Polar coordinates (Sect. 11.3)
I
Review: Arc-length of a curve.
I
Polar coordinates definition.
I
Transformation rules Polar-Cartesian.
I
Examples of curves in polar coordinates.
Review: Arc-length of a curve
Definition
A curve on the plane is given
in parametric form iff it is given by
the set of points x(t), y (t) , where the parameter t ∈ I ⊂ R.
Review: Arc-length of a curve
Definition
A curve on the plane is given
in parametric form iff it is given by
the set of points x(t), y (t) , where the parameter t ∈ I ⊂ R.
Remark: If the curve y = f (x) can be described by the parametric
functions x(t), y (t) , for t ∈ I ⊂ R, and if x 0 (t) 6= 0 for t ∈ I ,
then holds
df
(dy /dt)
=
.
dx
(dx/dt)
Review: Arc-length of a curve
Definition
A curve on the plane is given
in parametric form iff it is given by
the set of points x(t), y (t) , where the parameter t ∈ I ⊂ R.
Remark: If the curve y = f (x) can be described by the parametric
functions x(t), y (t) , for t ∈ I ⊂ R, and if x 0 (t) 6= 0 for t ∈ I ,
then holds
df
(dy /dt)
=
.
dx
(dx/dt)
Remark: The
arc-length of a continuously differentiable curve
x(t), y (y ) , for t ∈ [a, b] is the number
Z bq
2 2
L=
x 0 (t) + y 0 (t) dt.
a
Review: Arc-length of a curve
Remark:
I
The formula for the arc-length
Z
L=
b
q
2 2
x 0 (t) + y 0 (t) dt.
a
can be used in a curve of the form y = f (x).
Review: Arc-length of a curve
Remark:
I
The formula for the arc-length
Z
L=
b
q
2 2
x 0 (t) + y 0 (t) dt.
a
can be used in a curve of the form y = f (x).
I
Just choose the trivial parametrization:
x(t) = t,
Review: Arc-length of a curve
Remark:
I
The formula for the arc-length
Z
L=
b
q
2 2
x 0 (t) + y 0 (t) dt.
a
can be used in a curve of the form y = f (x).
I
Just choose the trivial parametrization:
x(t) = t,
y (t) = f (t).
Review: Arc-length of a curve
Remark:
I
The formula for the arc-length
Z
L=
b
q
2 2
x 0 (t) + y 0 (t) dt.
a
can be used in a curve of the form y = f (x).
I
Just choose the trivial parametrization:
x(t) = t,
I
Then x 0 (t) = 1, y 0 (t) = f 0 (t),
y (t) = f (t).
Review: Arc-length of a curve
Remark:
I
The formula for the arc-length
Z
b
L=
q
2 2
x 0 (t) + y 0 (t) dt.
a
can be used in a curve of the form y = f (x).
I
Just choose the trivial parametrization:
x(t) = t,
I
y (t) = f (t).
Then x 0 (t) = 1, y 0 (t) = f 0 (t), and the arc-length formula is
Z
L=
a
b
q
2
1 + f 0 (t) dt.
Polar coordinates (Sect. 11.3)
I
Review: Arc-length of a curve.
I
Polar coordinates definition.
I
Transformation rules Polar-Cartesian.
I
Examples of curves in polar coordinates.
Polar coordinates definition
Definition
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
P = ( r, 0 ) = (x,y)
y
r
0
x
Polar coordinates definition
Definition
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
Example
Graph the points P1 = (1, π/4),
P2 = (1, 3π/4).
P = ( r, 0 ) = (x,y)
y
r
0
x
Polar coordinates definition
Definition
P = ( r, 0 ) = (x,y)
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
y
x
Graph the points P1 = (1, π/4),
P2 = (1, 3π/4).
P1
y
P2
Example
r
0
r2 = 1
0 = 3pi/4
2
02
r1 = 1
0 1 = pi/4
x
Polar coordinates definition
Definition
P = ( r, 0 ) = (x,y)
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
y
x
Graph the points P1 = (1, π/4),
P2 = (1, 3π/4).
Example
Graph the points P1 = (1, π/4),
P3 = (1/2, 5π/4).
P1
y
P2
Example
r
0
r2 = 1
0 = 3pi/4
2
02
r1 = 1
0 1 = pi/4
x
Polar coordinates definition
Definition
P = ( r, 0 ) = (x,y)
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
y
x
02
r2 = 1
Graph the points P1 = (1, π/4),
P2 = (1, 3π/4).
Graph the points P1 = (1, π/4),
P3 = (1/2, 5π/4).
r1 = 1
0 1 = pi/4
x
0 = 3pi/4
2
y
Example
P1
y
P2
Example
r
0
P1
r1 = 1
0 = 5pi/4
3
r = 1/2
3
P3
0 1 = pi/4
x
Polar coordinates definition
Remark: The polar coordinates (r , θ)
are restricted to r > 0 and θ ∈ [0, 2π).
P = ( r, 0 ) = (x,y)
y
r
0
x
Polar coordinates definition
Remark: The polar coordinates (r , θ)
are restricted to r > 0 and θ ∈ [0, 2π).
P = ( r, 0 ) = (x,y)
y
r
0
x
Remark:
I
This restriction implies that for every point P 6= (0, 0) there is
a unique pair (r , θ) to label that point.
Polar coordinates definition
Remark: The polar coordinates (r , θ)
are restricted to r > 0 and θ ∈ [0, 2π).
