
00 0
1 1 1
0 .
2 2 2 4
d d
 


  

 
    
 
 
 
P
roblems involving Change of variables
i
n double integrals: jacobian

38. APPLICATION OF DOUBLE
INTEGRATION
1. VOLUME OF A
REGION
SOLUTION: We know that the volume V beneath
the surface z=f(x,y) (>0) and above the region R in
the XY plane is given by ( , )
R
V f x y dxdy
10/28/25 LINE INTEGRAL CONTI..........

39. ( , )
R
V f x y dxdy

 










2
0
3
0
2
32
0
3
0
22
9
3
4
)94(
yy x
dyxy
x
dydxyx
 


2
0
2
0
3
2
1447272
3
27
362736
y
y
ydyy
ANS
10/28/25 LINE INTEGRAL CONTI..........
APPLICATION OF DOUBLE INTEGRATION

40. 2. Center of Gravity
Find the coordinates of the center of gravity of a
mass of density f(x ,y)=1 in a region R. where R is the
triangle with the vertices (0,0), (b,0), (b, h)
SOLUTION: We know that if f(x,y) be the density
of a distribution of mass in the in the XY plane.
Then the total mass M in a region R is given by
(, ) .
R
M f xydxdy
yx,
APPLICATION OF DOUBLE
INTEGRATION

41. Given that f(x,y)=1 in the given region R in xy
plane x varies from 0 to b and y varies from 0 to



0 0 0 0
So 1.
hx hx
y y
b bb b
x y x y
M dydx dx dy
 
   
   
0,0
(b,h)
0,b
hx
b
10/28/25 LINE INTEGRAL CONTI..........
APPLICATION
OF DOUBLE
INTEGRATION

42. 2 2
0 0
2 2 2
b
b
x
hx hx hb hb
M dx
b b b

   
Let the coordinates of the center of gravity of the
mass in R
1 1
( , ) and ( , )
R R
x xf x y dxdy y yf x y dxdy
M M
  
10/28/25 LINE INTEGRAL CONTI..........
APPLICATION OF DOUBLE INTEGRATION

0
0 0 0 0 0
2 2 2
hx hx
b b bb b hx
bx xdydx xdy dx xy dx
hb hb hb
 
 
   
 
 
 
  

43. ANS
10/28/25 LINE INTEGRAL CONTI..........
APPLICATION OF DOUBLE INTEGRATION
2 3 3
2
2 2
0 0 0
2 2 2 2 2
3 3 3
b
b b
hx h x b b
dx x dx
hb b hb b b b
    
    
     
     
 
2
0 0 0 0 0 0
2 2 2
2
hx hx
hx
b b bb b
b
y
y ydydx ydy dx dx
hb hb hb
 
  
   
  
  
 
  
2 2 2 3 3
2
2 2 3 3
0 0 0
2 2
2 2 3 3 3
b
b b
h x h h x h b h
dx x dx
hb b hb b b b
     
          
     
 

44. APPLICATION DOUBLE
INTEGRATION
3. Moment of Inertia
Find the moment of inertia and polar moment of inertia
of a mass of density f(x ,y)=1 in a region R.
where R is the triangle with the vertices (0,0), (b,0), (b, h).
SOLUTION: We know that if f( x, y) be the
density of a distribution of mass in the in
the XY plane. Then moment of inertia in a
region R in x axis and y axis is given by

2 2
0
( , ) , ( , ) and
x y x y
R R
I y f x y dxdy I x f x y dxdy I I I    
0
, ,
x y
I I I

45. Given that f(x,y)=1 in the given region R in
XY- plane y varies from 0 to and
x varies from 0 to b

So
So

3
2 2
0 0 0 0 0 0
.
3
hx hx
hx
y y
b b bb b
b
x
x y y
y
I y dydx ydy dx dx
 
  
 
  
  
  
  
 
  
0,0
b,h
0,b
hx
b
APPLICATION
OF
DOUBLEINTE
GRATION

46. 2 2 2 3
0
0 0 0 0
hx
b b bb hx
b
y
R
h
I x dxdy x dy dx x y dx xdx
b
 
 
   
 
 
 
    
4 4 3
0
4 4 4
b
hx bh bh
b b
  
10/28/25 LINE INTEGRAL CONTI..........
APPLICATION OF DOUBLE INTEGRATION
3 3 3 3 4 4 3 3
3
3 3 3 3
0 0 0
3 3 3 4 12 12
b
b b
hx h h x bh bh
dx xdx
b b b b
 
    
 
 
 

47.  
3 3 3 3
2 2 2
3
0 0 0 0
0
3 3
hx
hx
b
b b bb
y hx h x
x y dy dx x y dx dx
b b
 
  
        
 
 
 
  
4 3 4 4 3 4 3 3
3 3
0
4 12 4 12 4 12
b
hx hx hb hb bh hb
b b b b
 
       
 
ANS10/28/25 LINE INTEGRAL CONTI..........
APPLICATION OF DOUBLE INTEGRATION
 
0
2 2
0
Thepolarmomentof inertiaI aboutoriginof themassin is
( , )
x y
R
R
I I I x y f x y dxdy   
 
2 2
(
R
x y dxdy 

48. TEST YOUR KNOWLEDGE
 
2
4
0 0
5
1 0
int
sin
.1:
.2: 1 2
y
x
x y
Evaluatethefollowingdouble egral
y
Q dxdy
y
Q x e dydx





10/28/25 LINE INTEGRAL CONTI..........

49. 2 2
.3: .
tan
1 , 1 .
Q Findthevolumeof the followingregionsinspace
The firstoc tregionboundedbythecoordinate planesand surfaces
y x z x   
2 2 2
.4:
( , ) 1 .
.
Q Findthecenterof theGravityof amassof density
f x y inaregionR whereRistheregion
x y a inthe firstquadrant

 
10/28/25 LINE INTEGRAL CONTI..........