1) The document discusses double integrals and methods for calculating them, including using iterated integrals and Riemann sums. Double integrals can represent volumes under surfaces. 2) Examples are provided to demonstrate calculating double integrals over rectangles and general regions using iterated integrals and partitioning the region. 3) There are two types of general regions: type I defined by ≤≤≤≤ and type II defined by ≤≤≤≤. The document provides methods for calculating double integrals over these region types.

1. GTU

2. CALCULUS
2110014
DEPARTMENT OF
MATHEMATICS

3. DOUBLE INTEGRALS

4. Double integrals over
rectangles
Suppose f(x) is defined on a interval [a,b].
Recall the definition of definite integrals of
functions of a single variable

5. Taking a partition P of [a, b] into subintervals:
bxxxxa nn
=<<<<= −110

letand],[inpointstheChoose 1 ii
xx− 1−
−=∆ iii
xxx
Using the areas of the small rectangles to approximate the
areas of the curve sided echelons

6. i
n
i
i
xxf ∆∑=1
*
)(
}max{ i
xP ∆=
i
n
i
i
b
a
P
xxfdxxf ∆∑∫ =
=→ 1
*
0
)(lim)(
and summing them, we have
(1)
(2)

7. Double integral of a function of two variables defined
on a closed rectangle like the following
},|),{(],[],[ 2
dycbxaRyxdcbaR ≤≤≤≤∈=×=
Taking a partition of the rectangle
dyyyyc
bxxxxa
nn
mm
=<<<<=
=<<<<=
−
−
110
110



9. ∑ ∑= =
∆
m
i
n
j
ijijij
Ayxf1 1
**
),(
),( **
ijij
yxChoosing a point in Rij and form the
double Riemann sum
(3)

10. (4) DEFINITION The double integral of f over the
rectangle R is defined as
∑ ∑∫∫ = =→
∆=
m
i
n
j
ijijijP
AyxfdAyxf 1 1
**
0
),(lim),(
if this limit exists

11. Using Riemann sum can be approximately evaluate a double
integral as in the following example.
EXAMPLE 1 Find an approximate value for the integral
},20,20|),{(where,)3( 2
≤≤≤≤=∫∫ − yxyxRdAyxR by computing
the double Riemann sum with partition pines x=1 and x=3/2
and taking ),( **
ijij
yx to be the center of each rectangle.

12. Solution The partition is shown as above Figure. The area of
each subrectangle is ,
2
1
=∆ ij
A ),( **
ijij
yx is the center Rij,
and f(x,y)=x-3y2
. So the corresponding Riemann sum is
( ) ( ) ( ) ( )
875.11
),(),(),(),(
),(),(),(),(
),(
8
95
2
1
16
123
2
1
16
51
2
1
16
139
2
1
16
67
4
7
2
3
4
5
2
3
4
7
2
1
4
5
2
1
22211211
22
*
22
*
2221
*
21
*
2112
*
12
*
1211
*
11
*
11
2
1
2
1
**
−=−=
−+−−−=
∆+∆+∆+∆=
∆+∆+∆+∆=
∑∑ ∆
++
= =
AfAfAfAf
AyxfAyxfAyxfAyxf
Ayxf
i j
ijijij
∫∫ −≈−R
dAyx 875.11)3(
haveweThus
2

13. ∑ ∑ ∑ ∑= = = =
∆=
m
i
n
j
m
i
n
j
ijijijij
Ayxfv1 1 1 1
**
),(
Interpretation of double integrals as volumes
∑∑ ==
∆ ≈
m
i
n
j
ij ij ij
A y x f V11
* *
) , ((5)

14. (6) THEOREM If and f is continuous on
the rectangle R, then the volume of the solid that
lies above R and under the surface is
0),( ≥yxf
) , (y x f z=
∫∫=
R
dAyxfV ),(

15. EXAMPLE 2
Estimate the volume of the solid that lies above the square
and below the elliptic paraboloid .
Use the partition of R into four squares and choose
to be the upper right corner of .
Sketch the solid and the approximating rectangle boxes.
]2,0[]2,0[ ×=R 22
216 yxz −−=
),( **
ijij
yx ij
R

16. the volume by the Riemann sum, we have
34)1(4)1(10)1(7)1(13
)2,2()1,2()2,1()1,1( 22211211
=+++=
∆+∆+∆+∆≈ AfAfAfAfV
This is the volume of the approximating rectangular
boxes shown as above.
Solution The partition and the graph of the function are
as the above. The area of each square is 1.Approximating