P = ( r, 0 ) = (x,y)
y
r
0
x
Remark:
I
This restriction implies that for every point P 6= (0, 0) there is
a unique pair (r , θ) to label that point.
I
Usually this restriction is not applied, and r ∈ R, θ ∈ R.
Polar coordinates definition
Remark: The polar coordinates (r , θ)
are restricted to r > 0 and θ ∈ [0, 2π).
P = ( r, 0 ) = (x,y)
y
r
0
x
Remark:
I
This restriction implies that for every point P 6= (0, 0) there is
a unique pair (r , θ) to label that point.
I
Usually this restriction is not applied, and r ∈ R, θ ∈ R.
I
This means that infinitely many ordered pairs (r , θ) label the
same point P.
Polar coordinates definition
Remark: The polar coordinates (r , θ)
are restricted to r > 0 and θ ∈ [0, 2π).
P = ( r, 0 ) = (x,y)
y
r
0
x
Remark:
I
This restriction implies that for every point P 6= (0, 0) there is
a unique pair (r , θ) to label that point.
I
Usually this restriction is not applied, and r ∈ R, θ ∈ R.
I
This means that infinitely many ordered pairs (r , θ) label the
same point P.
Example
Graph the points P1 = (1, π/4) and
P2 = (1, −7π/4).
Polar coordinates definition
Remark: The polar coordinates (r , θ)
are restricted to r > 0 and θ ∈ [0, 2π).
P = ( r, 0 ) = (x,y)
y
r
0
x
Remark:
I
This restriction implies that for every point P 6= (0, 0) there is
a unique pair (r , θ) to label that point.
I
Usually this restriction is not applied, and r ∈ R, θ ∈ R.
I
This means that infinitely many ordered pairs (r , θ) label the
same point P.
y
Graph the points P1 = (1, π/4) and
P2 = (1, −7π/4).
P1 = P2
r=1
Example
0 = pi/4
x
0 = −7pi/4
Polar coordinates definition
Example
Graph the points P1 = (1, π/4),
P2 = (−1/2, π/4), and
P3 = (1/2, 5π/4).
Polar coordinates definition
P1
y
Example
Graph the points P1 = (1, π/4),
P2 = (−1/2, π/4), and
P3 = (1/2, 5π/4).
r1 = 1
0 = 5pi/4
0 1 = pi/4
3
x
r = 1/2
3
r = −1/2
2
P3 = P
2
Polar coordinates definition
P1
y
Example
r1 = 1
Graph the points P1 = (1, π/4),
P2 = (−1/2, π/4), and
P3 = (1/2, 5π/4).
0 = 5pi/4
0 1 = pi/4
3
x
r = 1/2
3
r = −1/2
2
P3 = P
2
Remark: Polar coordinates are well adapted to describe circular
curves and disk sections.
y
x
Polar coordinates (Sect. 11.3)
I
Review: Arc-length of a curve.
I
Polar coordinates definition.
I
Transformation rules Polar-Cartesian.
I
Examples of curves in polar coordinates.
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
The polar coordinates of a point P = (x, y ) in the first and fourth
quadrants are given by
y p
r = x 2 + y 2 , θ = arctan
.
x
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
The polar coordinates of a point P = (x, y ) in the first and fourth
quadrants are given by
y p
r = x 2 + y 2 , θ = arctan
.
x
Proof: x 2 + y 2
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
The polar coordinates of a point P = (x, y ) in the first and fourth
quadrants are given by
y p
r = x 2 + y 2 , θ = arctan
.
x
Proof: x 2 + y 2 = r 2 cos2 (θ) + r 2 sin2 (θ)
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
The polar coordinates of a point P = (x, y ) in the first and fourth
quadrants are given by
y p
r = x 2 + y 2 , θ = arctan
.
x
Proof: x 2 + y 2 = r 2 cos2 (θ) + r 2 sin2 (θ) = r 2 ; r > 0 implies
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
The polar coordinates of a point P = (x, y ) in the first and fourth
quadrants are given by
y p
r = x 2 + y 2 , θ = arctan
.
x
2
2
2
2
2
2
2
Proof:
p x + y = r cos (θ) + r sin (θ) = r ; r > 0 implies
r=
x 2 + y 2.
Transformation rules Polar-Cartesian.
Remark: The polar coordinates (r , θ), with
r > 0 and θ ∈ (−π, π] can be related to
Cartesian coordinates.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
The polar coordinates of a point P = (x, y ) in the first and fourth
quadrants are given by
y p
r = x 2 + y 2 , θ = arctan
.
x
2
2
2
2
2
2
2
Proof:
p x + y = r cos (θ) + r sin (θ) = r ; r > 0 implies
r=
x 2 + y 2 . Finally, x/y = tan(θ).
Transformation rules Polar-Cartesian.
Remark:
I
If (x, y ) satisfies either x > 0, y > 0, or x 6 0, y 6 0, then
θ = arctan(x/y ) is in the first quadrant.
Transformation rules Polar-Cartesian.
Remark:
I
If (x, y ) satisfies either x > 0, y > 0, or x 6 0, y 6 0, then
θ = arctan(x/y ) is in the first quadrant.
I
If (x, y ) satisfies either x > 0, y 6 0, or x 6 0, y > 0, then
θ = arctan(x/y ) is in the fourth quadrant.
Transformation rules Polar-Cartesian.
Remark:
I
If (x, y ) satisfies either x > 0, y > 0, or x 6 0, y 6 0, then
θ = arctan(x/y ) is in the first quadrant.
