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Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
Areas and Lengths in Polar Coordinates
In this section we develop the formula for the area of a region whose boundary is given by a polar equation.
We need to use the formula for the area of a sector of a circle
1
A = r2θ
2
(1)
where r is the radius and θ is the radian measure of the central angle. Formula 1 follows from the fact
that the area of a sector is proportional to its central angle:
A=
θ
1
· πr 2 = r 2 θ
2π
2
Let R be the region bounded by the polar curve r = f (θ) and by the rays θ = a and θ = b, where f is a
positive continuous function and where 0 < b − a ≤ 2π.
We divide the interval [a, b] into subintervals with endpoints θ0 , θ1 , θ2 , . . . , θn and equal width ∆θ. The
rays θ = θi then divide R into n smaller regions with central angle ∆θ = θi − θi−1 . If we choose θi∗ in the
ith subinterval [θi−1 , θi ], then the area ∆Ai of the ith region is approximated by the area of the sector of
a circle with central angle ∆θ and radius f (θi∗ ). Thus from Formula 1 we have
1
∆Ai ≈ [f (θi∗ )]2 ∆θ
2
and so an approximation to the total area A of R is A ≈
n
X
1
i=1
2
(2)
[f (θi∗ )]2 ∆θ. One can see that the approxima-
1
tion in (2) improves as n → ∞. But the sums in (2) are Riemann sums for the function g(θ) = [f (θ)]2 ,
2
so
Z
n
b
X1
1
lim
[f (θi∗ )]2 ∆θ =
[f (θ)]2 dθ
n→∞
2
a 2
i=1
It therefore appears plausible (and can in fact be proved) that the formula for the area A of the polar
region R is
Z b
1
(3)
A=
[f (θ)]2 dθ
2
a
This formula is often written as
A=
Z
b
a
1
1 2
r dθ
2
(4)
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the area of each of the following regions:
(a)
(b)
(c)
(d)
Solution:
(a) We have
A=
Z
3π/4
π/4
1 2
1
· 1 dθ =
2
2
Z
3π/4
π/4
1
dθ =
2
3π π
−
4
4
1
=
2
2π
4
=
π
4
(b) We have
A=
Z
2π+π/4
3π/4
1
1 2
· 1 dθ =
2
2
Z
2π+π/4
3π/4
1
dθ =
2
π 3π
2π + −
4
4
1
=
2
2π
π
3π
2π −
=π− =
4
4
4
(c) We have
A=
Z
7π/4
5π/4
1
1 2
· 1 dθ =
2
2
Z
7π/4
5π/4
1
dθ =
2
7π 5π
−
4
4
1
=
2
2π
4
=
π
4
(d) We have
A=
Z
2π+π/4
7π/4
1 2
1
· 1 dθ =
2
2
or
A=
Z
π/4
−π/4
Z
2π+π/4
7π/4
1
1 2
· 1 dθ =
2
2
1
dθ =
2
Z
π 7π
2π + −
4
4
π/4
dθ =
−π/4
1
=
2
6π
3π
π
2π −
=π−
=
4
4
4
1 π π 1 π π π
=
=
− −
+
2 4
4
2 4
4
4
EXAMPLE: Find the area of the inner loop of r = 2 + 4 cos θ.
2
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the area of the inner loop of r = 2 + 4 cos θ.
Solution: We first find a and b:
2 + 4 cos θ = 0
Therefore the area is
Z
A =
4π/3
2π/3
=
Z
4π/3
Z
4π/3
=⇒
1
(2 + 4 cos θ)2 dθ =
2
cos θ = −
Z
2
=⇒
θ=
2π 4π
,
3
3
1
(4 + 16 cos θ + 16 cos2 θ)dθ
2
2π/3
(2 + 8 cos θ + 8 cos θ)dθ =
2π/3
=
4π/3
1
2
Z
4π/3
2π/3
(2 + 8 cos θ + 4(1 + cos 2θ)dθ =
2π/3
1 + cos 2θ
dθ
2 + 8 cos θ + 8 ·
2
Z
4π/3
(6 + 8 cos θ + 4 cos 2θ)dθ
2π/3
h
i4π/3
√
= 6θ + 8 sin θ + 2 sin 2θ
= 4π − 6 3 ≈ 2.174
2π/3
EXAMPLE: Find the area enclosed by one loop of the four-leaved rose r = cos 2θ.
3
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the area enclosed by one loop of the four-leaved rose r = cos 2θ.
Solution: Notice that the region enclosed by the right loop is swept out by a ray that rotates from θ = −π/4