17. ∫∫∫∫∫∫ +=+
RRR
dAyxgdAyxfdAyxgyxf ),(),()],(),([
∫∫∫∫ =
RR
dAyxfcdAyxcf ),(),(
),(),( yxgyxf ≥ ,, Ryx ∈
∫∫∫∫ ≥
RR
dAyxgdAyxf ),(),(
（ 7 ）
（ 8 ）
（ 9 ）If for all
The properties of the double integrals
then

18. EXERCISES 13.1
Page 837
1. 3. 15.16

19. 13.2 Iterated Integrals
to calculate
∫∫R
dAyxf ),( :],[],[ dcbaR ×=
x dyyxfxA
d
c∫= ),()( .y
dxxA
b
a∫ )(
The double integral can be obtained by evaluating two
single integrals.
The steps to calculate , where
Then calculate
with respect toFix

20. dxdyyxfdxxA
b
a
d
c
b
a ∫ ∫∫ = ]),([)(
dxdyyxfdxdyyxf
b
a
d
c
b
a
d
c ∫ ∫∫ ∫ = ]),([),(
dydxyxfdydxyxf
d
c
b
a
d
c
b
a ∫ ∫∫ ∫ = ]),([),(
(1) (called iterated integral)
(2)
(3) Similarly

21. EXAMPLE 1 Evaluate the iterated integrals
(See the blackboard)
dydxyxdxdyyxa ∫ ∫∫ ∫
2
1
3
0
23
0
2
1
2
(b))(

22. (4) Fubini’s Theorem If is continuous on the
rectangle then
More generally, this is true if we assume that
is bounded on , is discontinuous only on
a finite number of smooth curves, and the iterated
integrals exist.
f
},,|),{( dycbxayxR ≤≤≤≤=
dydxyxfdxdyyxfdAyxf
d
c
b
a
b
a
d
c
R
∫ ∫∫ ∫∫∫ == ),(),(),(
f
R f

23. Interpret the double integral as the
volume V of the solid
∫∫R
dAyxf ),(

24. where
is the area of a cross-section of S in the
plane through x perpendicular to the x-axis.
Similarly
dyyxfxA
d
c∫= ),()(
)(xA
dxxAV
b
a∫= )(

25. EXAMPLE 2
Evaluate the double integral dAyxR∫∫ − )3( 2
where }.21,20|),{( ≤≤≤≤= yxyxR
(See the blackboard)

26. EXAMPLE 3
∫∫R
dAxyy )sin( ],0[]2,1[ π×=REvaluate ,where

27. Solution 1 If we first integrate with respect to x,
we get
∫∫ =R
dAxyy )sin( dxdyxyy∫ ∫
π
0
2
1
)sin(
∫ −= =
=
π
0
2
1
)]cos([ dyxy x
x
∫ +−=
π
0
)cos2cos( dyyy
] 0sin2sin 02
1
=+−= π
yy

28. Solution 2 If we first integrate with respect to y, then
∫∫ =R
dAxyy )sin( dydxxyy∫ ∫
2
1 0
)sin(
π
( )dxxyydx∫ ∫ −=
2
1 0
))(cos(1π
dxdyxyxyy xx∫ ∫+−=
2
1 00
])cos(|)cos([ 11 ππ
dxxyx xx∫ +−=
2
1 02 ]|)sin()cos([ 11 π
ππ
dxx
x
x
x
∫ −=
2
1 2 ][ )cos()sin( πππ
∫−∫=
2
1
2
1 2
)cos()sin(
dxdx x
x
x
x πππ
∫−∫=
2
1
2
1 2
)sin()sin(
x
xd
x
x
dx
ππ
∫−∫= −
 2
1 2
2
1 2
)sin(2
1
)sin()sin(
dxdx x
x
x
x
x
x πππ 0
2
1
)sin(
=−= 

x
xπ

29. EXAMPLE 4 Find the volume of the solid S that is
bounded by the elliptic paraboloid , the
plane and , and three coordinate planes.2=x2=y
162 22
=++ zyx

30. We first observe that S is the solid that lies
under the surface 22
216 yxz −−= and the above the
Square ].2,0[]2,0[ ×=R Therefore,
∫∫ −−=
R
dAyxV )216( 22
∫ ∫ −−=
2
0
2
0
22
)216( dxdyyx
[ ]∫ −−=
=
=
2
0
23
2
03
1
216 dyxyxx
x
x
( )∫ −=
2
0
2
43
88
dyy
[ ] 48
2
03
4
3
88 3
=−= yy
Solution