I
If (x, y ) satisfies either x > 0, y 6 0, or x 6 0, y > 0, then
θ = arctan(x/y ) is in the fourth quadrant.
y
y
y
−y
−x
x
−y
x
x
−x
y
x
Polar coordinates (Sect. 11.3)
I
Review: Arc-length of a curve.
I
Polar coordinates definition.
I
Transformation rules Polar-Cartesian.
I
Examples of curves in polar coordinates.
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Solution: In Cartesian coordinates the equation is
x 2 + y 2 = 32 ,
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Solution: In Cartesian coordinates the equation is
x 2 + y 2 = 32 ,
r=
p
x2 + y2
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Solution: In Cartesian coordinates the equation is
(
p
r = 3,
x 2 + y 2 = 32 , r = x 2 + y 2 ⇒
θ ∈ [0, 2π).
C
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Solution: In Cartesian coordinates the equation is
(
p
r = 3,
x 2 + y 2 = 32 , r = x 2 + y 2 ⇒
θ ∈ [0, 2π).
Example
Find the equation in polar coordinates of the line y =
√
3 x.
C
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Solution: In Cartesian coordinates the equation is
(
p
r = 3,
x 2 + y 2 = 32 , r = x 2 + y 2 ⇒
θ ∈ [0, 2π).
Example
Find the equation in polar coordinates of the line y =
Solution: From the transformation laws,
θ = arctan(y /x)
√
3 x.
C
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Solution: In Cartesian coordinates the equation is
(
p
r = 3,
x 2 + y 2 = 32 , r = x 2 + y 2 ⇒
θ ∈ [0, 2π).
Example
Find the equation in polar coordinates of the line y =
Solution: From the transformation laws,
θ = arctan(y /x) = arctan
√ 3
√
3 x.
C
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of a circle radius 3 at (0, 0).
Solution: In Cartesian coordinates the equation is
(
p
r = 3,
x 2 + y 2 = 32 , r = x 2 + y 2 ⇒
θ ∈ [0, 2π).
Example
Find the equation in polar coordinates of the line y =
√
C
3 x.
Solution: From the transformation laws,
√ θ = arctan(y /x) = arctan 3
(
⇒
θ = π/3,
r ∈ R,
C
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
x 2 + y 2 − 6y + 9 = 9
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
x 2 + y 2 − 6y + 9 = 9
⇒
x 2 + y 2 = 6y .
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
x 2 + y 2 − 6y + 9 = 9
⇒
Recall: x = r cos(θ), and y = r sin(θ),
x 2 + y 2 = 6y .
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
x 2 + y 2 − 6y + 9 = 9
⇒
Recall: x = r cos(θ), and y = r sin(θ),
therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 6y .
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
x 2 + y 2 − 6y + 9 = 9
⇒
Recall: x = r cos(θ), and y = r sin(θ),
therefore x 2 + y 2 = r 2 ,
r 2 = 6r sin(θ)
x 2 + y 2 = 6y .
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
x 2 + y 2 − 6y + 9 = 9
⇒
Recall: x = r cos(θ), and y = r sin(θ),
therefore x 2 + y 2 = r 2 ,
r 2 = 6r sin(θ)
⇒
r = 6 sin(θ),
x 2 + y 2 = 6y .
Examples of curves in polar coordinates
Example
Find the equation in polar coordinates of the circle
x 2 + (y − 3)2 = 9.
Solution: Expand the square in the equation of the circle,
x 2 + y 2 − 6y + 9 = 9
⇒
Recall: x = r cos(θ), and y = r sin(θ),
therefore x 2 + y 2 = r 2 ,
r 2 = 6r sin(θ)
⇒
r = 6 sin(θ),
x 2 + y 2 = 6y .
y
r(0)=6 sin(0)
6
0
and θ ∈ [0, π].
C
x
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation,
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ),
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
Complete the square:
⇒
x 2 − 4x + y 2 = 0.
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
(x − 2)2 + y 2 = 4.
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
(x − 2)2 + y 2 = 4.
This is the equation of a circle radius
r = 2 with center at (2, 0).
Examples of curves in polar coordinates
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
y
r(0) = 4 cos(0)
(x − 2)2 + y 2 = 4.
0
This is the equation of a circle radius
C
r = 2 with center at (2, 0).
2
x
Graphing in polar coordinates (Sect. 11.4)
I
Review: Polar coordinates.
I
Review: Transforming back to Cartesian.
I
Computing the slope of tangent lines.
I
Using symmetry to graph curves.
Examples:
I
I
I
I
Circles in polar coordinates.
Graphing the Cardiod.
Graphing the Lemniscate.
Review: POlar coordinates
Definition
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
P = ( r, 0 ) = (x,y)
y
r
0
x
Review: POlar coordinates
Definition
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
Review: POlar coordinates
Definition
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
P = ( r, 0 ) = (x,y)
y
r
0
x
Theorem (Cartesian-polar transformations)
The Cartesian coordinates of a point P = (r , θ) are given by
x = r cos(θ),
y = r sin(θ).
The polar coordinates of a point P = (x, y ) in the first and fourth
quadrants are given by
y p
r = x 2 + y 2 , θ = arctan
.
x
Graphing in polar coordinates (Sect. 11.4)
I
Review: Polar coordinates.
I
Review: Transforming back to Cartesian.
I
Computing the slope of tangent lines.
I
Using symmetry to graph curves.
Examples:
I
I
I
I
Circles in polar coordinates.
