1. Areas in Polar Coordinates

2. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
r=f(θ)
θ=b θ=a

3. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
r=f(θ)
θ=b θ=a

4. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
r=f(θ)
θ=b θ=a

5. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
r=f(θ)
θ=b θ=a

6. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
r=f(θ)
θ=b θ=a

7. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
it "sweeps" out an area r=f(θ)
between them.
θ=b θ=a

8. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
it "sweeps" out an area r=f(θ)
between them.
θ=b θ=a
(The Polar Area Formula)
The area swept out by r = f(θ) from θ = a to θ = b is
b b
1 1
A= 2 ∫r 2
dθ or 2 ∫ f(θ)2 dθ
θ=a θ=a

9. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
it "sweeps" out an area r=f(θ)
between them.
θ=b θ=a
(The Polar Area Formula)
The area swept out by r = f(θ) from θ = a to θ = b is
b b
1 1
A= 2 ∫r 2
dθ or 2 ∫ f(θ)2 dθ
θ=a θ=a
Example A.
Draw and find the area A swept by
r = f(θ) = k with 0 ≤ θ ≤ 2π.

10. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
it "sweeps" out an area r=f(θ)
between them.
θ=b θ=a
(The Polar Area Formula)
The area swept out by r = f(θ) from θ = a to θ = b is
b b
1 1
A= 2 ∫r 2
dθ or 2 ∫ f(θ)2 dθ
θ=a θ=a
Example A. r = f(θ) = k
Draw and find the area A swept by
r = f(θ) = k with 0 ≤ θ ≤ 2π. k

11. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
it "sweeps" out an area r=f(θ)
between them.
θ=b θ=a
(The Polar Area Formula)
The area swept out by r = f(θ) from θ = a to θ = b is
b b
1 1
A= 2 ∫r 2
dθ or 2 ∫ f(θ)2 dθ
θ=a θ=a
Example A. r = f(θ) = k
Draw and find the area A swept by
r = f(θ) = k with 0 ≤ θ ≤ 2π. k
b
1
A =2 ∫ r2 dθ =
θ=a

12. Areas in Polar Coordinates
Given a polar function
r = f(θ) with a ≤ θ ≤ b,
it "sweeps" out an area r=f(θ)
between them.
θ=b θ=a
(The Polar Area Formula)
The area swept out by r = f(θ) from θ = a to θ = b is
b b
1 1
A= 2 ∫r 2
dθ or 2 ∫ f(θ)2 dθ
θ=a θ=a
Example A. r = f(θ) = k
Draw and find the area A swept by
r = f(θ) = k with 0 ≤ θ ≤ 2π. k
b 2π 2π
1 1 k2θ = π k2
A =2 ∫a r dθ = 2
2
∫ k2 dθ = 2
θ= θ=0 θ=0

13. Areas in Polar Coordinates
We have the following formulas cos2(θ) = 1 + cos(2θ)
that express the squares of the 2
sine and cosine in terms of the sin2(θ) = 1 – cos(2θ)
cosine double-angle. 2

14. Areas in Polar Coordinates
We have the following formulas cos2(θ) = 1 + cos(2θ)
that express the squares of the 2
sine and cosine in terms of the sin2(θ) = 1 – cos(2θ)
cosine double-angle. 2
We simplify the algebra below c2(θ) = 1 + c(2θ)
by writing cos(θ) as c(θ) 2
s (θ) =
2 1 – c(2θ)
and sin(θ) as s(θ). 2

15. Areas in Polar Coordinates
We have the following formulas cos2(θ) = 1 + cos(2θ)
that express the squares of the 2
sine and cosine in terms of the sin2(θ) = 1 – cos(2θ)
cosine double-angle. 2
We simplify the algebra below c2(θ) = 1 + c(2θ)
by writing cos(θ) as c(θ) 2
s (θ) =
2 1 – c(2θ)
and sin(θ) as s(θ). 2
We use these formulas to integrate c2(θ) and s2(θ) that
shows up in f2(θ), the integrand in the polar area
formula. Hence to integrate
∫ c2(θ) dθ =