31. If on , then)()(),( yhxgyxf = ],[],[ dcbaR ×=
dyyhdxxgdAyhxg
d
c
b
a
R
∫∫∫∫ = )()()()(

32. ]
2
,0[]
2
,0[ ππ ×=R
111
][sin]cos[
cossincossin
22
2 2
00
0 0
=⋅=
−=
=∫∫ ∫ ∫
ππ
π π
yx
ydyxdxydAx
R
EXAMPLE 5 , thenIf

33. EXERCISES 13.2
Page 842
1(2), 6, 10, 16, 17,

34. 13.3 Double integrals over general regions



=
0
),(
),(
yxf
yxF
Dyx ∈),(
DRyx innotbutinis),(
R
To integrate over general regions like
which is bounded, being enclosed in a rectangular region
R .Then we define a new function F with domain R by
(1)
if
if

35. If F is integrable over R , then we say f is integrable
over D and we define the double integral of f over D by
where is given by Equation 1.
dAyxFdAyxf
RD
∫∫∫∫ = ),(),(
F
(2)

36. Geometric interpretation
0),( ≥yxfWhen
The volume under f and above D equals to that under
F and above R.
R

37. Type I regions
)}()(,|),{( 21
xgyxgbxayxD ≤≤≤≤=
(3) If f is continuous on a type I region D such that
)}()(,|),{( 21
xgyxgbxayxD ≤≤≤≤=
dydxyxfdAyxf
b
a
xg
xg
D
∫ ∫∫∫ =
)(
)(
2
1
),(),(
then

38. )}()(,|),{( 21
yhxyhdycyxD ≤≤≤≤=
(5)
dxdyyxfdAyxf
d
c
yh
yh
D
∫ ∫∫∫ =
)(
)(
2
1
),(),(
Type II regions
(4)
where D is a type II region given by Equation 4

39. ( )dxdyyx
x
x∫ ∫ += −
+1
1
1
2
2
2 )2(
( ) dxyxy
xy
xy
2
2
1
2
1
1
2
+=
=−∫ +=
( )dxxxxxx∫ −−+++= −
1
1
43222
42)1()1(
( )dxxxxx∫ +++−−= −
1
1
234
123
1
123
2
45
3
345
−






+++−−= x
xxxx
15
32
=
• •
it is Type I region!

40. 22
yxz +=
xy 2= .2
xy =
xy
Example 2 Find the volume of the solid that lies under the
paraboloid and above the region D in the
-plane bounded by the line and the parabola
•
•
•
•
}2,20|),{(
IType
2
xyxxyxD ≤≤≤≤= }
2
,40|),{(
IIType
yx
y
yyxD ≤≤≤≤=

42. Solution 1
∫∫ +D
dAyx )( 22
∫ ∫ +=
2
0
2 22
2 ))((
x
x
dxdyyx
∫
=
=






+=
2
0
2
32
23
1
dxyyx
xy
xy
∫ −−+=
2
0
6433
)2( 3
1
3
8
dxxxxx
∫ +−−=
2
0
346
)( 3
14
3
1
dxxxx ]2
06
7
5
1
21
1 457
xxx +−−= 35
216
=
Solution 2 ∫∫ +D
dAyx )( 22
∫ ∫ +=
4
0 2/
22
)(
y
y
dxdyyx
[ ]∫ +=
=
=
4
0
2/
23
3
1
dyxyx
yx
yx
( )∫ −−+=
4
0
332/52/3
2
1
24
1
3
1
dyyyyy
]4
096
13
7
2
15
2 42/72/5
yyy −+= 35
216
=
Type I
Type II

43. Example 3 Evaluate , where D is the region bounded
by the line and the parabola
∫∫D
xydA
1−= xy .622
+= xy
D as a type I D as a type II
•
•
•

44. { }13,42|),( 2
2
+≤≤−≤≤−= yxyyxD
y
Solution We prefer to express D as a type II
∫∫D
xydA ∫ ∫= −
+
−
4
2
1
3
2
2
y
y
xydxdy ∫ 



= −
+=
−=
4
2
1
3
2
2
2
2
dyy
x
yx
y
x
( )∫ −−+= −
4
2
22
)3()1( 2
1
2
1
dyyyy
( )∫ −++= −
−
4
2
245
8244
5
2
1
dyyyyy
4
23
2
24
1
2
1 2346
4
−




−++−= yyyy
36=

45. .0and,0,2,22 ====++ zxyxzyx






−≤≤≤≤=
2
1
2
,10|),(
x
y
x
xyxD
Example 4 Find the volume of the tetrahedron bounded by
the planes
D
D

46. Solution
dAyxV
D
∫∫ −−= )22( ∫ ∫ −−=
−1
0
2
1
2
)22(
x
x dydxyx
[ ]∫ −−=
−=
=
1
0
2
1
2
2
2 dxyxyy
x
y
x
y
( ) ( )( )∫ +−−−−−−= +
1
0
222
4222
112 dxxxx xxxx
( )∫ +−=
1
0
2
12 dxxx
1
0
2
3
3 



− −= xxx
3
1
=
Here is wrong in the book!