Graphing the Cardiod.
Graphing the Lemniscate.
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation,
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ),
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
Complete the square:
⇒
x 2 − 4x + y 2 = 0.
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
(x − 2)2 + y 2 = 4.
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
(x − 2)2 + y 2 = 4.
This is the equation of a circle radius
r = 2 with center at (2, 0).
Review: Transforming back to Cartesian
Example
Find the equation of the curve in Cartesian coordinates for
r = 4 cos(θ), for θ ∈ [−π/2, π/2].
Solution: Multiply by r the whole equation, r 2 = 4r cos(θ).
Recall: x = r cos(θ), and y = r sin(θ), therefore x 2 + y 2 = r 2 ,
x 2 + y 2 = 4x
⇒
x 2 − 4x + y 2 = 0.
Complete the square:
4
2
x −2
x + 4 − 4 + y2 = 0
2
y
r(0) = 4 cos(0)
(x − 2)2 + y 2 = 4.
0
This is the equation of a circle radius
C
r = 2 with center at (2, 0).
2
x
Graphing in polar coordinates (Sect. 11.4)
I
Review: Polar coordinates.
I
Review: Transforming back to Cartesian.
I
Computing the slope of tangent lines.
I
Using symmetry to graph curves.
Examples:
I
I
I
I
Circles in polar coordinates.
Graphing the Cardiod.
Graphing the Lemniscate.
Computing the slope of tangent lines
Recall: The slope of the line tangent to the curve y = f (x), can
be written in terms of (x(t), y (t)) as follows
df
dy /dt
=
.
dx
dx/dt
Computing the slope of tangent lines
Recall: The slope of the line tangent to the curve y = f (x), can
be written in terms of (x(t), y (t)) as follows
df
dy /dt
=
.
dx
dx/dt
Remark: If the curve is given in polar coordinates, r = r (θ),
Computing the slope of tangent lines
Recall: The slope of the line tangent to the curve y = f (x), can
be written in terms of (x(t), y (t)) as follows
df
dy /dt
=
.
dx
dx/dt
Remark: If the curve is given in polar coordinates, r = r (θ), then
x(θ) = r (θ) cos(θ)
y (θ) = r (θ) sin(θ).
Computing the slope of tangent lines
Recall: The slope of the line tangent to the curve y = f (x), can
be written in terms of (x(t), y (t)) as follows
df
dy /dt
=
.
dx
dx/dt
Remark: If the curve is given in polar coordinates, r = r (θ), then
x(θ) = r (θ) cos(θ)
The formula for the slope is then
df
y 0 (θ)
= 0
dx
x (θ)
y (θ) = r (θ) sin(θ).
Computing the slope of tangent lines
Recall: The slope of the line tangent to the curve y = f (x), can
be written in terms of (x(t), y (t)) as follows
df
dy /dt
=
.
dx
dx/dt
Remark: If the curve is given in polar coordinates, r = r (θ), then
x(θ) = r (θ) cos(θ)
y (θ) = r (θ) sin(θ).
The formula for the slope is then
df
y 0 (θ)
= 0
dx
x (θ)
⇒
r 0 (θ) sin(θ) + r (θ) cos(θ)
df
= 0
.
dx
r (θ) cos(θ) − r (θ) sin(θ)
Computing the slope of tangent lines
Recall: The slope of the line tangent to the curve y = f (x), can
be written in terms of (x(t), y (t)) as follows
df
dy /dt
=
.
dx
dx/dt
Remark: If the curve is given in polar coordinates, r = r (θ), then
x(θ) = r (θ) cos(θ)
y (θ) = r (θ) sin(θ).
The formula for the slope is then
df
y 0 (θ)
= 0
dx
x (θ)
⇒
r 0 (θ) sin(θ) + r (θ) cos(θ)
df
= 0
.
dx
r (θ) cos(θ) − r (θ) sin(θ)
If the curve passes through the origin, r (θ0 ) = 0, then
df r 0 (θ0 ) sin(θ0 )
.
= 0
dx θ0
r (θ0 ) cos(θ0 )
Graphing in polar coordinates (Sect. 11.4)
I
Review: Polar coordinates.
I
Review: Transforming back to Cartesian.
I
Computing the slope of tangent lines.
I
Using symmetry to graph curves.
Examples:
I
I
I
I
Circles in polar coordinates.
Graphing the Cardiod.
Graphing the Lemniscate.
Using symmetry to graph curves
Remark: If a curve is symmetric under reflections about the x-axis,
or the y -axis, or the origin, then the work needed to graph of the
curve can be reduced.
Using symmetry to graph curves
Remark: If a curve is symmetric under reflections about the x-axis,
or the y -axis, or the origin, then the work needed to graph of the
curve can be reduced.
I
x-axis symmetry: (r , θ) and (r , −θ) belong to the graph.
Using symmetry to graph curves
Remark: If a curve is symmetric under reflections about the x-axis,
or the y -axis, or the origin, then the work needed to graph of the
curve can be reduced.
I
x-axis symmetry: (r , θ) and (r , −θ) belong to the graph.
I
Origin symmetry: (r , θ) and (−r , θ) belong to the graph.
Using symmetry to graph curves
Remark: If a curve is symmetric under reflections about the x-axis,
or the y -axis, or the origin, then the work needed to graph of the
curve can be reduced.
I
x-axis symmetry: (r , θ) and (r , −θ) belong to the graph.
I
Origin symmetry: (r , θ) and (−r , θ) belong to the graph.
