Areas and Lengths in Polar Coordinates

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