Section 6.3: Polar Forms & Area
Polar Coordinates
Rectangular coordinates were horizontal/vertical
directions for reaching the point
Polar coordinates are "as the crow flies"directions
(x,y)
 
Eg:  3,  in polar
 4
3
x
cos   , so
3

4
x = r cos
y = r sin
x2  y 2  r
tan
1
y

x
Notation
Warnings
 
 3, 
 4
 r , 
3

4
r
3cis
rcis

3e
4
i

4
rei
5 
 

 9 
3,
and

3,
and




 3,

4
4
4






Same point
 0,  is always the origin
Graph r = 3
Graph  =

4
Graph r = 
Graph r = 4 sin
r
θ
sin
Graph r = 4 sin
r
θ
0
sin
Graph r = 4 sin
r
θ
0
0
sin
Graph r = 4 sin
r
θ
0
0
π/2
sin
Graph r = 4 sin
r
θ
0
0
4
π/2
sin
Graph r = 4 sin
r
θ
0
0
4
π/2
π
sin
Graph r = 4 sin
r
θ
0
0
4
π/2
0
π
sin
Graph r = 4 sin
r
θ
0
0
4
π/2
0
π
3π/2
sin
Graph r = 4 sin
r
θ
0
0
4
π/2
0
π
-4
3π/2
sin
Graph r = 4 sin
r
θ
0
0
4
π/2
0
π
-4
3π/2
2π
sin
Graph r = 4 sin
r
θ
0
0
4
π/2
0
π
-4
3π/2
0
2π
sin
Cardiod r = 2(1 - cos  )
r
θ
Cardiod r = 2(1 - cos  )
r
θ
0
Cardiod r = 2(1 - cos  )
r
θ
0
0
Cardiod r = 2(1 - cos  )
r
θ
0
0
π/2
Cardiod r = 2(1 - cos  )
r
θ
0
0
2
π/2
Cardiod r = 2(1 - cos  )
r
θ
0
0
2
π/2
π
Cardiod r = 2(1 - cos  )
r
θ
0
0
2
π/2
4
π
Cardiod r = 2(1 - cos  )
r
θ
0
0
2
π/2
4
π
3π/2
Cardiod r = 2(1 - cos  )
r
θ
0
0
2
π/2
4
π
2
3π/2
Cardiod r = 2(1 - cos  )
r
θ
0
0
2
π/2
4
π
2
3π/2
2π
Cardiod r = 2(1 - cos  )
r
θ
0
0
2
π/2
4
π
2
3π/2
0
2π
2
4
2
Limacon r = 1  2 cos
r
θ
Limacon r = 1  2 cos
r
θ
-1
0
Limacon r = 1  2 cos
r
θ
-1
0
1
π/2
Limacon r = 1  2 cos
r
θ
-1
0
1
π/2
3
π
Limacon r = 1  2 cos
1
r
θ
-1
0
1
π/2
3
π
1
3π/2
3
-1
1
Limacon r = 1  2 cos
1
r
θ
-1
0
1
π/2
3
π
1
3π/2
3
-1
1
r
θ
-1
0
1
π/2
Limacon r = 1  2 cos
1
r
θ
-1
0
1
π/2
3
π
1
3π/2
3
-1
1
r
θ
-1
0
1-√2
π/4
1
π/2
Limacon r = 1  2 cos
1
r
θ
-1
0
1
π/2
3
π
1
3π/2
3
-1
1
r
θ
-1
0
1-√2
π/4
0
π/3
1
π/2
Rose r = 4 cos(2 )
r
θ
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
0
π/4
-4
π/2
3π/4
π
5pi/4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
0
π/4
-4
π/2
0
3π/4
π
5pi/4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
0
π/4
-4
π/2
0
3π/4
4
π
5pi/4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
0
π/4
-4
π/2
0
3π/4
4
π
0
5pi/4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
0
π/4
-4
π/2
0
3π/4
4
π
0
5pi/4
-4
3π/2
7π/4
Rose r = 4 cos(2 )
r
θ
4
0
0
π/4
-4
π/2
0
3π/4
4
π
0
5pi/4
-4
3π/2
0
7π/4
4
Lemniscate r 2  4cos(2 )
r
θ
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
0
π/4
π/2
3π/4
π
5pi/4
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
0
π/4
Undefined
π/2
3π/4
π
5pi/4
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
0
π/4
Undefined
π/2
0
3π/4
π
5pi/4
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
0
π/4
Undefined
π/2
0
3π/4
2
π
5pi/4
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
0
π/4
Undefined
π/2
0
3π/4
2
π
0
5pi/4
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
0
π/4
Undefined
π/2
0
3π/4
2
π
0
5pi/4
Undefined
3π/2
7π/4
Or r  4 cos(2 )
Lemniscate r 2  4cos(2 )
r
θ
2
0
0
π/4
Undefined
π/2
0
3π/4
2
π
0
5pi/4
Undefined
3π/2
0
7π/4
Or r  4 cos(2 )
2
1.
2.
3.
4.
5.
Something is changing, so we can’t use the old
algebra formulas.
Break the problem into pieces.
Pretend everything is constant on each piece and
use the old formulas.
Add up the pieces. (This is called a Riemann Sum)
If we use more and more pieces, the limit is the
right answer! (This limit is a definite integral.)
Polar Area
Insted of breaking the area into little rectangles,
we use little sectors...
That's not surprising if we think of graphing by running 
Fact: The area of sector

r
is
1

r 2   
2

1 
That's easy if you remember the area of circle is r  2 
2 
2
Example: Find the area of one leaf of r = 4 cos(2 )


  
4
4
Important: Always graph first


2
1
Riemann sum    4 cos  2 k*   k
2

2
1
Integral     4 cos 2  d
 2

4
4


4
Integral
  4 cos 2 

2

1
  d
2
4
2
8
cos
2 d



4

4

1

  8  1  cos 4   d
2




4
Two important identities:
1
2
cos x  1  cos 2 x 
2
1
2
sin x  1  cos 2 x 
2
4

4
 4  1  cos (4 ) d


4
1


 4    sin(4 ) 
4



4


4
 2
Example: Find the area enclosed by r = 2sin and r = 2cos
Find Intersections
2 sin = 2 cos
tan = 1

5
 = or
4
4
r  2 or - 2
But that misses the interaction at origin
because we can only
divide by sin if it's not zero!
Always Graph !!!
Example: Find the area enclosed by r = 2sin and r = 2cos


4
2

0
1
2
 2sin   d 
2

4
1
2
 2 cos   d
2
Example: Find the area outside the lemniscate r = 2 cos 2
and inside the circle r = 3 cos 

2
2
3



1
2
 3cos   d
2
2
  

4

4

2
1
2 cos 2 d
2
Warning : This is not the same as the area
between curves in rectangular coordinates!
2
1
You can't do  3cos   2 cos 2 d
2