I
y -axis symmetry: (r , θ) and (−r , −θ) belong to the graph.
Using symmetry to graph curves
Remark: If a curve is symmetric under reflections about the x-axis,
or the y -axis, or the origin, then the work needed to graph of the
curve can be reduced.
I
x-axis symmetry: (r , θ) and (r , −θ) belong to the graph.
I
Origin symmetry: (r , θ) and (−r , θ) belong to the graph.
I
y -axis symmetry: (r , θ) and (−r , −θ) belong to the graph.
y
(−r, − 0)
(r, 0)
x
(−r, 0)
(r, − 0)
Graphing in polar coordinates (Sect. 11.4)
I
Review: Polar coordinates.
I
Review: Transforming back to Cartesian.
I
Computing the slope of tangent lines.
I
Using symmetry to graph curves.
Examples:
I
I
I
I
Circles in polar coordinates.
Graphing the Cardiod.
Graphing the Lemniscate.
Circles in polar coordinates
Remark: Circles centered at the origin are trivial to graph.
Circles in polar coordinates
Remark: Circles centered at the origin are trivial to graph.
Example
Graph the curve r = 2, θ ∈ [0, 2π).
Circles in polar coordinates
Remark: Circles centered at the origin are trivial to graph.
y
Example
Graph the curve r = 2, θ ∈ [0, 2π).
2
x
Circles in polar coordinates
Remark: Circles centered at the origin are trivial to graph.
y
Example
Graph the curve r = 2, θ ∈ [0, 2π).
2
x
Remark:
Circles not centered at the origin are more complicated to graph.
Circles in polar coordinates
Remark: Circles centered at the origin are trivial to graph.
y
Example
Graph the curve r = 2, θ ∈ [0, 2π).
2
x
Remark:
Circles not centered at the origin are more complicated to graph.
Example
Graph the curve
r = 4 cos(θ), θ ∈ [0, 2π).
Circles in polar coordinates
Remark: Circles centered at the origin are trivial to graph.
y
Example
Graph the curve r = 2, θ ∈ [0, 2π).
2
x
Remark:
Circles not centered at the origin are more complicated to graph.
Example
Graph the curve
r = 4 cos(θ), θ ∈ [0, 2π).
Solution: Back to Cartesian:
Circles in polar coordinates
Remark: Circles centered at the origin are trivial to graph.
y
Example
Graph the curve r = 2, θ ∈ [0, 2π).
2
x
Remark:
y
Circles not centered at the origin are more complicated
to graph.
Example
Graph the curve
r = 4 cos(θ), θ ∈ [0, 2π).
r(0) = 4 cos(0)
0
2
Solution: Back to Cartesian:
x
Circles in polar coordinates
Remark: We now use the graph of the function r = 4 cos(θ) to
graph the curve r = 4 cos(θ) in the xy -plane.
Circles in polar coordinates
Remark: We now use the graph of the function r = 4 cos(θ) to
graph the curve r = 4 cos(θ) in the xy -plane.
Example
Graph the curve r = 4 cos(θ), θ ∈ [0, 2π).
Circles in polar coordinates
Remark: We now use the graph of the function r = 4 cos(θ) to
graph the curve r = 4 cos(θ) in the xy -plane.
Example
Graph the curve r = 4 cos(θ), θ ∈ [0, 2π).
Solution:
Notice that r (θ) = r (−θ).
(Reflection about x-axis symmetry.)
Circles in polar coordinates
Remark: We now use the graph of the function r = 4 cos(θ) to
graph the curve r = 4 cos(θ) in the xy -plane.
Example
Graph the curve r = 4 cos(θ), θ ∈ [0, 2π).
Solution:
Notice that r (θ) = r (−θ).
(Reflection about x-axis symmetry.)
The graph of r = 4 cos(θ) is
Circles in polar coordinates
Remark: We now use the graph of the function r = 4 cos(θ) to
graph the curve r = 4 cos(θ) in the xy -plane.
Example
Graph the curve r = 4 cos(θ), θ ∈ [0, 2π).
Solution:
Notice that r (θ) = r (−θ).
(Reflection about x-axis symmetry.)
r
r = 4 cos(0)
4
pi
The graph of r = 4 cos(θ) is
−4
0
Circles in polar coordinates
Remark: We now use the graph of the function r = 4 cos(θ) to
graph the curve r = 4 cos(θ) in the xy -plane.
Example
Graph the curve r = 4 cos(θ), θ ∈ [0, 2π).
Solution:
Notice that r (θ) = r (−θ).
(Reflection about x-axis symmetry.)
r
pi
The graph of r = 4 cos(θ) is
−4
The graph above helps to do the curve on the
xy -plane.
r = 4 cos(0)
4
0
Circles in polar coordinates
Remark: We now use the graph of the function r = 4 cos(θ) to
graph the curve r = 4 cos(θ) in the xy -plane.
Example
Graph the curve r = 4 cos(θ), θ ∈ [0, 2π).
Solution:
Notice that r (θ) = r (−θ).
(Reflection about x-axis symmetry.)
r
r = 4 cos(0)
4
−4
The graph above helps to do the curve on the
xy -plane. We actually cover the circle twice!
0
pi
The graph of r = 4 cos(θ) is
y
y
r(0) = 4 cos(0)
2
x
Graphing in polar coordinates (Sect. 11.4)
I
Review: Polar coordinates.
I
Review: Transforming back to Cartesian.
I
Computing the slope of tangent lines.