47. dydxyx∫ ∫
1
0
1 2
)sin(
{ }1,10|),( ≤≤≤≤= yxxyxD { }yxyyx ≤≤≤≤= 0,10|),(
D as a type I D as a type II
Example 5 Evaluate the iterated integral

48. Solution If we try to evaluate the integral as it stands, we
.)sin( 2
dyy∫
impossible to do so in finite terms since dyy∫ )sin( 2
elementary function.(See the end of Section 7.6.) So we must
are faced with the task of first evaluating But it is
is not an
change the order of integration. This is accomplished by first
expressing the given iterated integral as a double integral.
∫∫=∫ ∫ D
x
dAydydxy )sin()sin( 21
0
1 2
Where { }1,10|),( ≤≤≤≤= yxxyxD
Using the alternative description of D,
we have { }yxyyxD ≤≤≤≤= 0,10|),(

49. This enables us to evaluate the integral in the reverse
order:
∫∫=∫ ∫ D
x
dAydydxy )sin()sin( 21
0
1 2
dxdyy
y
∫ ∫=
1
0 0
2
)sin(
[ ]∫=
=
=
1
0 0
2
)sin( dyyx
yx
x
∫=
1
0
2
)sin( dyyy
]1
02
1
)cos( 2
y−=
)1cos1(2
1
−=

50. Properties of double integrals
∫∫ ∫∫∫∫ +=+
D DD
dAyxgdAyxfdAyxgyxf ),(),()],(),([
∫∫∫∫ =
DD
dAyxfcdAyxcf ),(),(
∫∫∫∫ ≥
DD
dAyxgdAyxf ),(),( .in),(allfor),(),( Dyxyxgyxf ≥
∫∫ ∫∫∫∫ +=
1 2
),(),(),([
D DD
dAyxfdAyxfdAyxf
2121
andwhere DDDDD ∪=
(6)
(7)
(8) if
(9)
if do not overlap except
perhaps on their boundaries like the following:

51. (10)
(11)
∫∫ =D
DAdA )(1
then,),(allfor),( DyxMyxfm ∈≤≤
)(),()( DMAdAyxfDmA
D
∫∫ ≤≤
, the area of region D.
If

52. Example 6
2.radiusandregioncenter thedisk withtheiswhere
,integraltheestimateto11PropertyseU
cossin
D
dAeD
yx
∫∫
Solution Since ,1cos1and,1sin1 ≤≤−≤≤− xx we have
,1cossin1 ≤≤− xx and therefore
eeee xx
=≤≤− 11 cossin
thus, using m=e-1
=1/e, M=e, and A(D)=π(2)2
in Property 11
we obtain
πedAeD
yx
4
e
4 cossin
≤∫∫≤
π

53. Exercises 13.3
Page 850: 7, 9, 11, 33, 35

54. 13.4 DBOUBLE INTEGRALS IN POLAR
COORDINATE
Suppose we want to evaluate a double integral
where is one of regions shown in the following.
,),(∫∫R
dAyxf
R
}20,10|),{( πθθ ≤≤≤≤= rrR(a) }0,21|),{( πθθ ≤≤≤≤= rrR(b)

55. Recall from Section 9.4 that the polar coordinates
of a point related to the rectangular coordinates ),( θr
θθ sinycos222
rrxyxr ==+=
},|),{( βθαθ ≤≤≤≤= brarR
Do the following partition (called polar partition)
The regions in
the above
figure are
special cases
of a polar rectangle
by the equations:

57. },|),{( 11 jjiiij
rrrrR θθθθ ≤≤≤≤= −−
)(
2
1
)(
2
1
1
*
1
*
jjjiii rrr θθθ +=+= −−
jii
jiiii
jii
jijiij
rr
rrrr
rr
rrA
θ
θ
θ
θθ
∆∆=
∆−+=
∆−=
∆−∆=∆
−−
−
−
*
11
2
1
2
2
1
2
))((
)(
2
1
2
1
2
1
2
1
1−
−=∆ iii
rrr
The center of this subrectangle
is
and the area is
where

58. jii
m
i
n
j
jiji
ij
m
i
n
j
jiji
rrrrf
Arrf
θθθ
θθ
∆∆=
∆
∑ ∑
∑ ∑
= =
= =
*
1 1
****
1 1
****
)sin,cos(
)sin,cos(
The typical Riemann sum is
(1)

