1) The document discusses calculus concepts in polar coordinates, including area, arc length, and tangents of polar curves. 2) It provides examples of calculating the area bounded by a polar curve, such as finding the area inside a four-leafed rose curve and between two other curves. 3) Arc length of a polar curve is computed by treating it as a parametric curve in terms of the parameter θ, and calculating the derivative. An example finds the length of a cardioid curve. 4) Tangents of polar curves use derivatives to find the slope, and examples show finding the slope at a point and where the tangent is horizontal/vertical for a cardioid curve.

1. Chapter 10: Parametric Equations and Polar
Coordinates
Section 10.4: Calculus in Polar Coordinates
Alea Wittig
SUNY Albany

2. Outline
Area
Arc Length
Tangents

3. Area

4. Area in Polar Coordinates
▶ The area of a sector of a
circle is
A =
1
2
r2
θ
▶ We use circle sectors to approximate the area of a region
bounded by polar curve r = f (θ) in the same way we use
rectangles to approximate the area of a region bounded by
curve y = f (x).

5. Area in Polar Coordinates
▶ Let f (θ) be a positive
continuous function on
a ≤ θ ≤ b.
▶ Let R be the region
bounded by polar curve
r = f (θ) and the rays
θ = a and θ = b, where
0 < b − a ≤ 2π

6. Area in Polar Coordinates
▶ Divide [a, b] into
subintervals with endpoints
θ0, θ1, . . . , θn and equal
width △θ = θi − θi−1.
▶ Choosing θ∗
i in [θi−1, θi ],
△Ai is approximated by the
area of a circle with central
angle △θ and radius f (θ∗
i )
A ≈
n
X
i=1
1
2
[f (θ∗
i )]2
△θ =⇒
A =
Z b
a
1
2
[f (θ)]2
dθ =
Z b
a
1
2
r2
dθ

7. Example 1
1/3
Find the area enclosed within the four-leafed rose r = cos 2θ.

8. Example 1
2/3
▶ r is positive for
−π
4 ≤ θ ≤ π
4 , which
corresponds to one leaf of
the rose.
▶ By symmetry, the area
enclosed by the four-leafed
rose is 4 times the area
enclosed in one leaf.
▶ The area of one leaf is
Z π
4
−π
4
1
2
r2
dθ =
1
2
Z π
4
−π
4
cos2
(2θ)dθ

9. Example 1
3/3
Using the identity : cos2
θ =
1
2
1 + cos 2θ
=⇒
1
2
Z π
4
−π
4
cos2
(2θ)dθ =
1
2
Z π
4
−π
4
1
2
(1 + cos (4θ)dθ [let u = 4θ du = 4dθ]
=
1
4
Z π
−π
(1 + cos (u))
du
4
=
1
16
(u + sin u)

12. π
−π
=
1
16
(π + sin (π) − (−π + sin (−π))
=
π
16
+
π
16
=
π
8
=⇒
A = 4 ·
π
8
=
π
2

13. Example 2
1/5
Find the area of the region that lies inside the circle r = 3 sin θ and
outside the cardioid r = 1 + sin θ.

14. Example 2
2/5
▶ We want to find the area of
the shaded region in graph
to the right.
▶ By symmetry, it is twice the
area of the shaded region
from π
6 to π
2 .
▶ Both r = 3 sin θ and
r = 1 + sin θ are positive on
the entire region
π
6 ≤ θ ≤ 5π
6 . But it will be
simpler to compute area
from π
6 ≤ θ ≤ π
2 and then
multiply by 2.

15. Area Between Polar Curves
▶ In general, to find the area
of the region enclosed by
positive polar curves
r = f (θ) and r = g(θ) as
in the figure, we have just
subtract:
A =
Z b
a
1
2
[f (θ)]2
dθ −
Z b
a
1
2
[g(θ)]2
dθ
=
1
2
Z b
a
[f (θ)]2
− [g(θ)]2
dθ

16. Example 2
3/5
▶ So the area of the shaded
region is
2·
1
2
Z π
2
π
6
[3 sin θ]2
−[1+sin θ]2
dθ
=
Z π
2
π
6
9 sin2
θ − 1 + 2 sin θ + sin2
θ
dθ
=
Z π
2
π
6
8 sin2
θ − 1 − 2 sin θdθ
= 8
Z π
2
π
6
sin2
θdθ
| {z }
A
−
Z π
2
π
6
dθ
| {z }
B
− 2
Z π
2
π
6
sin θdθ
| {z }
C