I
Using symmetry to graph curves.
Examples:
I
I
I
I
Circles in polar coordinates.
Graphing the Cardiod.
Graphing the Lemniscate.
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 − cos(θ), θ ∈ [0, 2π).
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 − cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 − cos(θ).
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 − cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 − cos(θ).
r
r = − cos(0)
1
0
pi
−1
r = cos(0)
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 − cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 − cos(θ).
r
r
r = − cos(0)
1
1
0
pi
r = 1 − cos(0)
0
pi
r = cos(0)
−1
r = cos(0)
−1
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 − cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 − cos(θ).
r
r
r = − cos(0)
1
r = 1 − cos(0)
1
0
pi
0
pi
r = cos(0)
−1
r = cos(0)
−1
y
From the previous graph we
obtain the curve: on the
xy -plane:
r = 1 − cos(0)
1
2
x
−1
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 + cos(θ), θ ∈ [0, 2π).
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 + cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 + cos(θ).
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 + cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 + cos(θ).
r
r = 1 + cos(0)
2
1
0
pi
−1
r = cos(0)
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 + cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 + cos(θ).
r
r = 1 + cos(0)
2
1
1
0
pi
r = 1 + cos(0)
r
2
0
pi
r = cos(0)
−1
r = cos(0)
−1
Graphing the Cardiod
Example
Graph on the xy -plane the curve r = 1 + cos(θ), θ ∈ [0, 2π).
Solution: We first graph the function r = 1 + cos(θ).
r
r = 1 + cos(0)
r = 1 + cos(0)
r
2
2
1
1
0
pi
0
pi
r = cos(0)
−1
r = cos(0)
−1
y
From the previous graph we
obtain the curve: on the
xy -plane:
r = 1 + cos(0)
1
2
−1
x
Graphing in polar coordinates (Sect. 11.4)
I
Review: Polar coordinates.
I
Review: Transforming back to Cartesian.
I
Computing the slope of tangent lines.
I
Using symmetry to graph curves.
Examples:
I
I
I
I
Circles in polar coordinates.
Graphing the Cardiod.
Graphing the Lemniscate.
Graphing the Lemniscate
Example
Graph on the xy -plane the curve r 2 = sin(2θ), θ ∈ [0, 2π).
Graphing the Lemniscate
Example
Graph on the xy -plane the curve r 2 = sin(2θ), θ ∈ [0, 2π).
p
Solution: We first graph the function r = ± sin(2θ).
Graphing the Lemniscate
Example
Graph on the xy -plane the curve r 2 = sin(2θ), θ ∈ [0, 2π).
p
Solution: We first graph the function r = ± sin(2θ).
r
r = sin(20)
1
pi
−1
2 pi
r = sin(0)
0
Graphing the Lemniscate
Example
Graph on the xy -plane the curve r 2 = sin(2θ), θ ∈ [0, 2π).
p
Solution: We first graph the function r = ± sin(2θ).
r
r
r = sin(20)
1
pi
−1
2 pi
r = sin(0)
1
r=+
0
sin(20)
pi
−1
r=−
sin(20)
2 pi
0
Graphing the Lemniscate
Example
Graph on the xy -plane the curve r 2 = sin(2θ), θ ∈ [0, 2π).
p
Solution: We first graph the function r = ± sin(2θ).
r
r
r = sin(20)
1
pi
−1
2 pi
r = sin(0)
1
r=+
0
sin(20)
pi
−1
r=−
2 pi
0
sin(20)
y
From the previous graph we
obtain the curve: on the
xy -plane:
r=+
cos(0)
x
r=−
cos(0)
Area of regions in polar coordinates (Sect. 11.5)
I
Review: Few curves in polar coordinates.
I
Formula for the area or regions in polar coordinates.
I
Calculating areas in polar coordinates.
Transformation rules Polar-Cartesian.
Definition
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
P = ( r, 0 ) = (x,y)
y
r
0
x
Transformation rules Polar-Cartesian.
Definition
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
Example
y
r = −2cos(0)
r = 2sin(0)
r = 2cos(0)
r=1
x
r = −2sin(0)
P = ( r, 0 ) = (x,y)
y
r
0
x
Transformation rules Polar-Cartesian.
Definition
P = ( r, 0 ) = (x,y)
The polar coordinates of a point P ∈ R2 is the
ordered pair (r , θ), with r > 0 and θ ∈ [0, 2π)
defined by the picture.
y
r
0
x
Example
y
r = −2cos(0)
r = 2sin(0)
y
r = 2cos(0)
r = 1 + sin(0)
r = 1 + cos(0)
1
r=1
x
r = −2sin(0)
−2
2
x
−1
r = 1 − cos(0)
r = 1 − sin(0)
Area of regions in polar coordinates (Sect. 11.5)
I
Review: Few curves in polar coordinates.
I
Formula for the area or regions in polar coordinates.
I
Calculating areas in polar coordinates.
Formula for the area or regions in polar coordinates
Theorem
If the functions r1 , r2 : [α, β] → R are continuous and 0 6 r1 6 r2 ,
then the area of a region D ⊂ R2 given by
D = (r , θ) ∈ R2 : r ∈ [r1 (θ), r2 (θ)], θ ∈ [α, β] .
is given by the integral
Z
β
A(D) =
α
2 2 1 r2 (θ) − r1 (θ)
dθ.
