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This gives us the following theorem.

Arc length of a curve defined by a polar function

Let f be a function whose derivative is continuous on an interval α θ β . The length of the graph of r = f ( θ ) from θ = α to θ = β is

L = α β [ f ( θ ) ] 2 + [ f ( θ ) ] 2 d θ = α β r 2 + ( d r d θ ) 2 d θ .

Finding the arc length of a polar curve

Find the arc length of the cardioid r = 2 + 2 cos θ .

When θ = 0 , r = 2 + 2 cos 0 = 4. Furthermore, as θ goes from 0 to 2 π , the cardioid is traced out exactly once. Therefore these are the limits of integration. Using f ( θ ) = 2 + 2 cos θ , α = 0 , and β = 2 π , [link] becomes

L = α β [ f ( θ ) ] 2 + [ f ( θ ) ] 2 d θ = 0 2 π [ 2 + 2 cos θ ] 2 + [ 2 sin θ ] 2 d θ = 0 2 π 4 + 8 cos θ + 4 cos 2 θ + 4 sin 2 θ d θ = 0 2 π 4 + 8 cos θ + 4 ( cos 2 θ + sin 2 θ ) d θ = 0 2 π 8 + 8 cos θ d θ = 2 0 2 π 2 + 2 cos θ d θ .

Next, using the identity cos ( 2 α ) = 2 cos 2 α 1 , add 1 to both sides and multiply by 2. This gives 2 + 2 cos ( 2 α ) = 4 cos 2 α . Substituting α = θ / 2 gives 2 + 2 cos θ = 4 cos 2 ( θ / 2 ) , so the integral becomes

L = 2 0 2 π 2 + 2 cos θ d θ = 2 0 2 π 4 cos 2 ( θ 2 ) d θ = 2 0 2 π | cos ( θ 2 ) | d θ .

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from 0 to π and double the answer. This strategy works because cosine is positive between 0 and π 2 . Thus,

L = 4 0 2 π | cos ( θ 2 ) | d θ = 8 0 π cos ( θ 2 ) d θ = 8 ( 2 sin ( θ 2 ) | 0 π = 16.
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Find the total arc length of r = 3 sin θ .

s = 3 π

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Key concepts

  • The area of a region in polar coordinates defined by the equation r = f ( θ ) with α θ β is given by the integral A = 1 2 α β [ f ( θ ) ] 2 d θ .
  • To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
  • The arc length of a polar curve defined by the equation r = f ( θ ) with α θ β is given by the integral L = α β [ f ( θ ) ] 2 + [ f ( θ ) ] 2 d θ = α β r 2 + ( d r d θ ) 2 d θ .

Key equations

  • Area of a region bounded by a polar curve
    A = 1 2 α β [ f ( θ ) ] 2 d θ = 1 2 α β r 2 d θ
  • Arc length of a polar curve
    L = α β [ f ( θ ) ] 2 + [ f ( θ ) ] 2 d θ = α β r 2 + ( d r d θ ) 2 d θ

For the following exercises, determine a definite integral that represents the area.

Region enclosed by r = 4

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Region enclosed by r = 3 sin θ

9 2 0 π sin 2 θ d θ

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Region in the first quadrant within the cardioid r = 1 + sin θ

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Region enclosed by one petal of r = 8 sin ( 2 θ )

32 0 π / 2 sin 2 ( 2 θ ) d θ

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Region enclosed by one petal of r = cos ( 3 θ )

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Region below the polar axis and enclosed by r = 1 sin θ

1 2 π 2 π ( 1 sin θ ) 2 d θ

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Region in the first quadrant enclosed by r = 2 cos θ

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Region enclosed by the inner loop of r = 2 3 sin θ

sin −1 ( 2 / 3 ) π / 2 ( 2 3 sin θ ) 2 d θ

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Region enclosed by the inner loop of r = 3 4 cos θ

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Region enclosed by r = 1 2 cos θ and outside the inner loop

0 π ( 1 2 cos θ ) 2 d θ 0 π / 3 ( 1 2 cos θ ) 2 d θ

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Region common to r = 3 sin θ and r = 2 sin θ

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Region common to r = 2 and r = 4 cos θ

4 0 π / 3 d θ + 16 π / 3 π / 2 ( cos 2 θ ) d θ

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Region common to r = 3 cos θ and r = 3 sin θ

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For the following exercises, find the area of the described region.