to θ = π/4. Therefore, Formula 4 gives
A =
Z
π/4
Z
π/4
−π/4
=
0
1 2
1
r dθ =
2
2
Z
π/4
2
cos 2θdθ =
−π/4
Z
π/4
cos2 2θdθ
0
π/4
π
1
1
1
=
θ + sin 4θ
(1 + cos 4θ)dθ =
2
2
4
8
0
Let R be the region bounded by curves with polar equations r = f (θ), r = g(θ), θ = a, and θ = b, where
f (θ) ≥ g(θ) ≥ 0 and 0 < b − a ≤ 2π. Then the area A of R is
A=
Z
b
a
1
[f (θ)]2 − [g(θ)]2 dθ
2
EXAMPLE: Find the area that lies inside r = 3 + 2 sin θ and outside r = 2.
4
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the area that lies inside r = 3 + 2 sin θ and outside r = 2.
Solution: We first find a and b:
3 + 2 sin θ = 2
=⇒
1
sin θ = −
2
=⇒
7π
π
θ=
, −
6
6
11π
6
Therefore the area is
Z 7π/6
Z 7π/6
1
1
2
2
A =
(3 + 2 sin θ) − 2 dθ =
(5 + 12 sin θ + 4 sin2 θ)dθ
−π/6 2
−π/6 2
=
=
Z
Z
7π/6
−π/6
7π/6
−π/6
1
2
Z 7π/6
1 − cos 2θ
1
5 + 12 sin θ + 4 ·
dθ =
(5 + 12 sin θ + 2(1 − cos 2θ))dθ
2
−π/6 2
i7π/6
1h
1
(7 + 12 sin θ − 2 cos 2θ)dθ = 7θ − 12 cos θ − sin 2θ
−π/6
2
2
√
11 3 14π
=
+
≈ 24.187
2
3
EXAMPLE: Find the area of the region outside r = 3 + 2 sin θ and inside r = 2.
5
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the area of the region outside r = 3 + 2 sin θ and inside r = 2.
Solution: We have
Z 11π/6
1 2
2 − (3 + 2 sin θ)2 dθ
A =
2
7π/6
=
Z
11π/6
Z
11π/6
7π/6
=
7π/6
1
(−5 − 12 sin θ − 4 sin2 θ)dθ
2
i11π/6 11√3 7π
1h
1
=
− 7θ + 12 cos θ + sin 2θ
(−7 − 12 sin θ + 2 cos 2θ)dθ =
−
≈ 2.196
7π/6
2
2
2
3
1
EXAMPLE: Find all points of intersection of the curves r = cos 2θ and r = .
2
1
1
Solution: If we solve the equations r = cos 2θ and r = , we get cos 2θ = and, therefore,
2
2
2θ = π/3, 5π/3, 7π/3, 11π/3
Thus the values of θ between 0 and 2π that satisfy both equations are
θ = π/6, 5π/6, 7π/6, 11π/6
We have found four points of intersection:
1
1
1
1
, π/6 ,
, 5π/6 ,
, 7π/6 , and
, 11π/6
2
2
2
2
However, you can see from the above figure that the curves have four other points of intersection — namely,
1
1
1
1
, π/3 ,
, 2π/3 ,
, 4π/3 , and
, 5π/3
2
2
2
2
1
These can be found using symmetry or by noticing that another equation of the circle is r = − and then
2
1
solving the equations r = cos 2θ and r = − .
2
6
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
Arc Length
To find the length of a polar curve r = f (θ), a ≤ θ ≤ b, we regard θ as a parameter and write the
parametric equations of the curve as
x = r cos θ = f (θ) cos θ
y = r sin θ = f (θ) sin θ
Using the Product Rule and differentiating with respect to θ, we obtain
dx
dr
=
cos θ − r sin θ
dθ
dθ
dy
dr
=
sin θ + r cos θ
dθ
dθ
So, using cos2 θ + sin2 θ = 1, we have
2 2 2
dx
dy
dr
dr
+
=
cos2 θ − 2r cos θ sin θ + r 2 sin2 θ
dθ
dθ
dθ
dθ
+
dr
dθ
2
dr
sin θ + 2r sin θ cos θ + r 2 cos2 θ =
dθ
2
dr
dθ
2
+ r2
Assuming that f ′ is continuous, we can use one of the formulas from Section 9.2 to write the arc length as
s
Z b 2 2
dy
dx
L=
+
dθ
dθ
dθ
a
Therefore, the length of a curve with polar equation r = f (θ), a ≤ θ ≤ b, is
L=
Z
b
a
s
r2
+
dr
dθ
EXAMPLE: Find the length of the curve r = θ, 0 ≤ θ ≤ 1.
7
2
dθ
(5)
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the length of the curve r = θ, 0 ≤ θ ≤ 1.
Solution: We have
L =
=
Z
Z
1
√
0

