76
Area in Polar Coordinates F r a n k M a 2 0 0 6
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Objective:
Integral formula of the area swept
out by a polar function between
two angles
Area in Polar Coordinates
Given a polar function
r = f() with A < < B,
it "sweeps" out an
area between them;
Area in Polar Coordinates
Given a polar function
r = f() with A < < B,
it "sweeps" out an
area between them;
r=f()
=A
=B
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Given a polar function
r = f() with A < < B,
it "sweeps" out an
area between them;
r=f()
=A
The Polar Area Formula:
The area swept out by r = f() from
=A to =B is:
B
A
B
Adfordr )(
2
1
2
1 22
=B
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = k with
0 < < 2π.
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = k with
0 < < 2π.
k
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = k with
0 < < 2π.
We have:
2
0
2
2
1dk
k
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = k with
0 < < 2π.
We have:
2
0
2
2
0
2
|2
1
2
1
k
dk
k
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = k with
0 < < 2π.
We have:
22
2
0
2
2
0
2
)2(2
1
|2
1
2
1
kk
k
dk
k
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
cos2(A) =
From cosine double-angle formulas, we get the following:
1 + cos(2A)
2
sin2(A) =1 – cos(2A)
2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
d)(cos2
cos2(A) =
From cosine double-angle formulas, we get the following:
1 + cos(2A)
2
sin2(A) =1 – cos(2A)
2
We use these formulas for the integral:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
dd 1)2cos(2
1)(cos2
cos2(A) =
From cosine double-angle formulas, we get the following:
1 + cos(2A)
2
sin2(A) =1 – cos(2A)
2
We use these formulas for the integral:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
c
dd
)))2sin(2
1(
2
1
1)2cos(2
1)(cos2
cos2(A) =
From cosine double-angle formulas, we get the following:
1 + cos(2A)
2
sin2(A) =1 – cos(2A)
2
We use these formulas for the integral:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
cc
dd
2
1)2sin(
4
1)))2sin(
2
1(
2
1
1)2cos(2
1)(cos2
cos2(A) =
From cosine double-angle formulas, we get the following:
1 + cos(2A)
2
sin2(A) =1 – cos(2A)
2
We use these formulas for the integral:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
cc
dd
2
1)2sin(
4
1)))2sin(
2
1(
2
1
1)2cos(2
1)(cos2
cos2(A) =
From cosine double-angle formulas, we get the following:
1 + cos(2A)
2
sin2(A) =1 – cos(2A)
2
We use these formulas for the integral:
We integrate sin2(x) similarly.
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
cc
dd
2
1)2sin(
4
1)))2sin(
2
1(
2
1
1)2cos(2
1)(cos2
cos2(A) =
From cosine double-angle formulas, we get the following:
1 + cos(2A)
2
sin2(A) =1 – cos(2A)
2
We use these formulas for the integral:
We integrate sin2(x) similarly. These are useful in polar
area calculations due to f2() in the integrand.
Area in Polar CoordinatesExample:
Find the area enclosed
by r = f() = 2sin().
Area in Polar CoordinatesExample:
Find the area enclosed
by r = f() = 2sin().
1
Area in Polar CoordinatesExample:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π. =π =0
1
Area in Polar CoordinatesExample:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
1
=π =0
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
So We have:
0
2))sin(2(2
1d
1
=π =0
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
So We have:
0
2
0
2 )(sin42
1))sin(2(
2
1dd
1
=π =0
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
So We have:
0
0
2
0
2
2
)2cos(
2
12
)(sin42
1))sin(2(
2
1
d
dd
1
=π =0
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
So We have:
00
0
2
0
2
|)2sin(2
1
2
)2cos(
2
12
)(sin42
1))sin(2(
2
1
d
dd
1
=π =0
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
So We have:
00
0
2
0
2
|)2sin(2
1
2
)2cos(
2
12
)(sin42
1))sin(2(
2
1
d
dd
1
=π =0
Area of a circle with r=1
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
So We have:
00
0
2
0
2
|)2sin(2
1
2
)2cos(
2
12
)(sin42
1))sin(2(
2
1
d
dd
1
=π =0
If we integrate from 0 to 2π, we get 2 for the area.
Area of a circle with r=1
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed
by r = f() = 2sin(). The
circle is traced once
round with from 0 to π.
