16
MULTIPLE INTEGRALS
MULTIPLE INTEGRALS
16.4
Double Integrals
in Polar Coordinates
In this section, we will learn:
How to express double integrals
in polar coordinates.
2
DOUBLE INTEGRALS IN POLAR
COORDINATES
Suppose that we want to evaluate a double
integral
, where R is one of
f
(
x
,
y
)
d
A

the regions
R shown here.
3
Fig. 16.4.1, p. 1010
DOUBLE INTEGRALS IN POLAR
COORDINATES
In either case, the description of R in terms of
rectangular coordinates is rather complicated but R
is easily described by polar coordinates.
4
Fig. 16.4.1, p. 1010
DOUBLE INTEGRALS IN POLAR COORDINATES
Recall from this figure that the polar
coordinates (r, θ) of a point are related
to the rectangular coordinates (x, y) by
the equations
r2 = x 2 + y 2
x = r cos θ
y = r sin θ
5
Fig. 16.4.2, p. 1010
POLAR RECTANGLE
The regions in the first figure are special
cases of a polar rectangle
R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β}
shown here.
6
Fig. 16.4.3, p. 1010
POLAR RECTANGLE
To compute the double integral

f ( x, y ) dA
R
where R is a polar rectangle, we divide:
 The interval [a, b] into m subintervals [ri–1, ri]
of equal width ∆r = (b – a)/m.
 The interval [α ,β] into n subintervals [θj–1, θi]
of equal width ∆θ = (β – α)/n.
7
POLAR RECTANGLES
Then, the circles r = ri and the rays θ = θi
divide the polar rectangle R into the small
polar rectangles shown here.
8
Fig. 16.4.4, p. 1010
POLAR SUBRECTANGLE
The “center” of the polar subrectangle
Rij = {(r, θ) | ri–1 ≤ r ≤ ri, θj–1 ≤ θ ≤ θi}
has polar coordinates
ri* = ½ (ri–1 + ri)
θj* = ½ (θj–1 + θj)
9
Fig. 16.4.4, p. 1010
POLAR SUBRECTANGLE
We compute the area of Rij using the fact
that the area of a sector of a circle with
radius r and central angle θ is ½r2θ.
10
POLAR SUBRECTANGLE
Subtracting the areas of two such sectors,
each of which has central angle ∆θ = θj – θj–1,
we find that the area of Rij is:
ri  

 Ai 
1
2

1
2
( ri  r )  

1
2
( ri  ri 1 )( ri  ri  1 )  
2
2
1
2
2
i 1
r
2
i 1
 ri  r  
*
11
POLAR RECTANGLES
We have defined the double integral

in terms of ordinary rectangles.
f ( x, y ) dA
R
However, it can be shown that, for
continuous functions f, we always obtain
the same answer using polar rectangles.
12
POLAR RECTANGLES
Equation 1
The rectangular coordinates of the center
of Rij are (ri* cos θj*, ri* sin θj*).
So, a typical Riemann sum is:
m
n

i 1
*
*
*
*
j 1
m

f ( ri cos  j , ri sin  j )  Ai
n

i 1
f ( ri cos  , ri sin  ) ri  r  
*
*
j
*
*
j
*
j 1
13
POLAR RECTANGLES
If we write g(r, θ) = r f(r cos θ, r sin θ),
the Riemann sum in Equation 1 can be
written as:
m
n
  g (r
i
i 1
*
, )  r  
*
j
j 1
 This is a Riemann sum for the double integral

 
b
g ( r ,  ) dr d 
a
14
POLAR RECTANGLES
Thus, we have:
m

f ( x , y ) dA  lim
m ,n  
R
 lim
m ,n  



b
 
a

b
 
n

i 1
j 1
m
n

i 1
f ( ri cos  j , ri sin  j )  Ai
*
*
*
*
g ( ri ,  j )  r  
*
*
j 1
g ( r ,  ) dr d 
f ( r cos  , r sin  ) r dr d 
a
15
CHANGE TO POLAR COORDS.
Formula 2
If f is continuous on a polar rectangle R
given by
0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β
where 0 ≤ β – α ≤ 2π, then

f ( x , y ) dA 

 
b
f ( r cos  , r sin  ) r dr d 
a
R
16
CHANGE TO POLAR COORDS.
Formula 2 says that we convert
from rectangular to polar coordinates
in a double integral by:
 Writing x = r cos θ and y = r sin θ
 Using the appropriate limits of integration
for r and θ
 Replacing dA by dr dθ
17
CHANGE TO POLAR COORDS.
Be careful not to forget
the additional factor r on
the right side of Formula 2.
18
CHANGE TO POLAR COORDS.
A classical method for remembering
the formula is shown here.
 The “infinitesimal” polar rectangle can be thought of
as an ordinary rectangle
with dimensions
r dθ and dr.
 So, it has “area”
dA = r dr dθ.
19
Fig. 16.4.5, p. 1012
CHANGE TO POLAR COORDS.
Evaluate
 (3 x  4 y
2
Example 1
) dA
R
where R is the region in
the upper half-plane bounded by
the circles x2 + y2 = 1 and x2 + y2 = 4.
20
CHANGE TO POLAR COORDS.
Example 1
The region R can be described as:
R = {(x, y) | y ≥ 0, 1 ≤ x2 + y2 ≤ 4}
21
CHANGE TO POLAR COORDS.
Example 1
It is the half-ring shown here.
In polar coordinates,
it is given by:
1 ≤ r ≤ 2, 0 ≤ θ ≤ π
22
Fig. 16.4.1b, p. 1010
CHANGE TO POLAR COORDS.
Example 1
Hence, by Formula 2,
 (3 x  4 y ) dA
2
R


 
2

2
0


1
 
0


0
(3 r cos   4 r sin  ) r dr d 
2
2
(3 r cos   4 r sin  ) dr d 
3
2
2
1
r2
[ r cos   r sin  ] r 1 d 
3
4
2
23
CHANGE TO POLAR COORDS.