16. Areas in Polar Coordinates
We have the following formulas cos2(θ) = 1 + cos(2θ)
that express the squares of the 2
sine and cosine in terms of the sin2(θ) = 1 – cos(2θ)
cosine double-angle. 2
We simplify the algebra below c2(θ) = 1 + c(2θ)
by writing cos(θ) as c(θ) 2
s (θ) =
2 1 – c(2θ)
and sin(θ) as s(θ). 2
We use these formulas to integrate c2(θ) and s2(θ) that
shows up in f2(θ), the integrand in the polar area
formula. Hence to integrate
∫ c2(θ) dθ = ∫ 1 + c(2θ) dθ =
2

17. Areas in Polar Coordinates
We have the following formulas cos2(θ) = 1 + cos(2θ)
that express the squares of the 2
sine and cosine in terms of the sin2(θ) = 1 – cos(2θ)
cosine double-angle. 2
We simplify the algebra below c2(θ) = 1 + c(2θ)
by writing cos(θ) as c(θ) 2
s (θ) =
2 1 – c(2θ)
and sin(θ) as s(θ). 2
We use these formulas to integrate c2(θ) and s2(θ) that
shows up in f2(θ), the integrand in the polar area
formula. Hence to integrate
∫ c2(θ) dθ = ∫ 1 + c(2θ) dθ = ½ ∫1 + c(2θ) dθ
2
= ½ (θ + ½ s(2θ)) + C or ½ θ + ¼ s(2θ) + C

18. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C

19. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π.

20. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
1

21. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ

22. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k

23. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] |
θ=0 θ=0

24. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π,
θ=0 θ=0

25. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is
θ=0 θ=0
the area of a circle with radius 1, swept over twice.

26. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is
θ=0 θ=0
the area of a circle with radius 1, swept over twice.

27. Areas in Polar Coordinates
We summarize the
above integrals here.
∫ c2(θ) dθ = ½ θ + ¼ s(2θ) + C
Example B. ∫ s2(θ) dθ = ½ θ – ¼ s(2θ) + C
Draw and find the area swept by r = f(θ) = 2sin(θ) with
θ goes from 0 to 2π. r = f(θ) = 2sin(θ)
The anti–derivative ∫ f 2(θ) dθ 1
=∫ [2s(θ)]2dθ = 4 ∫s2 (θ) dθ
= 4(½ θ – ¼ s(2θ)) + k
= 2θ – s(2θ) + k
The area swept over by r as θ goes from 0 to 2π is
2π 2π
½ ∫ f 2(θ) dθ = ½ [2θ – s(2θ)] | = ½ (4π) = 2π, that is
θ=0 θ=0
the area of a circle with radius 1, swept over twice.

28. Areas in Polar Coordinates
Example C.
Draw and find the area
enclosed by r = f(θ) = 1 – cos(θ)
in the first quadrant.

29. Areas in Polar Coordinates
Example C.
Draw and find the area θ=π/2
enclosed by r = f(θ) = 1 – cos(θ)
in the first quadrant. 2
The region in question is θ=0
shown.

30. Areas in Polar Coordinates
Example C.
Draw and find the area θ=π/2
enclosed by r = f(θ) = 1 – cos(θ)
in the first quadrant. 2
The region in question is θ=0
shown. It’s the swept area as
π
θ goes from 0 to 2 which is

31. Areas in Polar Coordinates
Example C.
Draw and find the area θ=π/2
enclosed by r = f(θ) = 1 – cos(θ)
in the first quadrant. 2
The region in question is θ=0
shown. It’s the swept area as
θ goes from 0 to 2 π

32. Areas in Polar Coordinates
Example C.
Draw and find the area θ=π/2
enclosed by r = f(θ) = 1 – cos(θ)
in the first quadrant. 2
The region in question is θ=0
shown. It’s the swept area as
π
θ goes from 0 to 2 which is
1 π /2 1 π /2
2 ∫0
(1 − cos(θ )) dθ = ∫ 1 − 2 cos(θ ) + cos 2 (θ )dθ
2
2 0
1 1 1 π /2
= (θ − 2 sin(θ ) + sin( 2θ ) + θ ) |
2 4 2 0
1 3 1 π /2 1 3 π 3π
= ( θ − 2 sin(θ ) + sin( 2θ )) | = ( * − 2 *1 + 0) = −1
2 2 4 0 2 2 2 8