17. Example 2
4/5
B :
Z π
2
π
6
dθ = θ

20. π
2
π
6
=
π
2
−
π
6
=
π
3
C : 2
Z π
2
π
6
sin θdθ = −2 cos θ

23. π
2
π
6
= −2 0 −
√
3
2
=
√
3
A : 8
Z π
2
π
6
sin2
θdθ = 4
Z π
2
π
6
1 − cos (2θ)
dθ
let u = 2θ, du = 2dθ
= 4
Z π
π
3
1 − cos u
du
2
= 2(u − sin u)

26. π
π
3
= 2
π − 0
−
π
3
−
√
3
2
= 2π −
2π
3
+
√
3
=
4π
3
+
√
3

27. Example 2
5/5
Area = A − B − C
=
4π
3
+
√
3 −
π
3
−
√
3
=
3π
3
= π

28. Arc Length

29. Arc Length
▶ Recall from 10.2: A curve C described by the parametric
equations
x = f (t) y = g(t) α ≤ t ≤ β
where f ′ and g′ are continuous and C is traversed once from
left to right as t increases from α to β, has length
L =
Z β
α
rdx
dt
2
+
dy
dt
2
dt
▶ A polar curve r = f (θ) on α ≤ θ ≤ β, is described by the
parametric equations
x = f (θ) cos θ y = f (θ) sin θ α ≤ θ ≤ β
by the conversion formulas x = r cos θ and y = r sin θ.

30. Arc Length
▶ So a polar curve r = f (θ) can be viewed as a parametric curve
with parameter θ.
▶ To find arc length of r = f (θ), we want to find dx
dθ and dx
dθ in
terms of θ.
x = f (θ) cos θ =⇒
dx
dθ
= −f (θ) sin θ + f ′
(θ) cos θ
by product rule
= −r sin θ +
dr
dθ
cos θ
since f (θ) = r
▶ And similarly,
y = f (θ) sin θ =⇒
dy
dθ
= f (θ) cos θ + f ′
(θ) sin θ
by product rule
= r cos θ +
dr
dθ
sin θ
since f (θ) = r

31. Arc Length
▶ Therefore we have
dx
dθ
2
+
dy
dθ
2
=
− r sin θ +
dr
dθ
cos θ
2
+
r cos θ +
dr
dθ
sin θ
2
= r2
(sin2
θ + cos2
θ) +
dr
dθ
2
(cos2
θ + sin2
θ)
= r2
+
dr
dθ
2
▶ Assuming f ′ continuous,
L =
Z β
α
r
r2 +
dr
dθ
2
dθ

32. Example 3
1/5
Find the length of the cardioid r = 1 + sin θ.

33. Example 3
2/5
L =
Z 2π
0
r
r2 +
dr
dθ
2
dθ
r = 1 + sin θ =⇒
dr
dθ
= cos θ
L =
Z 2π
0
q
(1 + sin θ)2 + cos2 θdθ
=
Z 2π
0
p
1 + 2 sin θ + sin2
θ + cos2 θdθ
=
Z 2π
0
√
2 + 2 sin θdθ
▶ Now we have an integral which is a little tricky.

34. Example 3
3/5
▶ We can make the substitution u = 2 + 2 sin θ and get
du = 2 cos θdθ =⇒ dθ = du
2 cos θ .
Z
√
2 + 2 sin θdθ =
Z u2
u1
√
u
du
2 cos θ
▶ But then we have to write 2 cos θ in terms of u, where
2 sin θ = u − 2, so we need an identity like:
4 cos2
θ + 4 sin2
θ = 4 =⇒ 2 cos θ = ±
q
4 − (2 sin θ)2
▶ cos θ is positive on the interval [−π
2 , π
2 ] which corresponds to
the right half of the cardioid.
▶ By symmetry, the length of the cardioid is twice that of its
length on [−π
2 , π
2 ]

35. Example 3
4/5
▶ So we have
Z 2π
0
√
2 + 2 sin θdθ = 2
Z π
2
−π
2
√
2 + 2 sin θ dθ
= 2
Z u2
u1
√
u
du
2 cos θ
and on [−π
2 , π
2 ], cos θ ≥ 0 so we have:
2 cos θ =
q
4 − (2 sin θ)2
=
q
4 − (u − 2)2
=
p
4 − u2 + 4u − 4
=
p
4u − u2 =⇒
2
Z u2
u1
√
u
du
2 cos θ
= 2
Z u2
u1
√
u
du
√
4u − u2