2
Formula for the area or regions in polar coordinates
Theorem
If the functions r1 , r2 : [α, β] → R are continuous and 0 6 r1 6 r2 ,
then the area of a region D ⊂ R2 given by
D = (r , θ) ∈ R2 : r ∈ [r1 (θ), r2 (θ)], θ ∈ [α, β] .
is given by the integral
Z
β
A(D) =
α
y
D
r2(0)
β
r1(0)
α
x
2 2 1 r2 (θ) − r1 (θ)
dθ.
2
Formula for the area or regions in polar coordinates
Theorem
If the functions r1 , r2 : [α, β] → R are continuous and 0 6 r1 6 r2 ,
then the area of a region D ⊂ R2 given by
D = (r , θ) ∈ R2 : r ∈ [r1 (θ), r2 (θ)], θ ∈ [α, β] .
is given by the integral
Z
β
A(D) =
α
y
2 2 1 r2 (θ) − r1 (θ)
dθ.
2
D
r2(0)
β
r1(0)
α
x
Remark: This result includes the
case of r1 = 0, which are fan-shaped
regions.
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ,
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ, with
k = 1, · · · , n,
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ, with
k = 1, · · · , n, and ∆θ =
β−α
n
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ, with
k = 1, · · · , n, and ∆θ =
y
D
Ak
r (0)
β
α
x
β−α
n
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ, with
k = 1, · · · , n, and ∆θ =
y
D
Ak
r (0)
β
α
x
β−α
n
The area of each fan-shaped region on
the figure is,
2
1
Ak = r (θk ) ∆θ.
2
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ, with
k = 1, · · · , n, and ∆θ =
y
D
Ak
β−α
n
The area of each fan-shaped region on
the figure is,
2
1
Ak = r (θk ) ∆θ.
2
r (0)
β
α
x
A Riemann sum that approximates the green region area is
n
n
X
X
2
1
Ak =
r (θk ) ∆θ.
2
k=1
k=1
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ, with
k = 1, · · · , n, and ∆θ =
y
D
Ak
β−α
n
The area of each fan-shaped region on
the figure is,
2
1
Ak = r (θk ) ∆θ.
2
r (0)
β
α
x
A Riemann sum that approximates the green region area is
n
n
X
X
2
1
Ak =
r (θk ) ∆θ.
2
k=1
k=1
Refining the partition and taking a limit n → ∞
Formula for the area or regions in polar coordinates
Idea of the Proof: Introduce a partition θk = k ∆θ, with
k = 1, · · · , n, and ∆θ =
y
D
Ak
β−α
n
The area of each fan-shaped region on
the figure is,
2
1
Ak = r (θk ) ∆θ.
2
r (0)
β
α
x
A Riemann sum that approximates the green region area is
n
n
X
X
2
1
Ak =
r (θk ) ∆θ.
2
k=1
k=1
Refining the partition and taking a limit n → ∞ one can prove
that the Riemann sum above converges and the limit is called
Z β
2
1
A(D) =
r (θ) dθ.
α 2
Area of regions in polar coordinates (Sect. 11.5)
I
Review: Few curves in polar coordinates.
I
Formula for the area or regions in polar coordinates.
I
Calculating areas in polar coordinates.
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Solution:
y
r=1
1
x
−1
r = 1 − sin(0)
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Solution:
y
r=1
The Theorem implies
1
Z
x
−1
r = 1 − sin(0)
β
A=
α
2 1
1 − 1 − sin(θ) dθ.
2
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Solution:
y
r=1
The Theorem implies
1
Z
x
β
A=
α
2 1
1 − 1 − sin(θ) dθ.
2
We need to find α and β.
−1
r = 1 − sin(0)
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Solution:
y
r=1
The Theorem implies
1
Z
x
−1
r = 1 − sin(0)
β
A=
α
2 1
1 − 1 − sin(θ) dθ.
2
We need to find α and β. They
are the intersection of the circle
and the cardiod:
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Solution:
y
r=1
The Theorem implies
1
Z
x
−1
r = 1 − sin(0)
1 = 1 − sin(θ)
β
A=
α
2 1
1 − 1 − sin(θ) dθ.
2
We need to find α and β. They
are the intersection of the circle
and the cardiod:
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Solution:
y
r=1
The Theorem implies
1
Z
x
α
2 1
1 − 1 − sin(θ) dθ.
2
We need to find α and β. They
are the intersection of the circle
and the cardiod:
−1
r = 1 − sin(0)
1 = 1 − sin(θ)
β
A=
⇒
sin(θ) = 0
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Solution:
y
r=1
The Theorem implies
1
Z
x
α
−1
r = 1 − sin(0)
1 = 1 − sin(θ)
β
A=
⇒
2 1
1 − 1 − sin(θ) dθ.
2
We need to find α and β. They
are the intersection of the circle
and the cardiod:
(
α = 0,
sin(θ) = 0 ⇒
β = π.
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Z π
2 1
Solution: Therefore: A =
1 − 1 − sin(θ) dθ.
0 2
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Z π
2 1
Solution: Therefore: A =
1 − 1 − sin(θ) dθ.
0 2
Z π
1
2 sin(θ) − sin2 (θ) dθ
A=
2 0
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Z π
2 1
Solution: Therefore: A =
1 − 1 − sin(θ) dθ.
0 2
Z π
1
2 sin(θ) − sin2 (θ) dθ
A=
2 0
Z
1 π
1
2 sin(θ) − 1 − cos(2θ) dθ
A=
2 0
2
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Z π
2 1
Solution: Therefore: A =
1 − 1 − sin(θ) dθ.