59. If we write , then the
above Riemann sum can be written as
which is the Riemann sum of the double integral
Therefore we have
)sin,cos(),( θθθ rrrfrg =
ji
m
i
n
j
ji
rrg θθ ∆∆∑ ∑= =1 1
**
),(
θθβ
α
ddrrg
b
a∫ ∫ ),(

60. θθθ
θθ
θθ
θθ
β
α
β
α
rdrdrrf
drdrg
rrg
ArrfdAyxf
b
a
b
a
ji
m
i
n
j
jiP
ij
m
i
n
j
jijiR P
∫ ∫
∫ ∫
∑ ∑
∑ ∑∫∫
=
=
∆∆=
∆=
= =→
= =→
)sin,cos(
),(
),(lim
)sin,cos(lim),(
1 1
**
0
1 1
****
0

61. (2) Change to polar coordinates in a double
integral If is continuous on a polar
rectangle given by
where then
f
R ,,0 βθα ≤≤≤≤≤ bra
,20 παβ ≤−≤
θθθ
β
α
rdrdrrfdAyxf
b
aR ∫ ∫=∫∫ )sin,cos(),(

62. Caution: Do not forget the factor r in (2)!

63. Example 1 Evaluate , where is the region
in the upper half-plane bounded by the circles
dAyxR∫∫ + )43( 2
R
.4and,1 2222
=+=+ yxyx
Solution The region R can be described as
}41,0|),{( 22
≤+≤≥= yxyyxR }0,21|),{( πθθ ≤≤≤≤= rr
dAyxR∫∫ + )43( 2
θθθ
π
rdrdrr∫ ∫ += 0
2
1
22
)sin4cos3(
θθθ
π
drdrr∫ ∫ += 0
2
1
232
)sin4cos3(
[ ] θθθ
π
drr
r
r
2
10
243
sincos
=
=∫ += [ ] θθθ
π
d∫ += 0
2
sin15cos7
θθθ
π
d∫ 



−+= 0
)2cos1(
2
15
cos7
2
15
2sin
4
15
2
15
sin7
0
π
θ
θ
θ
π
=

−+=

64. Example 2 Find the volume of the solid bounded
by the xy-plane and the paraboloid
22
1 yxz −−=
}1|),{( 22
≤+= yxyxD
}20,10|),{( πθθ ≤≤≤≤= rr
dAyxV
D
∫∫ −−= )1( 22
θ
π
rdrdr∫ ∫ −=
2
0
1
0
2
)1(
drrrd∫ ∫ −=
π
θ
2
0
1
0
3
)(
242
2
1
0
42
π
π =



−=
rr

65. What we have done so far can be extended to the
complicated type of region shown in the following.
(3) If f is continuous on a polar
region of the form
)}()(,|),{( 21
θθβθαθ hrhrD ≤≤≤≤=
θθθ
β
α
θ
θ
rdrdrrf
dAyxf
h
h
D
∫ ∫
∫∫ =
)(2
)(1
)sin,cos(
),(
then

66. Example 3 Use a double integral to find the area enclosed
by one loop of the four-leaved rose θ2cos=r
}2cos0,
44
|),{( θ
π
θ
π
θ ≤≤≤≤−= rrD
∫ ∫=∫∫= −
4/
4/
2cos
0
)(
π
π
θ
θrdrddADA
D
∫ 



= −
4/
4/
2cos
0
2
2
1π
π
θ
θdr
∫= −
4/
4/
2
2cos
2
1 π
π
θθd
( )∫ −= −
4/
4/
4cos1
4
1 π
π
θθ d
8
4sin
4
1
4
1
4/
4/
π
θθ
π
π
=



+=
−

67. Example 4 Find the volume of the solid that lies under the
paraboloid , above the plane, and inside
the cylinder
22
yxz += −xy
.222
xyx =+
Solution The solid lies above the disk, whose boundary circle
}cos20,
22
|),{( θ
π
θ
π
θ ≤≤≤≤−= rrD

68. =∫∫ +=
D
dAyxV )( 22
∫ ∫−
2/
2/
cos2
0
2π
π
θ
θrdrdr
∫ 



= −
2/
2/
cos2
0
4
4
π
π
θ
θd
r
∫= −
2/
2/
4
cos4
π
π
θθd
∫=
2/
0
4
cos8
π
θθd ∫ 




 −
=
2/
0
2
2
2cos1
8
π
θ
θ
d
( )∫ +++=
2/
0
]4cos12cos21[2 2
1π
θθθ d
2
3
22
3
24sin2sin
2
3
2
2/
0
8
1 ππ
θθθ
π
=











=



++=

69. Exercises 13.4
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