Enclosed by r = 6 sin θ

9 π

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Above the polar axis enclosed by r = 2 + sin θ

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Below the polar axis and enclosed by r = 2 cos θ

9 π 4

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Enclosed by one petal of r = 4 cos ( 3 θ )

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Enclosed by one petal of r = 3 cos ( 2 θ )

9 π 8

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Enclosed by r = 1 + sin θ

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Enclosed by the inner loop of r = 3 + 6 cos θ

18 π 27 3 2

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Enclosed by r = 2 + 4 cos θ and outside the inner loop

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Common interior of r = 4 sin ( 2 θ ) and r = 2

4 3 ( 4 π 3 3 )

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Common interior of r = 3 2 sin θ and r = −3 + 2 sin θ

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Common interior of r = 6 sin θ and r = 3

3 2 ( 4 π 3 3 )

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Inside r = 1 + cos θ and outside r = cos θ

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Common interior of r = 2 + 2 cos θ and r = 2 sin θ

2 π 4

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For the following exercises, find a definite integral that represents the arc length.

r = 4 cos θ on the interval 0 θ π 2

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r = 1 + sin θ on the interval 0 θ 2 π

0 2 π ( 1 + sin θ ) 2 + cos 2 θ d θ

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r = 2 sec θ on the interval 0 θ π 3

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r = e θ on the interval 0 θ 1

2 0 1 e θ d θ

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For the following exercises, find the length of the curve over the given interval.

r = 6 on the interval 0 θ π 2

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r = e 3 θ on the interval 0 θ 2

10 3 ( e 6 1 )

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r = 6 cos θ on the interval 0 θ π 2

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r = 8 + 8 cos θ on the interval 0 θ π

32

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r = 1 sin θ on the interval 0 θ 2 π

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For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.

[T] r = 3 θ on the interval 0 θ π 2

6.238

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[T] r = 2 θ on the interval π θ 2 π

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[T] r = sin 2 ( θ 2 ) on the interval 0 θ π

2

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[T] r = 2 θ 2 on the interval 0 θ π

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[T] r = sin ( 3 cos θ ) on the interval 0 θ π

4.39

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For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.

r = 3 sin θ on the interval 0 θ π

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r = sin θ + cos θ on the interval 0 θ π

A = π ( 2 2 ) 2 = π 2 and 1 2 0 π ( 1 + 2 sin θ cos θ ) d θ = π 2

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r = 6 sin θ + 8 cos θ on the interval 0 θ π

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For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.

r = 3 sin θ on the interval 0 θ π

C = 2 π ( 3 2 ) = 3 π and 0 π 3 d θ = 3 π

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r = sin θ + cos θ on the interval 0 θ π

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r = 6 sin θ + 8 cos θ on the interval 0 θ π

C = 2 π ( 5 ) = 10 π and 0 π 10 d θ = 10 π

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Verify that if y = r sin θ = f ( θ ) sin θ then d y d θ = f ( θ ) sin θ + f ( θ ) cos θ .

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For the following exercises, find the slope of a tangent line to a polar curve r = f ( θ ) . Let x = r cos θ = f ( θ ) cos θ and y = r sin θ = f ( θ ) sin θ , so the polar equation r = f ( θ ) is now written in parametric form.

Use the definition of the derivative d y d x = d y / d θ d x / d θ and the product rule to derive the derivative of a polar equation.

d y d x = f ( θ ) sin θ + f ( θ ) cos θ f ( θ ) cos θ f ( θ ) sin θ

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r = 1 sin θ ; ( 1 2 , π 6 )

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r = 4 cos θ ; ( 2 , π 3 )

The slope is 1 3 .

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r = 8 sin θ ; ( 4 , 5 π 6 )

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r = 4 + sin θ ; ( 3 , 3 π 2 )

The slope is 0.

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r = 6 + 3 cos θ ; ( 3 , π )

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r = 4 cos ( 2 θ ) ; tips of the leaves

At ( 4 , 0 ) , the slope is undefined. At ( −4 , π 2 ) , the slope is 0.

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r = 2 sin ( 3 θ ) ; tips of the leaves

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r = 2 θ ; ( π 2 , π 4 )

The slope is undefined at θ = π 4 .

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Find the points on the interval π θ π at which the cardioid r = 1 cos θ has a vertical or horizontal tangent line.

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For the cardioid r = 1 + sin θ , find the slope of the tangent line when θ = π 3 .

Slope = −1.

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For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of θ .

r = 3 cos θ , θ = π 3

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r = θ , θ = π 2

Slope is −2 π .

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[T] Use technology: r = 2 + 4 cos θ at θ = π 6

Calculator answer: −0.836.

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For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line.

r 2 = 4 cos ( 2 θ )

Horizontal tangent at ( ± 2 , π 6 ) , ( ± 2 , π 6 ) .

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The cardioid r = 1 + sin θ

Horizontal tangents at π 2 , 7 π 6 , 11 π 6 . Vertical tangents at π 6 , 5 π 6 and also at the pole ( 0 , 0 ) .

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Show that the curve r = sin θ tan θ (called a cissoid of Diocles ) has the line x = 1 as a vertical asymptote.

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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