Area in polar form

1. Objective:
The Integral formula of the area swept
out by a polar function
r = f() between two angles
Area in Polar Coordinates

2. Area in Polar Coordinates
Given a polar function r = f(),
r = f()

3. Area in Polar Coordinates
Given a polar function r = f(), as  varies from A to B
it "sweeps" out an area between them.
r = f()
 = A = B  = A = B

4. Area in Polar Coordinates
r = f()
 = A = B
S1
S2S3
Sn
Given a polar function r = f(), as  varies from A to B
it "sweeps" out an area between them.
To find its area, slice the region radially into n slices
S1, S2, S3, .. Sn, with the radial angle Δ for each slice
. Δ
Δ Δ
. .
.

5. Area in Polar Coordinates
r = f()
 = A = B
S1
S2S3
Sn
Given a polar function r = f(), as  varies from A to B
it "sweeps" out an area between them.
To find its area, slice the region radially into n slices
S1, S2, S3, .. Sn, with the radial angle Δ for each slice
and let r1, r2, r3, .. rn be the radial lengths of the slices
as shown.
. Δ
Δ Δ
. .
.
r1
r2
r3

6. Area in Polar Coordinates
The area of
S1 ≈ ½ r1 Δ
= Area of a slice from
the circle of radius r1
with the radial angle Δ.
2
r = f()
 = A = B
S1
S2S3
r1
r2
r3
Sn
. Δ
Δ Δ
. .
.
Given a polar function r = f(), as  varies from A to B
it "sweeps" out an area between them.
To find its area, slice the region radially into n slices
S1, S2, S3, .. Sn, with the radial angle Δ for each slice
and let r1, r2, r3, .. rn be the radial lengths of the slices
as shown.

7. Area in Polar Coordinates
Given a polar function r = f(), as  varies from A to B
it "sweeps" out an area between them.
To find its area, slice the region radially into n slices
S1, S2, S3, .. Sn, with the radial angle Δ for each slice
and let r1, r2, r3, .. rn be the radial lengths of the slices
as shown .
The area of
S1 ≈ ½ r1 Δ
= Area of a slice from
the circle of radius r1
with the radial angle Δ.
2
S2 ≈ ½ r2 Δ
= Area of a slice from the circle of radius r2
with the radial angle Δ.
2
r = f()
 = A = B
S1
S2S3
r1
r2
r3
Sn
. Δ
Δ Δ
. .
.

8. Area in Polar Coordinates
Given a polar function r = f(), as  varies from A to B
it "sweeps" out an area between them.
To find its area, slice the region radially into n slices
S1, S2, S3, .. Sn, with the radial angle Δ for each slice
and let r1, r2, r3, .. rn be the radial lengths of the slices
as shown .
The area of
S1 ≈ ½ r1 Δ
= Area of a slice from
the circle of radius r1
with the radial angle Δ.
2
S2 ≈ ½ r2 Δ
= Area of a slice from the circle of radius r2
with the radial angle Δ. Similarly, S3 ≈ ½ r3 Δ, etc..
2
r = f()
 = A = B
S1
S2S3
r1
r2
r3
Sn
. Δ
Δ Δ
. .
.
2

9. Area in Polar Coordinates
In general Si ≈ ½ ri Δ where i = 1, 2,.. n.2

10. Area in Polar Coordinates
In general Si ≈ ½ ri Δ where i = 1, 2,.. n.
The area swept as  varies from A to B
2
= lim (S1 + S2 + S3 ..+ Sn)
Σ ½ ri Δ as n→∞,
i =1
2n
= lim

11. Area in Polar Coordinates
In general Si ≈ ½ ri Δ where i = 1, 2,.. n.
The area swept as  varies from A to B
2
= lim (S1 + S2 + S3 ..+ Sn)
Σ ½ ri Δ as n→∞,
i =1
2n
= lim
= ∫ r2 d½
=A
B
∫ f()2 d½
=A
B
or
by the Fundamental Theorem of Calculus.

12. Area in Polar Coordinates
In general Si ≈ ½ ri Δ where i = 1, 2,.. n.
The area swept as  varies from A to B
2
= lim (S1 + S2 + S3 ..+ Sn)
Σ ½ ri Δ as n→∞,
i =1
2n
= lim
= ∫ r2 d½
=A
B
∫ f()2 d½
=A
B
or
The Polar Area
Formula
by the Fundamental Theorem of Calculus.

13. Area in Polar Coordinates
In general Si ≈ ½ ri Δ where i = 1, 2,.. n.
The area swept as  varies from A to B
2
= lim (S1 + S2 + S3 ..+ Sn)
Σ ½ ri Δ as n→∞,
i =1
2n
= lim
= ∫ r2 d½
=A
B
∫ f()2 d½
=A
B
or
k
Example A.
Find the area enclosed by r = f() = k
0 ≤  ≤ 2π.
The Polar Area
Formula
by the Fundamental Theorem of Calculus.