So We have:
00
0
2
0
2
|)2sin(2
1
2
)2cos(
2
12
)(sin42
1))sin(2(
2
1
d
dd
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 of a circle with r=1
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < < π2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < < π2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < < π2
=0
=π/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < < π2
=0
=π/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < <
We have:
2/
0
2))cos(1(2
1
d
π2
=0
=π/2
Area in Polar Coordinates
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < <
We have:
2/
0
22/
0
2 )(cos)cos(212
1))cos(1(
2
1
dd
π2
=0
=π/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < <
We have:
2/
0
2/
0
22/
0
2
|)2
1)2sin(
4
1)sin(2(
2
1
)(cos)cos(212
1))cos(1(
2
1
dd
π2
=0
=π/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < <
We have:
2/
0
2/
0
2/
0
22/
0
2
|))2sin(4
1)sin(2
2
3(
2
1
|)2
1)2sin(
4
1)sin(2(
2
1
)(cos)cos(212
1))cos(1(
2
1
dd
π2
=0
=π/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the area enclosed by
r = f() = 1 – cos() with
0 < <
We have:
18
3)01*2
2*
2
3(
2
1|))2sin(
4
1)sin(2
2
3(
2
1
|)2
1)2sin(
4
1)sin(2(
2
1
)(cos)cos(212
1))cos(1(
2
1
2/
0
2/
0
2/
0
22/
0
2
dd
π2
=0
=π/2
Area in Polar CoordinatesGiven polar functions
R = f() and r = g()
where R > r for
between A and B ,
Area in Polar CoordinatesGiven polar functions
R = f() and r = g()
where R > r for
between A and B ,
R = f()
r = g()
= A
= B
outer
inner
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Given polar functions
R = f() and r = g()
where R > r for
between A and B , the
area between them is:
R = f()
r = g()
B
A
B
AdgfordrR )()(
2
1
2
1 2222
= A
= B
outer
inner
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Given polar functions
R = f() and r = g()
where R > r for
between A and B , the
area between them is:
R = f()
r = g()
B
A
B
AdgfordrR )()(
2
1
2
1 2222
= A
= B
To find area enclosed by two polar graphs,
the most important step is to determine the
angles A and B.
outer
inner
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Given polar functions
R = f() and r = g()
where R > r for
between A and B , the
area between them is:
R = f()
r = g()
B
A
B
AdgfordrR )()(
2
1
2
1 2222
= A
= B
To find area enclosed by two polar graphs,
the most important step is to determine the
angles A and B. This is done both
algebraically and graphically.
outer
inner
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
Example:
Find the shaded area
shown in the figure.
R = 2cos()r = 1
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
R = 2cos()r = 1
Example:
Find the shaded area
shown in the figure.
Method I: The area consists
of two regions:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
R = 2cos()r = 1
1. the red area with r=1 as boundary
2. the black area with R=2cos() as the boundary
Example:
Find the shaded area
shown in the figure.
Method I: The area consists
of two regions:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
R = 2cos()r = 1
1. the red area with r=1 as boundary
2. the black area with R=2cos() as the boundary
The angle of intersection is in the 1st quad where
1 = 2cos()
Example:
Find the shaded area
shown in the figure.
Method I: The area consists
of two regions:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
R = 2cos()r = 1
1. the red area with r=1 as boundary
2. the black area with R=2cos() as the boundary
The angle of intersection is in the 1st quad where
1 = 2cos() cos()=1/2 so =π/3.
Example:
Find the shaded area
shown in the figure.
Method I: The area consists
of two regions:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
R = 2cos()r = 1
1. the red area with r=1 as boundary
2. the black area with R=2cos() as the boundary
The angle of intersection is in the 1st quad where
1 = 2cos() cos()=1/2 so =π/3.
Therefore the red area is bounded by = 0 to =/3,
=0
=/3
Example:
Find the shaded area
shown in the figure.
Method I: The area consists
of two regions:
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
R = 2cos()r = 1
1. the red area with r=1 as boundary
2. the black area with R=2cos() as the boundary
The angle of intersection is in the 1st quad where
1 = 2cos() cos()=1/2 so =π/3.
Therefore the red area is bounded by = 0 to =/3,
=0
=/3
and the black area is bounded by =/3 to = /2.
=/2
Example:
Find the shaded area
shown in the figure.
Method I: The area consists
of two regions:
Area in Polar CoordinatesThe red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
R = 2cos()r = 1
=0
=/3=/2
Area in Polar CoordinatesThe red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
R = 2cos()r = 1
=0
=/3=/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
The red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
2/
3/
2))cos(2(2
1
d
R = 2cos()r = 1
=0
=/3=/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
The red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
2/
3/
2
2/
3/
2
)(cos2
))cos(2(2
1
d
d
R = 2cos()r = 1
=0
=/3=/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
The red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
2/
3/
2/
3/
2
2/
3/
2
|)2
1)2sin(
4
1(2)(cos2
))cos(2(2
1
d
d
R = 2cos()r = 1
=0
=/3=/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
The red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
2/
3/
2/
3/
2/
3/
2
2/
3/
2
|)2sin(2
1
|)2
1)2sin(
4
1(2)(cos2
))cos(2(2
1
d
d
R = 2cos()r = 1
=0
=/3=/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
The red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
)3
)3
2sin(
2
1()
20(|)2sin(
2
1
|)2
1)2sin(
4
1(2)(cos2
))cos(2(2
1
2/
3/
2/
3/
2/
3/
2
2/
3/
2
d
d
R = 2cos()r = 1
=0
=/3=/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
The red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
4
3
6)
3)
3
2sin(
2
1()
20(|)2sin(
2
1
|)2
1)2sin(
4
1(2)(cos2
))cos(2(2
1
2/
3/
2/
3/
2/
3/
2
2/
3/
2
d
d
R = 2cos()r = 1
=0
=/3=/2
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
The red area bounded from
= 0 to =/3 is 1/6 of the
unit circle. It's area is /6.
