Calculus 8.3 Polar Graphs Area: The area of a small sector of a curve (circle in geometry) was

given by the formula

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sector area =angle measure

2π⋅ π r2

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=angle measure

2⋅ r2 . Since we want the angle measure of the

sector to be very tiny, let

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dθ = the angle measure, giving the area of the sector

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= 12 r2dθ . Therefore the area of the entire region

will be 12r2

!

"#

$

%&dθ

θ=a

θ=b∫ = 1

2f θ( )( )2 dθ

θ=a

θ=b∫ where r = f θ( ) .

Example: Find the area inside r = 2+ 2sinθ .

Solution: Area is 12

2+ 2sinθ( )2 dθ0

2π∫ = 1

24+8sinθ + 4sin2θ  dθ

0

2π∫

= 2+ 4sinθ + 2sin2θ  dθ0

2π∫ = 2+ 4sinθ + 2 1−cos 2θ( )

2"#$

%&' dθ

0

2π∫ = 2+ 4sinθ +1− cos 2θ( ) dθ

0

2π∫

= 3+ 4sinθ − cos 2θ( ) dθ0

2π∫ = 3θ − 4cosθ − sin 2θ( )

2 0

2π

= 3 2π( )− 4cos 2θ( )− sin 4θ( )2

!"#

$%&− 3 0( )− 4cos 0( )− sin 0( )2!"#

$%&= 6π .

Example: Find the area in the first leaf of

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r = 2sin 3θ( )

Solution: To find the limits,

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2sin 3θ( ) = 0

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⇒ sin 3θ( ) = 0

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⇒ 3θ = 0, π, 2π, 3π

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⇒θ = 0, π3 , 2π3 , π . Therefore, the first

leaf (of three) starts at

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α = 0 and ends at

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β = π3 , then the area of

this leaf will be

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12

2sin 3θ( )( )2

0

π3∫ dθ

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= 2sin2 3θ( )0

π3∫ dθ , using

the half angle formula sin2θ = 12 1− cos 2θ( )( ) gives

2 12 1− cos 2 3θ( )( )( )( )

0

π3∫ dθ

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= 1− cos 6θ( )0

π3∫ dθ

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= θ −sin6θ6 0

π3

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=π3−sin6 π

3( )6

$

%

& &

'

(

) ) − 0 − sin6 0( )

6$

% &

'

( )

€

=π3

Example: Find the area shared by the graphs r = 2 and

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r = 2 1− cosθ( ) .

Solution: To find the angle values for the shared region set the equations equal to each other and solve for q:

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2 1− cosθ( ) = 2

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⇒1− cosθ =1

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⇒ cosθ = 0 , giving

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θ = − π2 and π2 . The shared

area equals the area of a semicircle plus the part of the cardioid

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12π 2( )2 +

12

2 1− cosθ( )( )2−π2

π2∫ dθ

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= 2π + 2 − 4cosθ + 2cos2θ−π2

π2∫ dθ , using the half angle gives

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2π + 2 − 4cosθ + 21+ cos2θ

2%

& '

(

) *

−π2

π2∫ dθ

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= 2π + 3 − 4cosθ + cos2θ−π2

π2∫ dθ

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= 2π + 3θ − 4sinθ +12sin2θ

%

& '

(

) * −π2

π2

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= 2π + 3 π2( ) − 4sin π

2( ) +12sin2 π

2( )$

% &

'

( ) − 3 − π2( ) − 4sin − π2( ) +

12sin2 − π2( )$

% &

'

( )

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= 2π + 32 π − 4 + 0 + 3

2 π − 4 + 0

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= 5π − 8

Example: Find the area inside the graph of

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r = 3+ 2cosθ , but outside the graph of r = 2.

Solution: To find the angle values for the non-shared region (shown as the white part in the diagram) set the equations equal to each other and solve for q:

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3+ 2cosθ = 2

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⇒ 2cosθ = −1

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⇒ cosθ = − 12 , giving

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θ = − 2π3 and 2π

3 . The non-shaded area

equals area of cardioid minus circle. Using symmetry, find the area of the top portion then multiply by 2.

Area

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= 212

3+ 2cosθ( )2 − 2( )2( )0

2π3∫ dθ

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= 9 +12cosθ + 4cos2θ − 4( )0

2π3∫ dθ

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= 5 +12cosθ + 4 cos2θ0

2π3∫ dθ , using the half angle,

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= 5 +12cosθ + 41+ cos2θ

2#

$ %

&

' (

0

2π3∫ dθ

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= 7 +12cosθ + 2cos2θ0

2π3∫ dθ

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= 7θ +12sinθ + sin2θ0

2π3

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= 7 2π3( ) +12sin 2π

3( ) + sin2 2π3( )( ) − 7 0( ) +12sin 0( ) + sin2 0( )( )

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=14π3

+12 32

#

$ %

&

' ( +

− 32

#

$ %

&

' (

#

$ % %

&

' ( (

€

=14π3

+11 32

≈ 24.187

Arc Length: Similar to a curve defined by parametric equations, arc length of polar equation can be approximated using the Pythagorean Theorem. As with function and parametric curves, the length of

the hypotenuse of the right triangle will be L = Δx( )2 + Δy( )2

= dx( )2 + dy( )2 . Using the polar conversion formula x = rcosθ ,

then dxdθ =drdθ cosθ + r −sinθ( ) = dr

dθ cosθ − rsinθ using the product

rule. Similarly, y = rsinθ giving dydθ =drdθ sinθ + rcosθ .

Therefore dx = r 'cosθ − rsinθ( )dθ and dy = r 'sinθ + rcosθ( )dθ , substituting in Pythagorean Theorem

gives L = r 'cosθ − rsinθ( )dθ( )2 + r 'sinθ + rcosθ( )dθ( )2

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= " r cosθ − rsinθ( )2 + " r sinθ + rcosθ( )2dθ

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= " r ( )2 cos2θ − 2 " r rcosθ sinθ + r2 sin2θ + " r ( )2 sin2θ + 2 " r rcosθ sinθ + r2 cos2θdθ

€

= " r ( )2 cos2θ + sin2θ( ) + r2 cos2θ + sin2θ( )dθ

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= " r ( )2 + r2dθ .

Therefore the total arc length is !r( )2 + r2 dθθ=a

θ=b∫ = f ' θ( )( )2 + f θ( )( )2 dθ

θ=a

θ=b∫ where

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r = f θ( ).

Using this example, the length around the entire graph of r = 2+ 2sinθ is

2cosθ( )2 + 2+ 2sinθ( )2 dθ0

2π∫ =16 .