Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
9.4
Areas and Lengths in Polar Coordinates
In this section, we develop the fomrulas for finding areas and arc lengths in
polar coordinates.
Area in Polar coordinates
Next, we are going to look at areas enclosed by polar curves. Figure 9.4.1
shows a sketch of what the area that we will be finding in this section looks
like.
Figure 9.4.1
To find the area we use the “thin slices” approach. Divide the range of θ
from a to b into lots of small parts of width ∆θ and the area consequently
being divided up into lots of thin pie slices. Any one of these slices, at
angle θ, is approximately a triangle with sides f (θ) and included angle ∆θ
as shown in Figure 9.4.2. The area of such a triangle is 12 [f (θ)]2 sin (∆θ).
Since ∆θ is assumed to be very small, sin (∆θ) ≈ ∆θ. So we approximate
the area of the pie-slice by 21 r2 ∆θ.
Figure 9.4.2
1
Adding up the thin slices to obtain
Area of region =
Pn
1 2
i=1 2 r ∆θ
Taking the limit as n → ∞ and ∆θ → 0 we obtain
Rb
Area of region = a 21 r2 dθ.
Example 9.4.1
Find the area inside the cardioid r = a(1 + cos θ).
Solution.
Let A denote the area inside the cardioid. Using symmetry we have
Z
π
A =2
0
=a2
Z
r2
dθ =
2
Z
π
a2 (1 + cos θ)2 dθ
0
θ
(1 + 2 cos θ + cos2 θ)dθ
0
θ sin 2θ π
3a2 π
2
=a θ + 2 sin θ + +
=
2
4
2
0
Example 9.4.2
Find the area of the region that lies inside the limaçon r = 3 + 2 sin θ and
outside the circle r = 2.
Solution.
The region is shown in Figure 9.4.3.
Figure 9.4.3
2
To find a and b is equivalent for finding the points of intersection of the
two curves. Solving the equation 3 + 2 sin θ = 2 we find a = θ = 7π
6 and
b = θ = 11π
.
Hence,
the
required
area
is
6
Z
A=
11π
6
7π
6
1
[(3 + 2 sin θ)2 − 22 ]dθ ≈ 24.187
2
Finding a and b in the previous example requires finding the intersection
of two polar curves. Due to multiple representations of a point in polar
coordinates, solving r1 = r2 for θ does not necessarily result in obtaining
all the intersection points. Thus, it is recommended that you draw the
graphs of both curves to see exactly the number of points of intersection.
We illustrate this point in the next example.
Example 9.4.3
Find all points of intersection of the curves r = cos 2θ and r = 12 .
Solution.
The two polar curves are shown in Figure 9.4.4.
Figure 9.4.4
The figure shows 8 points of intersection. Solving the equation cos 2θ =
we find
π 5π 7π 11π
2θ = , , ,
.
3 3 3
3
Thus, the values of θ between 0 and 2π that satisfy both equations are
θ=
π 5π 7π 11π
, , ,
6 6 6
6
3
1
2
resulting in the four points
1 π
1 5π
1 7π
1 11π
,
,
,
.
,
,
,
,
2 6
2 6
2 6
2 6
By symmetry, the remaining four points are
1 π
1 2π
1 4π
1 5π
,
,
,
,
,
,
,
2 3
2 3
2 3
2 3
Arc Length
Using the polar equation r = f (θ), the Cartesian coordinates can be expressed as
x = f (θ) cos θ and y = f (θ) sin θ.
Using the product rule of differentiation we find
dx
dθ
Hence,
dx
dθ
2
=
+
dr
dθ
dy
dθ
cos θ − r sin θ and
2
dr
dθ
2
dr
dθ
2
=
=
dy
dθ
=
dr
dθ
sin θ + r cos θ.
[cos2 θ + sin2 θ] + r2 [cos2 θ + sin2 θ]
+ r2 .
Hence, the length of the polar curve r = f (θ) where a ≤ θ ≤ b is
s
s
2
Z b 2 2
Z b
dr
dx
dy
2
r +
L=
+
dθ =
dθ.
dθ
dθ
dθ
a
a
Example 9.4.4
Find the length of the spiral r = θ where 0 ≤ θ ≤ 1.
Solution.
Using the substitution θ = tan x we find
Z
Z 1p
2
L=
θ + 1dθ =
0
π
4
sec3 xdx
0
π
4
1
= (sec x tan x + ln |secx + tan x|
2
0
√
1 √
= ( 2 + ln (1 + 2)]
2
4
Example 9.4.5
Find the length of the cardioid r = 1 + sin θ.
Solution.
We have
Z
2π
L=
p
(1 + sin θ)2 + cos2 θdθ
0
Z
=
2π
√
2 + 2 sin θdθ = 8
0
5