33. Areas in Polar Coordinates
Example C.
Draw and find the area θ=π/2
enclosed by r = f(θ) = 1 – cos(θ)
in the first quadrant. 2
The region in question is θ=0
shown. It’s the swept area as
π
θ goes from 0 to 2 which is
1 π /2 1 π /2
2 ∫0
(1 − cos(θ )) dθ = ∫ 1 − 2 cos(θ ) + cos 2 (θ )dθ
2
2 0
1 1 1 π /2
= (θ − 2 sin(θ ) + sin( 2θ ) + θ ) |
2 4 2 0
1 3 1 π /2 1 3 π 3π
= ( θ − 2 sin(θ ) + sin( 2θ )) | = ( * − 2 *1 + 0) = −1
2 2 4 0 2 2 2 8

34. Areas in Polar Coordinates
Example C.
Draw and find the area θ=π/2
enclosed by r = f(θ) = 1 – cos(θ)
in the first quadrant. 2
The region in question is θ=0
shown. It’s the swept area as
π
θ goes from 0 to 2 which is
1 π /2 1 π /2
2 ∫0
(1 − cos(θ )) dθ = ∫ 1 − 2 cos(θ ) + cos 2 (θ )dθ
2
2 0
1 1 1 π /2
= (θ − 2 sin(θ ) + sin( 2θ ) + θ ) |
2 4 2 0
1 3 1 π /2 1 3 π 3π
= ( θ − 2 sin(θ ) + sin( 2θ )) | = ( * − 2 *1 + 0) = −1
2 2 4 0 2 2 2 8

35. Areas in Polar Coordinates
Given polar functions R = f(θ) outer
R = f(θ) and r = g(θ)
where R > r for θ between θ=a
a and b, θ=b
inner
r = g(θ)

36. Areas in Polar Coordinates
Given polar functions R = f(θ) outer
R = f(θ) and r = g(θ)
where R > r for θ between θ=a
a and b, then the area θ=b
inner
r = g(θ)
between them is
b b
½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ
θ=a θ=a

37. Areas in Polar Coordinates
Given polar functions R = f(θ) outer
R = f(θ) and r = g(θ)
where R > r for θ between θ=a
a and b, then the area θ=b
inner
r = g(θ)
between them is
b b
½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ
θ=a θ=a
To find the area enclosed by two polar graphs, the
first step is to determine the angles θ = a and θ = b
that wedge in the area.

38. Areas in Polar Coordinates
Given polar functions R = f(θ) outer
R = f(θ) and r = g(θ)
where R > r for θ between θ=a
a and b, then the area θ=b
inner
r = g(θ)
between them is
b b
½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ
θ=a θ=a
To find the area enclosed by two polar graphs, the
first step is to determine the angles θ = a and θ = b
that wedge in the area. This should be done both
algebraically and graphically because the algebraic
solution might be incomplete.

39. Areas in Polar Coordinates
Given polar functions R = f(θ) outer
R = f(θ) and r = g(θ)
where R > r for θ between θ=a
a and b, then the area θ=b
inner
r = g(θ)
between them is
b b
½ ∫ R2– r2dθ = ½ ∫ f2(θ) – g2(θ) dθ
θ=a θ=a
To find the area enclosed by two polar graphs, the
first step is to determine the angles θ = a and θ = b
that wedge in the area. This should be done both
algebraically and graphically because the algebraic
solution might be incomplete. Next we have to split
the area into separate wedges radialIy from corner to
corner.

40. Areas in Polar Coordinates
Example D.
R = 2cos(θ)
Find the shaded area r=1
shown in the figure.

41. Areas in Polar Coordinates
Example D.
R = 2cos(θ)
Find the shaded area r=1
shown in the figure.
First set the two curves
equal to each other to find
the angles of the intersection.
1 = 2cos(θ)

42. Areas in Polar Coordinates
Example D.
R = 2cos(θ)
Find the shaded area r=1
shown in the figure.
First set the two curves
equal to each other to find
the angles of the intersection.
1 = 2cos(θ)  cos(θ)=1/2 so θ = π/3

43. Areas in Polar Coordinates
Example D. θ=π/2 θ=π/3
R = 2cos(θ)
Find the shaded area r=1
shown in the figure.
θ=0
First set the two curves
equal to each other to find
the angles of the intersection.
1 = 2cos(θ)  cos(θ)=1/2 so θ = π/3
The area consists of two regions.
i. the red area with the r = 1 as boundary with
θ goes from 0 to π/3.

