14.31. Evaluate
RR
D
(
x
8
sin
x
+
y
9
+
1
)
dA
, where
D
is the disk
x
2
+
y
2
≤
4.
115
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LECTURE 15
Double Integrals in Polar Coordinates
15.1. The formula
T
heorem
15.1.
Let f
(
x
,
y
)
be piecewise continuous in a region D in the (Cartesian) xyplane
and let R be the same region in polar coordinates, that is,
R
=
(
r
,
θ
) : (
r
cos
θ
,
r
sin
θ
)
∈
D
.
Then
ZZ
D
f
(
x
,
y
)
dA
=
ZZ
R
f
(
r
cos
θ
,
r
sin
θ
)
r dA
.
(15.1)
The formal proof of (15.1) lies beyond the scope of these notes, but we should give at least
some motivation for this formula.
Let
a
,
b
,
α
,
β
be nonnegative numbers such that
a
<
b
and
0
≤
α
<
β
≤
2
π
. Consider the case when
D
is the region bounded by the circles
x
2
+
y
2
=
a
2
and
x
2
+
y
2
=
b
2
and by the halflines through the origin forming angles
α
and
β
with the positive
x
axis
(see Figure 15.1). Note that
D
contains the points (
x
,
y
) in the plane whose polar coordinates (
r
,
θ
)
satisfy the inequalities
a
≤
r
≤
b
and
α
≤
θ
≤
β
. We recall that the area of
D
is given by the
formula
area(
D
)
=
1
2
(
β

α
)
(
b
2

a
2
)
=
(
β

α
)(
b

a
)
b
+
a
2
.
(15.2)
For each integer
n
≥
1, let
r
0
=
a
,
r
1
=
a
+
h
,
r
2
=
a
+
2
h
,
. . . ,
r
n
=
a
+
nh
=
b
,
h
=
(
b

a
)
/
n
,
be the points that divide the interval [
a
,
b
] into
n
subintervals of equal lengths; and let
θ
0
=
α
,
θ
1
=
α
+
δ
,
θ
2
=
α
+
2
δ
,
. . . ,
θ
n
=
α
+
n
δ
=
β
,
δ
=
(
β

α
)
/
n
,
θ
=
α
θ
=
β
θ
=
θ
j
r
=
a
r
=
r
i
r
=
b
(
x
*
i j
,
y
*
i j
)
α
θ
j
β
a
r
i
b
r
θ
(
r
*
i
,
θ
*
j
)
F
igure
15.1.
The regions
D
and
R
, with partitions and sample points
117
be the points that divide the interval [
α
,
β
] into
n
subintervals of equal lengths. We now subdivide
the region
D
into
n
2
subregions of similar shapes bounded by the circles of radii
a
,
r
1
,
r
2
, . . . ,
r
n

1
,
b
centered at the origin and the halflines through the origin forming angles
α
,
θ
1
,
θ
2
, . . . ,
θ
n

1
,
β
with
the positive
x
axis (see Figure 15.1). Let
D
i j
denote the subregion bounded by the circles of radii
r
i

1
and
r
i
and by the halflines at angles
θ
j

1
and
θ
j
with the positive
x
axis: in polar coordinates,
D
i j
is given by the inequalities
r
i

1
≤
r
≤
r
i
,
θ
j

1
≤
θ
≤
θ
j
.
We now pick a sample point (
x
*
i j
,
y
*
i j
) in each subregion
D
i j
: we choose (
x
*
i j
,
y
*
i j
) to be the point in
the plane whose polar coordinates are
r
*
i
=
(
r
i

1
+
r
i
)
/
2 and
θ
*
j
=
(
θ
j

1
+
θ
j
)
/
2. Using the subregions
D
i j
and the sample points (
x
*
i j
,
y
*
i j
), we form the sum
n
X
i
=
1
n
X
j
=
1
f
(
x
*
i j
,
y
*
i j
)

D
i j

,

D
i j

=
area(
D
i j
)
.
Note that this sum bears a strong resemblance to a Riemann sum for
f
(
x
,
y
). Indeed, it can be
proved that if
f
(
x
,
y
) is continuous, then
ZZ
D
f
(
x
,
y
)
dA
=
lim
n
→∞
n
X
i
=
1
n
X
j
=
1
f
(
x
*
i j
,
y
*
i j
)

D
i j

.
(15.3)
We now express the double sum in (15.3) in terms of the
r
i
’s and the
θ
j
’s. By (15.2) with
a
=
r
i

1
,
b
=
r
i
,
α
=
θ
j

1
and
β
=
θ
j
, we have

D
i j

=
(
θ
j

θ
j

1
)(
r
i

r
i

1
)
r
i
+
r
i

1
2
=
δ
hr
*
i
.
Furthermore, since the polar coordinates of (
x
*
i j
,
y
*
i j
) are (
r
*
i
,
θ
*
j
), we have
x
*
i j
=
r
*
i
cos
θ
*
j
,
y
*
i j
=
r
*
i
sin
θ
*
j
,
whence
f
(
x
*
i j
,
y
*
i j
)
=
f
(
r
*
i
cos
θ
*
j
,
r
*
i
sin
θ
*
j
). Therefore,
n
X
i
=
1
n
X
j
=
1
f
(
x
*
i j
,
y
*
i j
)

D
i j

=
n
X
i
=
1
n
X
j
=
1
f
(
r
*
i
cos
θ
*
j
,
r
*
i
sin
θ
*
j
)
r
*
i
δ
h
.