The black area are bounded
by =/3 to = /2 is:
4
3
6)
3)
3
2sin(
2
1()
20(|)2sin(
2
1
|)2
1)2sin(
4
1(2)(cos2
))cos(2(2
1
2/
3/
2/
3/
2/
3/
2
2/
3/
2
d
d
The total area is 3
– 34
R = 2cos()r = 1
=0
=/3=/2
6– 3
46
+ =
Area in Polar CoordinatesR = 2cos()r = 1Method II: The blue area is
¼ of a circle minus the red
area.
Area in Polar CoordinatesR = 2cos()r = 1Method II: The blue area is
¼ of a circle minus the red
area. The red area is
bounded by f()=1 and
g()=2cos() where goes
from /3 to /2.
=/3=/2
Area in Polar CoordinatesR = 2cos()r = 1Method II: The blue area is
¼ of a circle minus the red
area. The red area is
bounded by f()=1 and
g()=2cos() where goes
from /3 to /2. It's area is:
2/
3/
22 ))cos(2(12
1
d
=/3=/2
outer inner
Area in Polar CoordinatesR = 2cos()r = 1Method II: The blue area is
¼ of a circle minus the red
area. The red area is
bounded by f()=1 and
g()=2cos() where goes
from /3 to /2. It's area is:
2/
3/
22/
3/
2
2/
3/
22
)(cos212
1
))cos(2(12
1
dd
d
=/3=/2
outer inner
Area in Polar CoordinatesR = 2cos()r = 1Method II: The blue area is
¼ of a circle minus the red
area. The red area is
bounded by f()=1 and
g()=2cos() where goes
from /3 to /2. It's area is:
)4
3
6()
32(
2
1
)(cos212
1
))cos(2(12
1
2/
3/
22/
3/
2
2/
3/
22
dd
d
=/3=/2
outer inner
Area in Polar CoordinatesR = 2cos()r = 1Method II: The blue area is
¼ of a circle minus the red
area. The red area is
bounded by f()=1 and
g()=2cos() where goes
from /3 to /2. It's area is:
4
3
12)
4
3
6()
32(
2
1
)(cos212
1
))cos(2(12
1
2/
3/
22/
3/
2
2/
3/
22
dd
d
=/3=/2
outer inner
Area in Polar Coordinates
F
r
a
n
k
M
a
2
0
0
6
R = 2cos()r = 1Method II: The blue area is
¼ of a circle minus the red
area. The red area is
bounded by f()=1 and
g()=2cos() where goes
from /3 to /2. It's area is:
4
3
12)
4
3
6()
32(
2
1
)(cos212
1
))cos(2(12
1
2/
3/
22/
3/
2
2/
3/
22
dd
d
=/3=/2
So the blue area is
12+ 3
4
4– (– )
3
– 34
=
outer inner
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
We need to find the angles of intersections that bound
the area.
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
We need to find the angles of intersections that bound
the area.
Set 4sin(2) =2 or sin(2) = ½
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
We need to find the angles of intersections that bound
the area.
Set 4sin(2) =2 or sin(2) = ½
Hence one answer is 2 = /6, or = /12,
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
We need to find the angles of intersections that bound
the area.
Set 4sin(2) =2 or sin(2) = ½
Hence one answer is 2 = /6, or = /12, and the
other is = 5/12.
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
We need to find the angles of intersections that bound
the area.
Set 4sin(2) =2 or sin(2) = ½
Hence one answer is 2 = /6, or = /12, and the
other is = 5/12. Hnce the area is:
12/5
12/
22 2))2sin(4(2
1
d
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
We need to find the angles of intersections that bound
the area.
Set 4sin(2) =2 or sin(2) = ½
Hence one answer is 2 = /6, or = /12, and the
other is = 5/12. Hence the area is:
12/5
12/
212/5
12/
22 2)2(sin82))2sin(4(2
1
dd
Area in Polar CoordinatesFind the area that is outside
of r = 2 and inside of
R = 4sin(2).
We only need to find one of the
four equal regions as shaded.
We need to find the angles of intersections that bound
the area.
Set 4sin(2) =2 or sin(2) = ½
Hence one answer is 2 = /6, or = /12, and the
other is = 5/12. Hnce the area is:
33
22)2(sin82))2sin(4(
2
1 12/5
12/
212/5
12/
22
dd
The total area is 4*(2/3 + 3)