44. Areas in Polar Coordinates
Example D. θ=π/2 θ=π/3
R = 2cos(θ)
Find the shaded area r=1
shown in the figure.
θ=0
First set the two curves
equal to each other to find
the angles of the intersection.
1 = 2cos(θ)  cos(θ)=1/2 so θ = π/3
The area consists of two regions.
i. the red area with the r = 1 as boundary with
θ goes from 0 to π/3.
ii. the black area with R = 2cos(θ) as the boundary
with θ going from π/3 to θ = π/2.

45. Areas in Polar Coordinates
Example D. θ=π/2 θ=π/3
R = 2cos(θ)
Find the shaded area r=1
shown in the figure.
θ=0
First set the two curves
equal to each other to find
the angles of the intersection.
1 = 2cos(θ)  cos(θ)=1/2 so θ = π/3
The area consists of two regions.
i. the red area with the r = 1 as boundary with
θ goes from 0 to π/3.
ii. the black area with R = 2cos(θ) as the boundary
with θ going from π/3 to θ = π/2.
The answer is the sum of two areas

46. Areas in Polar Coordinates
The red area bounded from θ=π/2 θ=π/3
r=1 R = 2cos(θ)
θ = 0 to θ=π/3 is 1/6 of the
unit circle. It's area is π/6.
θ=0

47. Areas in Polar Coordinates
The red area bounded from θ=π/2 θ=π/3
r=1 R = 2cos(θ)
θ = 0 to θ=π/3 is 1/6 of the
unit circle. It's area is π/6.
θ=0
The black area are bounded
by θ = π/3 to θ = π/2 is

48. Areas in Polar Coordinates
The red area bounded from θ=π/2 θ=π/3
r=1 R = 2cos(θ)
θ = 0 to θ=π/3 is 1/6 of the
unit circle. It's area is π/6.
θ=0
The black area are bounded
by θ = π/3 to θ = π/2 is
1 π /2
2 ∫π / 3
(2 cos(θ )) 2 dθ
π /2 1 1 π /2
= 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) |
2
π /3 4 2 π /3
1 π /2 π 1 2π π π 3
= sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = −
2 π /3 2 2 3 3 6 4
The total area is π + π – √3 = π – √3
6 6 4 3 4

49. Areas in Polar Coordinates
The red area bounded from θ=π/2 θ=π/3
r=1 R = 2cos(θ)
θ = 0 to θ=π/3 is 1/6 of the
unit circle. It's area is π/6.
θ=0
The black area are bounded
by θ = π/3 to θ = π/2 is
1 π /2
2 ∫π / 3
(2 cos(θ )) 2 dθ
π /2 1 1 π /2
= 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) |
2
π /3 4 2 π /3
1 π /2 π 1 2π π π 3
= sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = −
2 π /3 2 2 3 3 6 4
The total area is π + π – √3 = π – √3
6 6 4 3 4

50. Areas in Polar Coordinates
The red area bounded from θ=π/2 θ=π/3
r=1 R = 2cos(θ)
θ = 0 to θ=π/3 is 1/6 of the
unit circle. It's area is π/6.
θ=0
The black area are bounded
by θ = π/3 to θ = π/2 is
1 π /2
2 ∫π / 3
(2 cos(θ )) 2 dθ
π /2 1 1 π /2
= 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) |
2
π /3 4 2 π /3
1 π /2 π 1 2π π π 3
= sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = −
2 π /3 2 2 3 3 6 4

51. Areas in Polar Coordinates
The red area bounded from θ=π/2 θ=π/3
r=1 R = 2cos(θ)
θ = 0 to θ=π/3 is 1/6 of the
unit circle. It's area is π/6.
θ=0
The black area are bounded
by θ = π/3 to θ = π/2 is
1 π /2
2 ∫π / 3
(2 cos(θ )) 2 dθ
π /2 1 1 π /2
= 2∫ cos (θ )dθ = 2( sin( 2θ ) + θ ) |
2
π /3 4 2 π /3
1 π /2 π 1 2π π π 3
= sin( 2θ ) + θ | = (0 + ) − ( sin( ) + ) = −
2 π /3 2 2 3 3 6 4
The total area is π + π – √3 = π – √3
6 6 4 3 4

52. Areas in Polar Coordinates
Method II: The blue area is θ=π/2 θ=π/3
R = 2cos(θ)
¼ of a circle minus the red r=1
area.

53. Areas in Polar Coordinates
Method II: The blue area is θ=π/2 θ=π/3
R = 2cos(θ)
¼ of a circle minus the red r=1
area. The red area is
bounded by f(θ)=1 and
g(θ)=2cos(θ) where θ goes
from π/3 to π/2.