14. Area in Polar Coordinates
In general Si ≈ ½ ri Δ where i = 1, 2,.. n.
The area swept as  varies from A to B
2
= lim (S1 + S2 + S3 ..+ Sn)
Σ ½ ri Δ as n→∞,
i =1
2n
= lim
= ∫ r2 d½
=A
B
∫ f()2 d½
=A
B
or
k
∫k2 d½  = 0
2π
Example A.
Find the area enclosed by r = f() = k
0 ≤  ≤ 2π.
We obtain the area of the circle from
the polar formula
The Polar Area
Formula
by the Fundamental Theorem of Calculus.

15. Area in Polar Coordinates
In general Si ≈ ½ ri Δ where i = 1, 2,.. n.
The area swept as  varies from A to B
2
= lim (S1 + S2 + S3 ..+ Sn)
Σ ½ ri Δ as n→∞,
i =1
2n
= lim
= ∫ r2 d½
=A
B
∫ f()2 d½
=A
B
or
k
∫k2 d =½  = 0
2π
Example A.
Find the area enclosed by r = f() = k
0 ≤  ≤ 2π.
We obtain the area of the circle from
k2 = π k2½
 = 0
2π
the polar formula
The Polar Area
Formula
by the Fundamental Theorem of Calculus.

16. Example B.
Find the area enclosed by
r = f() = 2sin().
1
Area in Polar Coordinates (2, π/2)

17. Example B.
Find the area enclosed by
r = f() = 2sin().
1
Area in Polar Coordinates
The main word here is “enclosed”.
(2, π/2)

18. Example B.
Find the area enclosed by
r = f() = 2sin().
1
=π =0
Area in Polar Coordinates
The main word here is “enclosed”.
The graph of r = 2sin() is a circle
which is traced once around with  going from 0 to π.
(2, π/2)

19. Example B.
Find the area enclosed by
r = f() = 2sin().
1
=π =0
Area in Polar Coordinates
The main word here is “enclosed”.
The graph of r = 2sin() is a circle
which is traced once around with  going from 0 to π.
So We have:
∫(2sin()) 2d½
∫sin2() d2
(2, π/2)
=

20. Example B.
Find the area enclosed by
r = f() = 2sin().
1
=π =0
Area in Polar Coordinates
The main word here is “enclosed”.
The graph of r = 2sin() is a circle
which is traced once around with  going from 0 to π.
So We have:
∫(2sin()) 2d
∫1 – cos(2)d,
½
∫sin2() d2 =
(2, π/2)
=
sin2() =
1 – cos(2)
2
from  = 0 to π ,

21. Example B.
Find the area enclosed by
r = f() = 2sin().
1
=π =0
Area in Polar Coordinates
The main word here is “enclosed”.
The graph of r = 2sin() is a circle
which is traced once around with  going from 0 to π.
So We have:
∫(2sin()) 2d
∫1 – cos(2)d,
=  – ½ sin(2)
½
∫sin2() d2 =
=0
π
(2, π/2)
=
sin2() =
1 – cos(2)
2
from  = 0 to π ,

22. Example B.
Find the area enclosed by
r = f() = 2sin().
1
=π =0
Area in Polar Coordinates
The main word here is “enclosed”.
The graph of r = 2sin() is a circle
which is traced once around with  going from 0 to π.
So We have:
∫(2sin()) 2d
∫1 – cos(2)d,
=  – ½ sin(2)
½
∫sin2() d2 =
=0
π
= π.
(2, π/2)
=
sin2() =
1 – cos(2)
2
from  = 0 to π ,

23. Example B.
Find the area enclosed by
r = f() = 2sin().
1
=π =0
If we integrate  from 0 to 2π, we get 2 for the area.
This because the graph traced over the circle twice
hence the answer for the area of two circles.
Area in Polar Coordinates
The main word here is “enclosed”.
The graph of r = 2sin() is a circle
which is traced once around with  going from 0 to π.
So We have:
∫(2sin()) 2d
∫1 – cos(2)d,
=  – ½ sin(2)
½
∫sin2() d2 =
=0
π
= π.
(2, π/2)
=
sin2() =
1 – cos(2)
2
from  = 0 to π ,

24. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
Area in Polar Coordinates

25. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.

26. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.

27. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
(0, 0)
Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.
(2, π)

28. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
(0, 0) =0
=π/2Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.
(2, π)
As  goes from 0 to π/2, r
swipes out an area in the first
quadrant as shown.

29. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
(0, 0) =0
=π/2Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.
(2, π)
∫(1 – cos())2d½
As  goes from 0 to π/2, r
swipes out an area in the first
quadrant as shown. Its area is

30. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
(0, 0) =0
=π/2Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.
(2, π)
∫(1 – cos())2d½ ∫1 – 2cos() + cos2() d½=
cos2() =
1 + cos(2)
2
As  goes from 0 to π/2, r
swipes out an area in the first
quadrant as shown. Its area is

31. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
(0, 0) =0
=π/2Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.
(2, π)
∫(1 – cos())2d½ ∫1 – 2cos() + cos2() d½=
= ½( – 2sin() + ¼ sin(2) + ½ ),
cos2() =
1 + cos(2)
2
As  goes from 0 to π/2, r
swipes out an area in the first
quadrant as shown. Its area is

32. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
(0, 0) =0
=π/2Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.
(2, π)
∫(1 – cos())2d½ ∫1 – 2cos() + cos2() d½
=0
=
setting  = 0 to π/2,= ½( – 2sin() + ¼ sin(2) + ½ ),
½(3/2 – 2sin() + ¼ sin(2))
π/2
cos2() =
1 + cos(2)
2
As  goes from 0 to π/2, r
swipes out an area in the first
quadrant as shown. Its area is

33. Example C.
Find the area enclosed by
r = f() = 1 – cos() with 0 ≤  ≤
π
2
(0, 0) =0
=π/2Area in Polar Coordinates
The function r = 1 – cos() is a cardioid.
The furthest point on the graph to the origin is (2, π)
hence it's the cardioid that protrudes to the left.
(2, π)
∫(1 – cos())2d½ ∫1 – 2cos() + cos2() d½
=0
= 3π/8 – 1 .
=
setting  = 0 to π/2,= ½( – 2sin() + ¼ sin(2) + ½ ),
½(3/2 – 2sin() + ¼ sin(2))
π/2
cos2() =
1 + cos(2)
2
As  goes from 0 to π/2, r
swipes out an area in the first
quadrant as shown. Its area is

34. Given polar functions
R = f() and r = g() where
R ≥ r for  between A and B,
the area between them is:
Outer R = f()
 = A = B
Inner r = g()
Area in Polar Coordinates
∫ f2() – g2() dor
=A
B
∫R2 – r2 d½
=A
B
½

35. Given polar functions
R = f() and r = g() where
R ≥ r for  between A and B,
the area between them is:
Outer R = f()
 = A = B
Inner r = g()
Area in Polar Coordinates
∫ f2() – g2() dor
=A
B
∫R2 – r2 d½
=A
B
½
Example D. Find the area in the set quadrant that is
outside of r = 2 and inside the rose R = 4sin(2).

36. Given polar functions
R = f() and r = g() where
R ≥ r for  between A and B,
the area between them is:
Outer R = f()
 = A = B
Inner r = g()
Area in Polar Coordinates
∫ f2() – g2() dor
=A
B
∫R2 – r2 d½
=A
B
½
Example D. Find the area in the set quadrant that is
outside of r = 2 and inside the rose R = 4sin(2).
R = 4sin(2)
r = 2
The graph of R = 4sin(2) is a 4-petal
rose and r = 2 is the circle of radius 2
as shown.

37. Given polar functions
R = f() and r = g() where
R ≥ r for  between A and B,
the area between them is:
Outer R = f()
 = A = B
Inner r = g()
Area in Polar Coordinates
∫ f2() – g2() dor
=A
B
∫R2 – r2 d½
=A
B
½
Example D. Find the area in the set quadrant that is
outside of r = 2 and inside the rose R = 4sin(2).
R = 4sin(2)
r = 2
The graph of R = 4sin(2) is a 4-petal
rose and r = 2 is the circle of radius 2
as shown. The important step here is
to determine the angles A and B
of the points of intersection.
 = A
 = B

38. Setting 4sin(2) = 2 or sin(2) = ½,
hence 2 = /6 and 5/6,
Area in Polar Coordinates
R = 4sin(2)
r = 2
 = A
 = B

39. Setting 4sin(2) = 2 or sin(2) = ½,
hence 2 = /6 and 5/6,
and that A = /12 and B = 5/12.
Area in Polar Coordinates
R = 4sin(2)
r = 2
 = A
 = B

40. Setting 4sin(2) = 2 or sin(2) = ½,
hence 2 = /6 and 5/6,
and that A = /12 and B = 5/12.
Area in Polar Coordinates
R = 4sin(2)
r = 2
 = A
 = B
Have the “outer” area for one petal is
∫(4sin(2))2 – 22 d½
/12
5/12

41. Setting 4sin(2) = 2 or sin(2) = ½,
hence 2 = /6 and 5/6,
and that A = /12 and B = 5/12.
Area in Polar Coordinates
R = 4sin(2)
r = 2
 = A
 = B
Have the “outer” area for one petal is
∫(4sin(2))2 – 22 d½
= ∫ 8sin2(2) – 2 d
/12
5/12
/12
5/12

42. Setting 4sin(2) = 2 or sin(2) = ½,
hence 2 = /6 and 5/6,
and that A = /12 and B = 5/12.
Area in Polar Coordinates
R = 4sin(2)
r = 2
 = A
 = B
and the total outer area is 4*(2/3 + 3).
Have the “outer” area for one petal is
∫(4sin(2))2 – 22 d½
= ∫ 8sin2(2) – 2 d = 2/3 + 3
/12
5/12
/12
5/12