LECTURE 22: POLAR COORDINATES AND INTEGRATION
This roughly covers Section 8.3 in the Hughes-Hallett text.
22.1. Polar Coordinates and Integration.
(1) Learn the difference between polar and rectangular coordinates and functions.
(2) Learn how to find the area of a region bounded by a polar equation.
22.1.1. Polar Coordinates. The set R2 has more than one set of coordinates. A popular alternative
is called polar coordinates. Draw a pair of perpendicular axes as usual. Given a point in the plane,
connect the point to the origin of the line segment. The length of the line segment is r and the angle
it makes with the righthand part of the horizontal axis is θ. The polar coordinates of the point are
(r, θ). Here, we may take −∞ < r < ∞ and −∞ < θ < ∞.
Example: Plot the points in polar coordinates:
(1) (1, 0), (1, π), (1/2, π/2)
(2) (2, π/3)
(3) (−2, π/3). This is the antipodal point of (2, π/3). Draw a circle of radius two. Locate
(2, π/3). The point on the opposite side of the diameter is (−2, π/3). This is also (2, π/3+π).
Example: It is important to note that a representations of a point in polar coordinates is not
unique: (1, π/4) = (1, 9π/4) = (1, −7π/4) = (−1, −3π/4)
A point given in polar coordinates may be converted into a point in cartesian coordinates using
trigonometry.
x
y
=
=
tan(θ)
=
x2 + y 2
=
r cos(θ)
r sin(θ)
y
x
r2
22.1.2. Polar Equations. The graphs of polar equations are often very pretty. We’ll start off with
some easy ones and proceed to fancier ones.
Example: Let r0 be a constant. The equation r = r0 is a circle of radius r0 centered at the
origin if r0 > 0. If r0 = 0, it is a point.
Example: Let θ0 be a constant. The equation θ = θ0 is a line through the origin.
Example: r · cos(θ) = 1. This is a vertical line.
Example: Let a 6= 0 be a constant. r = 2a sin(θ). This is a circle of radius a centered at
1
2
LECTURE 22: POLAR COORDINATES AND INTEGRATION
(0, a).
x2 + (y − a)2
= a2
r2 cos2 (θ) + r2 sin2 (θ) − 2ar sin(θ) + a2
= a2
r(r − 2a sin(θ))
= 0
This gives the two solutions r = 0 (the origin) and r = 2a sin(θ).
Example: Let a > 0 be a constant. The equation r = aθ is a spiral. As theta increases, r
also increases.
To convert rectangular coordinates to polar coordinates, one uses the substitutions given above.
Example: Convert y 2 − 8x − 16 = 0 to polar coordinates.
y 2 − 8x − 16 =
0
r sin (θ) − 8r cos(θ) − 16 =
0
2
2
r
=
r
=
=
=
8 cos(θ) ±
q
64 cos2 (θ) + 64 sin2 (θ)
2 sin2 (θ)
4 cos(θ) ± 4
1 − cos2 (θ)
4(cos(θ) ± 1)
(1 − cos(θ))(1 + cos(θ))
4
4
,−
1 − cos(θ) 1 + cos(θ)
22.1.3. Integration in Polar coordinates. We consider a function of the form r = f (θ). For example,
we have the curve r = 3 + 2 cos(θ), 0 ≤ θ ≤ 2π that is discussed in the text. Suppose we wish to
find the area of the region bounded by θ = θ1 , θ = θ2 and the curve r = f (θ). In this case, we break
up the region into tiny sectors. The area of a sector of a circle which sweeps out an angle of θ is:
1
θ
πr2 = r2 θ
2π
2
Thus, when we have a small bit of sector, the tiny bit of area is given by:
∆A ≈
1 2
r ∆θ
2
Adding up all of the contributions gives:
A≈
X
small sectors
∆A ≈
X
small sectors
1 2
r ∆θ
2
In the limit as ∆θ goes to zero, the approximation becomes an equality and the right hand side is
given by the integral below:
Z
1 θ2
[f (θ)]2 dθ
A=
2 θ1
Example: Find the area of the region bounded by one petal of the rose curve r = 3 cos(3θ).
First sketch the graph of the curve. We see that the angle goes from −π/6 to π/6. This gives the
LECTURE 22: POLAR COORDINATES AND INTEGRATION
3
following integral:
A =
=
1
2
Z
π/6
9
2
Z
π/6
9 cos2 (3θ)dθ
−π/6
−π/6
1
(1 + cos(6θ))dθ
2
Example:(In-Class) Find the area of the region bounded by the cardioid r = f (θ) = 2 − 2 cos(θ)
where θ ranges from 0 to 2π.