54. Areas in Polar Coordinates
Method II: The blue area is θ=π/2 θ=π/3
R = 2cos(θ)
¼ of a circle minus the red r=1
area. The red area is
bounded by f(θ)=1 and
g(θ)=2cos(θ) where θ goes
from π/3 to π/2. It's area is
(outer curve)2 (inner curve)2
1 π/2 2
2 ∫ /3
1 − (2 cos(θ )) 2 dθ
π
1 π/2 2 π/2
= ∫ 1 dθ − 2 ∫ cos 2 (θ ) dθ
2 π /3 π /3
1 π π π 3 π 3
= ( − ) −( − ) =− +
2 2 3 6 4 12 4
So the blue area is π – (– π + √3 ) = π – √3
4 12 4 3 4

55. Areas in Polar Coordinates
Method II: The blue area is θ=π/2 θ=π/3
R = 2cos(θ)
¼ of a circle minus the red r=1
area. The red area is
bounded by f(θ)=1 and
g(θ)=2cos(θ) where θ goes
from π/3 to π/2. It's area is
(outer curve)2 (inner curve)2
1 π/2 2
2 ∫ /3
1 − (2 cos(θ )) 2 dθ
π
1 π/2 2 π/2
= ∫ 1 dθ − 2 ∫ cos 2 (θ ) dθ
2 π /3 π /3
1 π π π 3 π 3
= ( − ) −( − ) =− +
2 2 3 6 4 12 4

56. Areas in Polar Coordinates
Method II: The blue area is θ=π/2 θ=π/3
R = 2cos(θ)
¼ of a circle minus the red r=1
area. The red area is
bounded by f(θ)=1 and
g(θ)=2cos(θ) where θ goes
from π/3 to π/2. It's area is
(outer curve)2 (inner curve)2
1 π/2 2
2 ∫ /3
1 − (2 cos(θ )) 2 dθ
π
1 π/2 2 π/2
= ∫ 1 dθ − 2 ∫ cos 2 (θ ) dθ
2 π /3 π /3
1 π π π 3 π 3
= ( − ) −( − ) =− +
2 2 3 6 4 12 4
So the blue area is π – (– π + √3 ) = π – √3
4 12 4 3 4

57. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2

58. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2
We only need to find one of
the four equal regions as
shaded.

59. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2
We only need to find one of
the four equal regions as
shaded. We need to find
the angles of intersections that wedge in the area.

60. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2
We only need to find one of
the four equal regions as
shaded. We need to find
the angles of intersections that wedge in the area.
Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer
is 2θ = π/6, or that θ = π/12.

61. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2
We only need to find one of
the four equal regions as
shaded. We need to find
the angles of intersections that wedge in the area.
Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer
is 2θ = π/6, or that θ = π/12. From the symmetry of
the graph the other θ is 5π/12.

62. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2
We only need to find one of
the four equal regions as
shaded. We need to find
the angles of intersections that wedge in the area.
Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer
is 2θ = π/6, or that θ = π/12. From the symmetry of
the graph the other θ is 5π/12.

63. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2
We only need to find one of
the four equal regions as
shaded. We need to find
the angles of intersections that wedge in the area.
Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer
is 2θ = π/6, or that θ = π/12. From the symmetry of
the graph the other θ is 5π/12. Hence one piece is
(outer curve)2 (inner curve)2
1 5π /12 5π / 12 2π
∫π /12 (4 sin(2θ )) −2 dθ = ∫π /12 8 sin (2θ ) − 2dθ = 3 + 3
2 2 2
2

64. Areas in Polar Coordinates
Example E. Find the area that R = 4sin(2θ)
is outside of r = 2 and inside of
R = 4sin(2θ).
r=2
We only need to find one of
the four equal regions as
shaded. We need to find
the angles of intersections that wedge in the area.
Set 4sin(2θ) = 2 or sin(2θ) = ½. Hence one answer
is 2θ = π/6, or that θ = π/12. From the symmetry of
the graph the other θ is 5π/12. Hence one piece is
(outer curve)2 (inner curve)2
1 5π /12 5π / 12 2π
∫π /12 (4 sin(2θ )) −2 dθ = ∫π /12 8 sin (2θ ) − 2dθ = 3 + 3
2 2 2
2
So the total area is 4*(2π/3 + √3)