Section 6.3: Polar Forms & Area

Rectangular coordinates were horizontal/vertical

directions for reaching the point

Eg: 3, in polar4

Polar coordinates are "as the crow flies"directions

(x,y)

4

3cos ,

3

xso

2 2

1

x = r cos

y = r sin tan

x y r

y

x

Polar Coordinates

Notation

43, 3 3cis 3e4 4 4

i

, r rcis reir

Warnings 5 93, 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

r = 2(1 - cos )Cardiod

r θ

r = 2(1 - cos )Cardiod

r θ

0

r = 2(1 - cos )Cardiod

r θ

0 0

r = 2(1 - cos )Cardiod

r θ

0 0

π/2

r = 2(1 - cos )Cardiod

r θ

0 0

2 π/2

r = 2(1 - cos )Cardiod

r θ

0 0

2 π/2

π

r = 2(1 - cos )Cardiod

r θ

0 0

2 π/2

4 π

r = 2(1 - cos )Cardiod

r θ

0 0

2 π/2

4 π

3π/2

r = 2(1 - cos )Cardiod

r θ

0 0

2 π/2

4 π

2 3π/2

r = 2(1 - cos )Cardiod

r θ

0 0

2 π/2

4 π

2 3π/2

2π

r = 2(1 - cos )Cardiod

r θ

0 0

2 π/2

4 π

2 3π/2

0 2π

4

2

2

r θ

r = 1 2 cosLimacon

r θ

-1 0

r = 1 2 cosLimacon

r θ

-1 0

1 π/2

r = 1 2 cosLimacon

r θ

-1 0

1 π/2

3 π

r = 1 2 cosLimacon

r θ

-1 0

1 π/2

3 π

1 3π/2

r = 1 2 cosLimacon

3 -1

1

1

r θ

-1 0

1 π/2

3 π

1 3π/2

r = 1 2 cosLimacon

3 -1

1

1

r θ-1 0

1 π/2

r θ

-1 0

1 π/2

3 π

1 3π/2

r = 1 2 cosLimacon

3 -1

1

1

r θ-1 0

1 - √ 2 π/4

1 π/2

r θ

-1 0

1 π/2

3 π

1 3π/2

r = 1 2 cosLimacon

3 -1

1

1

r θ-1 0

1 - √ 2 π/4

0 π/3

1 π/2

r = 4 cos(2 )Rose

r θ

0

π/4

π/2

3π/4

π

5pi/4

3π/2

7π/4

r = 4 cos(2 )Rose

r θ

4 0

π/4

π/2

3π/4

π

5pi/4

3π/2

7π/4

r = 4 cos(2 )Rose

r θ

4 0

0 π/4

π/2

3π/4

π

5pi/4

3π/2

7π/4

r = 4 cos(2 )Rose

r θ

4 0

0 π/4

-4 π/2

3π/4

π

5pi/4

3π/2

7π/4

r = 4 cos(2 )Rose

r θ

4 0

0 π/4

-4 π/2

0 3π/4

π

5pi/4

3π/2

7π/4

r = 4 cos(2 )Rose

r θ

4 0

0 π/4

-4 π/2

0 3π/4

4 π

5pi/4

3π/2

7π/4

r = 4 cos(2 )Rose

r θ

4 0

0 π/4

-4 π/2

0 3π/4

4 π

0 5pi/4

3π/2

7π/4

r = 4 cos(2 )Rose

r θ

4 0

0 π/4

-4 π/2

0 3π/4

4 π

0 5pi/4

-4 3π/2

7π/4

r = 4 cos(2 )Rose

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

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

0

π/4

π/2

3π/4

π

5pi/4

3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

π/4

π/2

3π/4

π

5pi/4

3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

0 π/4

π/2

3π/4

π

5pi/4

3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

0 π/4

Undefined π/2

3π/4

π

5pi/4

3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

0 π/4

Undefined π/2

0 3π/4

π

5pi/4

3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

0 π/4

Undefined π/2

0 3π/4

2 π

5pi/4

3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

0 π/4

Undefined π/2

0 3π/4

2 π

0 5pi/4

3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

0 π/4

Undefined π/2

0 3π/4

2 π

0 5pi/4

Undefined 3π/2

7π/4

2 r 4cos(2 )Lemniscate 4cos r (2 )Or

r θ

2 0

0 π/4

Undefined π/2

0 3π/4

2 π

0 5pi/4

Undefined 3π/2

0 7π/4

2

1. Something is changing, so we can’t use the old algebra formulas.

2. Break the problem into pieces.

3. Pretend everything is constant on each piece and use the old formulas.

4. Add up the pieces. (This is called a Riemann Sum)

5. If we use more and more pieces, the limit is the right answer! (This limit is a definite integral.)

Big Idea

Insted of breaking the area into little rectangles,

we use little sectors...

Polar Area

That's not surprising if we think of graphing by running

The area of seFact: ctor 2 1is

2r

r

2 1That's easy if you remember the area of circle is 2

2r

Example: Find the area of o rn e = leaf 4 co of s(2 )

Always grImportant: aph first

2

*1Riemann sum

24cos 2 kk

4

2

4

1Integral

24cos 2 d

4 4

4

4

2cos 28 d

4

2

4

1Integral 4cos 2

2d

4

4

11 c8 os 4

2d

2

2

1cos

Two impor

1 cos 22

tant identi

1sin

tie s

1 c s2

:

o 2

x x

x x

4

4

4 1 cos (4 ) d

4

4

14 sin( )4

4

2

Example: Find the area enclosed by and rr = 2s = in 2cos

Find Intersections

= 2 2 s

in cos

But that misses the interaction at origin

because we can only

divide by sin if it's not zero!

!!!Always Graph

tan = 1

5=

4 4

2 - 2

or

r or

4 2

2 2

04

1 12cos2sin

2 2d d

Example: Find the area enclosed by and rr = 2s = in 2cos

r = 2 cosExample: Find the area outside the lemniscate

and inside the circl r = 3 c

2

e os

2

: This is not the same as the area

between curves in rectangular coordinates!

You can't do 3cos co1

22 s 2

Warn

d

ing

2 4 22

2 4

1 12 cos 23cos

2 2d d

2 3