Example 1
(7 cos   15 sin  ] d 
2
0

[7 cos  
0
15
2
15
(1  cos 2 )] d 


 7 sin  

sin 2 
2
4
0

15
1 5
2
24
CHANGE TO POLAR COORDS.
Example 2
Find the volume of the solid bounded
by:
 The plane z = 0
 The paraboloid z = 1 – x2 – y2
25
CHANGE TO POLAR COORDS.
Example 2
If we put z = 0 in the equation of
the paraboloid, we get x2 + y2 = 1.
 This means that the plane intersects
the paraboloid in the circle x2 + y2 = 1.
26
CHANGE TO POLAR COORDS.
Example 2
So, the solid lies under the paraboloid
and above the circular disk D given by
x2 + y2 ≤ 1.
27
Fig. 16.4.1a, p. 1010
Fig. 16.4.6, p. 1012
CHANGE TO POLAR COORDS.
Example 2
In polar coordinates, D is given by
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.
 As 1 – x2 – y2 = 1 – r2, the volume is:
V 
 (1  x
2
 y ) dA 
2
2
 
0
1
(1  r ) r dr d 
2
0
D


2
d
0

1
( r  r ) dr
3
0
1
r
r 

 2 

 
4 0
2
 2
2
4
28
CHANGE TO POLAR COORDS.
Example 2
Had we used rectangular coordinates instead,
we would have obtained:
V 
 (1  x
2
 y ) dA
2
D

1
 
1
1 x
2
 1 x
(1  x  y ) dy dx
2
2
2
 This is not easy to evaluate because
it involves finding ∫ (1 – x2)3/2 dx
29
CHANGE TO POLAR COORDS.
What we have done so far can be extended
to the more complicated type of region
shown here.
 It’s similar to
the type II
rectangular regions
considered in
Section 15.3
30
Fig. 16.4.7, p. 1013
CHANGE TO POLAR COORDS.
In fact, by combining Formula 2 in this
section with Formula 5 in Section 16.3,
we obtain the following formula.
31
CHANGE TO POLAR COORDS.
Formula 3
If f is continuous on a polar region
of the form
D = {(r, θ) | α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)}
then

D
f ( x , y ) dA 

 
h2 ( )
h1 (  )
f ( r cos  , r sin ) r dr d 
32
CHANGE TO POLAR COORDS.
In particular, taking f(x, y) = 1, h1(θ) = 0,
and h2(θ) = h(θ) in the formula, we see that
the area of the region D bounded by θ = α,
θ = β, and r = h(θ) is:
A( D ) 

 1 dA   
h ( )
r dr d  
0


D

 This agrees with Formula 3
in Section 10.4


h ( )
r 
 
 2 0
2
1
2
d
[ h ( )] d 
2
33
CHANGE TO POLAR COORDS.
Example 3
Use a double integral to find the area
enclosed by one loop of the four-leaved
rose r = cos 2θ.
34
CHANGE TO POLAR COORDS.
Example 3
From this sketch of the curve, we see that
a loop is given by the region
D = {(r, θ) | –π/4 ≤ θ ≤ π/4, 0 ≤ r ≤ cos 2θ}
35
Fig. 16.4.8, p. 1013
CHANGE TO POLAR COORDS.
Example 3
So, the area is:
A( D ) 
cos 2 
 /4
 dA    

/4
r dr d 
0
D




 /4


1
2
1
2
2 cos 2 
0
[ r ]
/4
 /4



1
4


cos 2 d 
2
/4
 /4
1
4
d
(1  cos 4 ) d 
/4

1
4
 /4
sin 4    / 4 

8
36
CHANGE TO POLAR COORDS.
Example 4
Find the volume of the solid that
lies:
 Under the paraboloid z = x2 + y2
 Above the xy-plane
 Inside the cylinder x2 + y2 = 2x
37
CHANGE TO POLAR COORDS.
Example 4
The solid lies above the disk D whose
boundary circle has equation x2 + y2 = 2x.
 After completing the square, that is: (x – 1)2 + y2 = 1
38
Fig. 16.4.9, p. 1014
Fig. 16.4.10, p. 1014
CHANGE TO POLAR COORDS.
Example 4
In polar coordinates, we have:
x2 + y2 = r2 and x = r cos θ
So, the boundary circle becomes:
r2 = 2r cos θ
or
r = 2 cos θ
39
CHANGE TO POLAR COORDS.
Example 4
Thus, the disk D is given by:
D=
{(r, θ) | –π/2 ≤ θ ≤ π/2 , 0 ≤ r ≤ 2 cos θ}
40
CHANGE TO POLAR COORDS.
Example 4
So, by Formula 3, we have:
V

 ( x
2
 y ) dA
2
D

 /2
 

/2
2 co s 
r r dr d 
2
0
2 co s 
r 

 
 / 2
 4 0
 /2
 4
 /2
 / 2
4
d
cos  d 
4
41
CHANGE TO POLAR COORDS.
 8
 8
 /2
Example 4
cos  d 
4
0
 /2
0
 2
 /2
2
 1  cos 2 

 d
2


[1  2 cos 2 
0
1
2
(1  cos 4 )] d 
 /2
 2[ 32   sin 2  81 sin 4 ] 0
 3     3
 2   
2
 2  2 
42