±
9.5
AREA IN POLAR COORDINATES
Here we develop a technique for calculating the area of a region the boundary of which
is given in polar coordinates.
As a start, we suppose that
α
and
β
are two real numbers with
α<β
≤
α
+
2
π
.
We take
ρ
as a function that is continuous on [
α
,
β
] and keeps a constant sign on that
interval. We want the area of the polar region
±
generated by the curve
r
=
ρ
(
θ
),
α
≤
θ
≤
β
.
Such a region is portrayed in Figure 9.5.1.
In the ±gure
ρ
(
θ
) remains nonnegative. If
ρ
(
θ
) were negative, the region
±
would
appear on the opposite side of the pole. In either case, the area of
±
is given by the
formula
=
θβ
polar axis
=
θα
r
=
ρθ
(
)
Γ
O
Figure 9.5.1
(9.5.1)
A
=
³
β
α
1
2
[
ρ
(
θ
)]
2
d
θ
.
PROOF
We consider the case where
ρ
(
θ
)
≥
0. We take
P
={
θ
0
,
θ
1
,
...
,
θ
n
}
as a
partition of [
α
,
β
] and direct our attention to what happens from
θ
i
−
1
to
θ
i
. We set
r
i
=
min value of
ρ
on [
θ
i
−
1
,
θ
i
]
and
R
i
=
max value of
ρ
on [
θ
i
−
1
,
θ
i
].
The part of
±
that lies between
θ
i
−
1
and
θ
i
contains a circular sector of radius
r
i
and
central angle
²θ
i
=
θ
i
−
θ
i
−
1
and is contained in a circular sector of radius
R
i
with
the same central angle
²θ
i
. (See Figure 9.5.2.) Its area
A
i
must therefore satisfy the
inequality
polar axis
R
i
r
i
i
–1
θ
i
=
αθ
o
=
βθ
n
O
Figure 9.5.2
1
2
r
2
i
²θ
i
≤
A
i
≤
1
2
R
2
i
²θ
i
. †
By summing these inequalities from
i
=
1to
i
=
n
, we can see that the total area
A
must satisfy the inequality
(1)
L
f
(
P
)
≤
A
≤
U
f
(
P
)
† The area of a circular sector of radius
r
and central angle
α
is
1
2
r
2
α
.
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View Full Document9.5
AREA IN POLAR COORDINATES
±
549
where
f
(
θ
)
=
1
2
[
ρ
(
θ
)]
2
.
Since
f
is continuous and (1) holds for every partition
P
of [
α
,
β
], we can conclude
that
A
=
±
β
α
f
(
θ
)
d
θ
=
±
β
α
1
2
[
ρ
(
θ
)]
2
d
θ
.
±
Example 1
Calculate the area enclosed by the cardioid
r
=
1
−
cos
θ
(Figure 9.5.3)
SOLUTION
The entire curve is traced out as
θ
increases from 0 to 2
π
, and 1
−
cos
θ
≥
0
on [0, 2
π
]. Therefore
polar axis
[2,
]
π
r
= 1 – cos
θ
1,
2
1,
2
3
[]
Figure 9.5.3
A
=
±
2
π
0
1
2
(1
−
cos
θ
)
2
d
θ
=
1
2
±
2
π
0
(1
−
2 cos
θ
+
cos
2
θ
)
d
θ
=
1
2
±
2
π
0
(
3
2
−
2 cos
θ
+
1
2
cos 2
θ
)
d
θ
.
↑
halfangle formula: cos
2
θ
=
1
2
+
1
2
cos 2
θ
Since
±
2
π
0
cos
θ
d
θ
=
sin
θ
²
²
²
2
π
0
=
0
and
±
2
π
0
cos 2
θ
d
θ
=
1
2
sin 2
θ
²
²
²
2
π
0
=
0,
we have
A
=
1
2
±
2
π
0
3
2
d
θ
=
3
4
±
2
π
0
d
θ
=
3
2
π
.
±
A slightly more complicated type of region is pictured in Figure 9.5.4. We approach
the problem of calculating the area of the region
±
in the same way that we calculated
the area between two curves in Section 5.5; that is, we calculate the area out to
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 Spring '10
 SMITH
 Equations, Conic Sections, Parametric Equations, Polar Coordinates, Cos, Polar coordinate system, Conic section, β α

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