0 2
Z π
1
2 sin(θ) − sin2 (θ) dθ
A=
2 0
Z
1 π
1
2 sin(θ) − 1 − cos(2θ) dθ
A=
2 0
2
π
1
A=
−2 cos(θ) −
2
0
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Z π
2 1
Solution: Therefore: A =
1 − 1 − sin(θ) dθ.
0 2
Z π
1
2 sin(θ) − sin2 (θ) dθ
A=
2 0
Z
1 π
1
2 sin(θ) − 1 − cos(2θ) dθ
A=
2 0
2
π i
π 1 h
1
1
A=
−2 cos(θ) − π − sin(2θ)
2
2
2
0
0
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Z π
2 1
Solution: Therefore: A =
1 − 1 − sin(θ) dθ.
0 2
Z π
1
2 sin(θ) − sin2 (θ) dθ
A=
2 0
Z
1 π
1
2 sin(θ) − 1 − cos(2θ) dθ
A=
2 0
2
π i
π 1 h
1
1
A=
−2 cos(θ) − π − sin(2θ)
2
2
2
0
0
1
π
A=
4−
2
2
Calculating areas in polar coordinates
Example
Find the area inside the circle r = 1 and outside the cardiod
r = 1 − sin(θ).
Z π
2 1
Solution: Therefore: A =
1 − 1 − sin(θ) dθ.
0 2
Z π
1
2 sin(θ) − sin2 (θ) dθ
A=
2 0
Z
1 π
1
2 sin(θ) − 1 − cos(2θ) dθ
A=
2 0
2
π i
π 1 h
1
1
A=
−2 cos(θ) − π − sin(2θ)
2
2
2
0
0
1
π
π
A=
4−
⇒ A=2− .
2
2
4
C
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Solution: We first review that these curves are actually circles.
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Solution: We first review that these curves are actually circles.
r = cos(θ)
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Solution: We first review that these curves are actually circles.
r = cos(θ)
⇔
r 2 = r cos(θ)
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Solution: We first review that these curves are actually circles.
r = cos(θ)
⇔
r 2 = r cos(θ)
⇔
x 2 + y 2 = x.
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Solution: We first review that these curves are actually circles.
r = cos(θ)
⇔
r 2 = r cos(θ)
Completing the square in x we obtain
x−
1 2
1 2
+ y2 =
.
2
2
⇔
x 2 + y 2 = x.
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Solution: We first review that these curves are actually circles.
r = cos(θ)
⇔
r 2 = r cos(θ)
Completing the square in x we obtain
x−
1 2
1 2
+ y2 =
.
2
2
Analogously, r = sin(θ) is the circle
1 2 1 2
x2 + y −
=
.
2
2
⇔
x 2 + y 2 = x.
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Solution: We first review that these curves are actually circles.
r = cos(θ)
⇔
r 2 = r cos(θ)
⇔
x 2 + y 2 = x.
Completing the square in x we obtain
y
r=sin(0)
1 2
1 2
x−
+ y2 =
.
2
2
1/2
r=cos(0)
Analogously, r = sin(θ) is the circle
1 2 1 2
x2 + y −
=
.
2
2
1/2
x
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Z π/4
1 2
Solution: The Theorem implies: A = 2
sin (θ) dθ;
2
0
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Z π/4
1 2
Solution: The Theorem implies: A = 2
sin (θ) dθ;
2
0
Z
A=
0
π/4
1
1 − cos(2θ) dθ
2
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Z π/4
1 2
Solution: The Theorem implies: A = 2
sin (θ) dθ;
2
0
Z
A=
0
π/4
π/4 i
1
1 h π
1
1 − cos(2θ) dθ =
− 0 − sin(2θ)
;
2
2 4
2
0
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Z π/4
1 2
Solution: The Theorem implies: A = 2
sin (θ) dθ;
2
0
Z
π/4
A=
0
A=
π/4 i
1
1 h π
1
1 − cos(2θ) dθ =
− 0 − sin(2θ)
;
2
2 4
2
0
i
1hπ 1
−
−0
2 4
2
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Z π/4
1 2
Solution: The Theorem implies: A = 2
sin (θ) dθ;
2
0
Z
π/4
A=
0
A=
π/4 i
1
1 h π
1
1 − cos(2θ) dθ =
− 0 − sin(2θ)
;
2
2 4
2
0
i π 1
1hπ 1
−
−0 = −
2 4
2
8 4
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Z π/4
1 2
Solution: The Theorem implies: A = 2
sin (θ) dθ;
2
0
Z
π/4
A=
0
A=
π/4 i
1
1 h π
1
1 − cos(2θ) dθ =
− 0 − sin(2θ)
;
2
2 4
2
0
i π 1
1hπ 1
−
−0 = −
2 4
2
8 4
⇒
1
A = (π − 2).
8
C
Calculating areas in polar coordinates
Example
Find the area of the intersection of the interior of the regions
bounded by the curves r = cos(θ) and r = sin(θ).
Z π/4
1 2
Solution: The Theorem implies: A = 2
sin (θ) dθ;
2
0
Z
π/4
A=
0
A=
π/4 i
1
1 h π
1
1 − cos(2θ) dθ =
− 0 − sin(2θ)
;
2
2 4
2
0
i π 1
1hπ 1
−
−0 = −
2 4
2
8 4
⇒
1
A = (π − 2).
8
C
Z
Also works: A =
0
π/4
1
sin2 (θ) dθ +
2
Z
π/2
π/4
1
cos2 (θ) dθ.
2