36. Example 3
5/5
. . . = 2
Z u2
u1
r
u
u(4 − u)
du
= 2
Z u2
u1
1
√
4 − u
du
let t = 4 − u, dt = −du
= 2
Z 4−u2
4−u1
t−1
2 dt
= −4
√
t

39. 4−u2
4−u1
= 4
√
4 − u1 − 2
√
4 − u2
= 4
p
4 − (2 + 2 sin θ1) − 4
p
4 − (2 + 2 sin θ2)
= 4
r
2 − 2 sin
−
π
2
− 4
r
2 − 2 sin
π
2
= 4
√
2 + 2 − 4
√
0
= 8

40. Tangents

41. Tangents
▶ We will again use the fact that a polar curve r = f (θ) is a
parametric curve described by equations
x = f (θ) cos θ y = f (θ) sin θ
which yields the derivatives
dx
dθ
=
dr
dθ
sin θ + r cos θ
dy
dθ
=
dr
dθ
cos θ − r sin θ =⇒
dy
dx
=
dr
dθ
sin θ + r cos θ
dr
dθ
cos θ − r sin θ

42. Example 4
1/7
(a) For the cardioid r = 1 + sin θ, find the slope of the tangent
line at θ = π
3 .
(b) Find the points on the cardioid where the tangent line is
horizontal or vertical.

43. Example 4
2/7
(a) We want to compute
dy
dx
=
dr
dθ
sin θ + r cos θ
dr
dθ
cos θ − r sin θ
where
r = 1 + sin θ =⇒
dr
dθ
= cos θ =⇒
dy
dx
=
cos θ sin θ + r cos θ
cos2 θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos2 θ − (1 + sin θ) sin θ
=
cos θ(1 + 2 sin θ)
1 − sin2
θ − (1 + sin θ) sin θ

44. Example 4
3/7
. . . =
cos θ(1 + 2 sin θ)
(1 − sin θ)(1 + sin θ) − (1 + sin θ) sin θ
=
cos θ(1 + 2 sin θ)
(1 + sin θ)(1 − 2 sin θ)
▶ Using sin π
3
=
√
3
2 , cos π
3
, and r = 1 + sin π
3
= 1 +
√
3
2 :
dy
dx
π
3
=
1
2(1 +
√
3)
(1 +
√
3
2 )(1 −
√
3)
=
1 +
√
3
(2 +
√
3)(1 −
√
3)
=
1 +
√
3
−1 −
√
3
= −1

45. Example 4
4/7
(b) To find the points where the cardioid has horizontal and
vertical tangents, we determine where the numerator and
denominator are 0.
▶ We will consider all possible values of θ between 0 and 2π.
▶ Horizontal tangent when numerator = 0 and denominator ̸= 0.
▶ Vertical tangent when denominator = 0 and numerator ̸= 0.
▶ If both numerator and denominator are 0, take the limit.
▶ Starting with the numerator we have
cos θ(1 + 2 sin θ) = 0 ⇐⇒
cos θ = 0
θ =
π
2
or
3π
2
or 1 + 2 sin θ = 0
sin θ = −
1
2
θ =
7π
6
or
11π
6

46. Example 4
5/7
▶ And the denominator we have:
(1 + sin θ)(1 − 2 sin θ) = 0 ⇐⇒
(1 + sin θ) = 0
sin θ = −1
θ =
3π
2
or (1 − 2 sin θ) = 0
sin θ =
1
2
θ =
π
6
or
5π
6
▶ Horizontal tangents at θ = π
2 , 7π
6 , and 11π
6 .
▶ Vertical tangents at θ = π
6 and 5π
6 .

47. Example 4
6/7
▶ At θ = 3π
2 , both numerator and denominator are zero.
▶ Let’s take the limit:
lim
θ→ 3π
2
cos θ(1 + 2 sin θ)
(1 + sin θ)(1 − 2 sin θ)
=
1 + 2 sin 3π
2
1 − 2 sin 3π
2
· lim
θ→3π
2
cos θ
1 + sin θ
= −
1
3
lim
θ→ 3π
2
cos θ
1 + sin θ
→
0
0
= −
1
3
lim
θ→ 3π
2
− sin θ
cos θ
by L’Hosp.
=
1
3
lim
θ→ 3π
2
sin θ
cos θ
= −∞
since sin 3π
2 = −1 and cos 3π
2 = 0. So the cardioid has a
vertical tangent at θ = 3π
2 .

48. Example 4
7/7