Areas and Lengths in PolarCoordinates

December 9, 2019

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Lengths in Polar Coordinates

Given a polar curve r = f(θ), we can use the relationship betweenCartesian coordinates and Polar coordinates to write parametricequations which describe the curve using the parameter θ

x(θ) = f(θ) cos(θ) y(θ) = f(θ) sin(θ)

To compute the arc length of such a curve between θ = a and θ = b,we need to compute the integral

L =

∫ b

a

√(dx

dθ

)2

+

(dy

dθ

)2

dθ

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We can simplify the formula L =

∫ b

a

√(dx

dθ

)2

+

(dy

dθ

)2

dθ because(dx

dθ

)2

+

(dy

dθ

)2

=(f ′(θ) cos(θ)− f(θ) sin(θ)

)2+(f ′(θ) sin(θ) + f(θ) cos(θ)

)2=(f ′(θ)

)2+(f(θ)

)2= r2 +

(dr

dθ

)2

= r2 + (r′)2.

L =

∫ b

a

√r2 + (r′)2 dθ

provided the parametrization moves steadily along the curve once.

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Example 1

Compute the length of the polar curve r = 6 sin θ for 0 6 θ 6 π.

I Last lecture, we saw that the graph of this equation is a circle ofradius 3 and as θ increases from 0 to π, the curve traces out thecircle exactly once.

0

Π

4

Π

2

3 Π

4

Π

5 Π

4

3 Π

2

7 Π

4

1.

2.

3.

4.

5.

6.

I We have L =

∫ π

0

√r2 + (r′)2 dθ

I r = 6 sin θ anddr

dθ= 6 cos θ. r2 + (r′)2 = 36 cos2 θ+ 36 sin2 θ = 36.

I L =

∫ π

0

√36 dθ = 6π.

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Areas in Polar Coordinates

Suppose we are given a polar curve r = f(θ) and wish to calculate thearea swept out by this polar curve between two given angles θ = aand θ = b. This is the region R in the picture on the left below:

Dividing this shape into smaller pieces (on right) and estimating theareas of the small pieces with pie-like shapes (center picture), we get aRiemann sum approximating A = the area of R of the form

A ≈n∑i=1

f(θ∗i )2

2∆θ

Letting n→∞ yields the following integral expression of the area

∫ b

a

f(θ)2

2dθ =

1

2

∫ b

a

r2dθ

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Example 2

Compute the area bounded by the curve r = sin 2θ for 0 6 θ 6 π2 .

I Using the formula A =1

2

∫ b

a

f(θ)2dθ =1

2

∫ π2

0

(sin 2θ)2dθ

I Using the half-angle formula, we get A =1

4

∫ π2

0

1− cos(4θ)dθ

=1

4

(θ − sin(4θ)

4

) ∣∣∣∣∣π/2

0

=π

8.

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Areas of Region between two curves

If instead we consider a region bounded between two polar curvesr = f(θ) and r = g(θ) then the equations becomes

1

2

∫ b

a

f(θ)2 − g(θ)2dθ

CAREFUL about parametrization!

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Example from Paul’s Online Notes

Find the area inside the circle r = 2 and the limacon r = 3 + 2 sin(θ).

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Find the area inside the circle r = 2 and the limacon r = 3 + 2 sin(θ).

The area we want is the sum of the light green and the green.

1

2

∫ 7π6

−π622dθ +

1

2

∫ 11π6

7π6

(3 + 2 sin(θ)

)2dθ

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Example 3

Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.

I We use the formula A =1

2

∫ b

a

f(θ)2 − g(θ)2dθ

I To find the angles where the curves intersect, we find the valuesof θ where 3 sin θ = 1 + sin θ or 2 sin θ = 1. That is sin θ = 1/2

I Therefore θ =π

6,

5π

6.

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Example 3 continued

Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.

I1

2

∫ b

a

f(θ)2 − g(θ)2dθ =1

2

∫ 5π6

π6

9 sin2 θ − (1 + sin θ)2dθ

I =1

2

∫ 5π6π6

8 sin2θ − 1 − 2 sin θdθ = 4

∫ 5π6π6

sin2θdθ −

1

2

∫ 5π6π6

1dθ −∫ 5π

6π6

sin θdθ

= 2

∫ 5π6

π6

1− cos(2θ)dθ − π

3+ cos θ

∣∣∣5π/6π/6

=4π

3− sin(2θ)

∣∣∣5π/6π/6− π

3−√

3 = π.

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Word of Warning

NOTE The fact that a single point has many representation in polarcoordinates makes it very difficult to find all the points of intersectionsof two polar curves. It is important to draw the two curves!!!Example 4 Find all possible intersection points of the curvesr = cos 2θ and r = 1

2 .

0

Π

6

Π

3

Π

22 Π

3

5 Π

6

Π

7 Π

6

4 Π

3 3 Π

2

5 Π

3

11 Π

6

1.

I Here we must solve two equations 12 = cos(2θ) and − 1

2 = cos(2θ)since the curve r = 1

2 is the same as the curve r = − 12 .

I From the first equation, we get 2θ = ±π3 , ±5π3 and from the

second we get 2θ = ± 2π3 , ±

4π3 .

I Therefore we get 8 points of intersectionθ = ±π6 , ±

π3 , ±

2π3 , ±

5π6 .

I There are infinitely many points of intersection since we may takeany of the eight points we just found and add any integralmultiple of 2π.

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Example 5

Find the area inside the circle and outside the four-leafed rose in thefirst quadrant. Equations for the curves are r = cos 2θ and r = 1

2 andwe computed the intersection points in Example 4.

From the previ-ous slide the twopoints of inter-section occur at

a =π

6and

b = −2π

3.

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r = cos(2θ)

r =1

2

Start at θ = 0 and eachpane shows a successive

interval of lengthπ

4.

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The intersection angles areπ

3and −2π

3.

1

2

∫ π4

π3

((1

4

)− cos2(2θ)

)dθ +

1

2

∫ 2π− 2π3

5π4

((1

4

)− cos2(2θ)

)dθ

OR

1

2

∫ π4

π3

((1

4

)− cos2(2θ)

)dθ +

1

2

∫ − 2π3

5π4 −2π

((1

4

)− cos2(2θ)

)dθ

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