=⇒
√
θ2 + 1 =
√
tan2 x + 1 =

θ2 + 1dθ = 
 dθ = d tan x
√
sec2 x = | sec x| = sec x
dθ = sec2 xdx
π/4
1
sec xdx = (sec x tan x + ln | sec x + tan x|)
2
3
0
θ = tan x
EXAMPLE: Find the length of the cardioid r = 1 − cos θ.
8
π/4
0
√
1 √
= ( 2 + ln(1 + 2))
2




Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the length of the cardioid r = 1 − cos θ.
Solution: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5 gives
s
2
Z 2π
Z 2π q
dr
L=
dθ =
r2 +
(1 − cos θ)2 + sin2 θdθ
dθ
0
0
=
=
Z
Z
2π
0
2π
0
2π
p
√
1 − 2 cos θ + cos2 θ + sin2 θdθ
2 − 2 cos θdθ
r
θ
4 sin2 dθ
2
0
Z 2π θ
=
2 sin dθ
2
0
Z 2π
θ
=
2 sin dθ
2
0
2π
θ
=4+4=8
= −4 cos
2 0
=
Z
EXAMPLE: Find the length of the cardioid r = 1 + sin θ.
9
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
EXAMPLE: Find the length of the cardioid r = 1 + sin θ.
Solution 1: Note that
π
r = 1 + sin θ = 1 − cos θ +
2
Therefore the graph of r = 1 + sin θ is the rotation of the graph of r = 1 − cos θ. Hence the length of the
cardioid r = 1 + sin θ is 8 by the previous Example.
Solution 2: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5
gives
s
2
Z 2π
dr
2
L=
dθ
r +
dθ
0
Z 2π p
=
(1 + sin θ)2 + cos2 θdθ
0
2π
=
Z
2π
=
Z
0
p
1 + 2 sin θ + sin2 θ + cos2 θdθ
√
2 + 2 sin θdθ
0


π
θ+ =u
r
Z 5π/2
2π
 2 π

√
π
=
=
dθ = 
2 − 2 cos θ +
2 − 2 cos udu
d
θ
+
=
du


2
0
π/2
2
dθ = du
Z 2π Z 5π/2 Z 5π/2 Z 5π/2 r
u
u
u
2 u
2 sin du +
2 sin du
2 sin du =
=
4 sin du =
2
2
2
2
π/2
2π
π/2
π/2
Z 2π
Z 5π/2
u
u
=
2 sin du −
2 sin du
2
2
π/2
2π
u i5π/2
u i2π
+ 4 cos
= −4 cos
2 π/2
2 2π
5π
π
+ 4 cos
− 4 cos π
= −4 cos π + 4 cos
4
4
√
√
= (4 + 2 2) + (−2 2 + 4) = 8
Z
10
Section 9.4 Areas and Lengths in Polar Coordinates
2010 Kiryl Tsishchanka
Solution 3: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5
gives
s
2
Z 2π
Z 2π p
Z 2π p
dr
2
2
2
L =
dθ =
r +
(1 + sin θ) + cos θdθ =
1 + 2 sin θ + sin2 θ + cos2 θdθ
dθ
0
0
0
√
√
Z 2π √
√ Z 2π √
√ Z 2π 1 + sin θ 1 − sin θ
√
2 + 2 sin θdθ = 2
1 + sin θdθ = 2
=
dθ
1 − sin θ
0
0
0
√
Z 2π p
√ Z 2π
√ Z 2π | cos θ|
√
1 − sin2 θ
cos2 θ
√
√
√
dθ = 2
dθ = 2
dθ
= 2
1 − sin θ
1 − sin θ
1 − sin θ
0
0
0
Z π/2
√ Z 3π/2
√ Z 2π
√
cos θ
cos θ
cos θ
√
√
√
= 2
dθ − 2
dθ + 2
dθ
1 − sin θ
1 − sin θ
1 − sin θ
0
π/2
3π/2
Note that
Z

1 − sin θ = u

 d(1 − sin θ) = du
cos θ
√
dθ = 
 − cos θdθ = du
1 − sin θ

cos θdθ = −du


Z
Z
−1/2+1

du
 = − √ = − u−1/2 du = − u
+C

u
−1/2 + 1

√
= −2 u + C
√
= −2 1 − sin θ + C
Therefore
iπ/2
i3π/2
i2π
√ √
√ √
√ √
+ 2 2 1 − sin θ
− 2 2 1 − sin θ
L = −2 2 1 − sin θ
π/2
0
√ p
√
1 − sin(π/2) − 1 − sin 0
= −2 2
p
√ p
1 − sin(3π/2) − 1 − sin(π/2)
+2 2
p
√ p
−2 2
1 − sin(2π) − 1 − sin(3π/2)
√
√ √
√
√ = −2 2 (0 − 1) + 2 2 2 − 0 − 2 2 1 − 2
√
√
=2 2+4−2 2+4 =8
11
3π/2