Area and Arc Length in Polar Coordinates

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function

y = f (x)

on this interval, the area between the curve and the x-axis is given by

A = \int_{a}^{b} f (x) d x .

This fact, along with the formula for evaluating this integral, is summarized in the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by

L = \int_{a}^{b} \sqrt{1 + {(f^{'} (x))}^{2}} d x .

In this section, we study analogous formulas for area and arc length in the polar coordinate system.

Areas of Regions Bounded by Polar Curves

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

into n equal-width subintervals. The width of each subinterval is given by the formula

Δ θ = (β - α) / n,

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

radians. Therefore a fraction of a circle can be measured by the central angle

θ .

Since the radius of a typical sector in [link] is given by

r_{i} = f (θ_{i}),

[link] involved finding the area inside one curve. We can also use [link] to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.

Finding the Area between Two Polar Curves

Find the area outside the cardioid $r = 2 + 2 sin θ$

and inside the circle $r = 6 sin θ .$

First draw a graph containing both curves as shown.

A cardioid with equation r = 2 + 2 sinθ is shown, so it has its upper heart part at the origin and the rest of the cardioid is pointed up. There is a circle with radius 6 centered at (3, π/2). The area above the cardioid but below the circle is shaded orange.

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for $θ :$

\begin{matrix} 6 sin θ & = & 2 + 2 sin θ \\ 4 sin θ & = & 2 \\ sin θ & = & \frac{1}{2} . \end{matrix}

This gives the solutions $θ = \frac{π}{6}$

and $θ = \frac{5 π}{6},$

which are the limits of integration. The circle $r = 3 sin θ$

is the red graph, which is the outer function, and the cardioid $r = 2 + 2 sin θ$

is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between $θ = \frac{π}{6}$

and $θ = \frac{5 π}{6},$

then subtract the area inside the cardioid between $θ = \frac{π}{6}$

and $θ = \frac{5 π}{6} :$

\begin{matrix} A & = circle - cardioid \\ = \frac{1}{2} \int_{π / 6}^{5 π / 6} {[6 sin θ]}^{2} d θ - \frac{1}{2} \int_{π / 6}^{5 π / 6} {[2 + 2 sin θ]}^{2} d θ \\ = \frac{1}{2} \int_{π / 6}^{5 π / 6} 36 {sin}^{2} θ d θ - \frac{1}{2} \int_{π / 6}^{5 π / 6} 4 + 8 sin θ + 4 {sin}^{2} θ d θ \\ = 18 \int_{π / 6}^{5 π / 6} \frac{1 - cos (2 θ)}{2} d θ - 2 \int_{π / 6}^{5 π / 6} 1 + 2 sin θ + \frac{1 - cos (2 θ)}{2} d θ \\ = 9 {[θ - \frac{sin (2 θ)}{2}]}_{π / 6}^{5 π / 6} - 2 {[\frac{3 θ}{2} - 2 cos θ - \frac{sin (2 θ)}{4}]}_{π / 6}^{5 π / 6} \\ = 9 (\frac{5 π}{6} - \frac{sin 2 (5 π / 6)}{2}) - 9 (\frac{π}{6} - \frac{sin 2 (π / 6)}{2}) \\ − (3 (\frac{5 π}{6}) - 4 cos \frac{5 π}{6} - \frac{sin 2 (5 π / 6)}{2}) + (3 (\frac{π}{6}) - 4 cos \frac{π}{6} - \frac{sin 2 (π / 6)}{2}) \\ = 4 π . \end{matrix}

In [link] we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for

θ

However, in the graph there are three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of

θ .

for any integer value of n. These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

Arc Length in Polar Curves

Here we derive a formula for the arc length of a curve defined in polar coordinates.

In rectangular coordinates, the arc length of a parameterized curve

(x (t), y (t))

In order to adapt the arc length formula for a polar curve, we use the equations

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

\begin{matrix} L & = \int_{α}^{β} \sqrt{{[f (θ)]}^{2} + {[f^{'} (θ)]}^{2}} d θ \\ = \int_{0}^{2 π} \sqrt{{[2 + 2 cos θ]}^{2} + {[- 2 sin θ]}^{2}} d θ \\ = \int_{0}^{2 π} \sqrt{4 + 8 cos θ + 4 {cos}^{2} θ + 4 {sin}^{2} θ} d θ \\ = \int_{0}^{2 π} \sqrt{4 + 8 cos θ + 4 ({cos}^{2} θ + {sin}^{2} θ)} d θ \\ = \int_{0}^{2 π} \sqrt{8 + 8 cos θ} d θ \\ = 2 \int_{0}^{2 π} \sqrt{2 + 2 cos θ} d θ . \end{matrix}

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

\begin{matrix} L & = 2 \int_{0}^{2 π} \sqrt{2 + 2 cos θ} d θ \\ = 2 \int_{0}^{2 π} \sqrt{4 {cos}^{2} (\frac{θ}{2})} d θ \\ = 2 \int_{0}^{2 π} 2 \| cos (\frac{θ}{2}) \| d θ . \end{matrix}

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 $\frac{π}{2} .$

Thus,

\begin{matrix} L & = 4 \int_{0}^{2 π} \| cos (\frac{θ}{2}) \| d θ \\ = 8 \int_{0}^{π} cos (\frac{θ}{2}) d θ \\ = 8 {(2 sin (\frac{θ}{2}))}_{0}^{π} \\ = 16. \end{matrix}

Key Concepts

Key Equations

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

Region enclosed by $r = 4$

Region enclosed by $r = 3 sin θ$

\frac{9}{2} \int_{0}^{π} {sin}^{2} θ d θ

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

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

32 \int_{0}^{π / 2} {sin}^{2} (2 θ) d θ

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

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

\frac{1}{2} \int_{π}^{2 π} {(1 - sin θ)}^{2} d θ

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

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

\int_{{sin}^{−1} (2 / 3)}^{π / 2} {(2 - 3 sin θ)}^{2} d θ

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

Region enclosed by $r = 1 - 2 cos θ$

and outside the inner loop

\int_{0}^{π} {(1 - 2 cos θ)}^{2} d θ - \int_{0}^{π / 3} {(1 - 2 cos θ)}^{2} d θ

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

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

4 \int_{0}^{π / 3} d θ + 16 \int_{π / 3}^{π / 2} ({cos}^{2} θ) d θ

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

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

Enclosed by $r = 6 sin θ$

9 π

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

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

\frac{9 π}{4}

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

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

\frac{9 π}{8}

Enclosed by $r = 1 + sin θ$

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

\frac{18 π - 27 \sqrt{3}}{2}

Enclosed by $r = 2 + 4 cos θ$

and outside the inner loop

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

\frac{4}{3} (4 π - 3 \sqrt{3})

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

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

\frac{3}{2} (4 π - 3 \sqrt{3})

Inside $r = 1 + cos θ$

and outside $r = cos θ$

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

2 π - 4

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

r = 4 cos θ on the interval 0 \leq θ \leq \frac{π}{2}

r = 1 + sin θ

on the interval $0 \leq θ \leq 2 π$

\int_{0}^{2 π} \sqrt{{(1 + sin θ)}^{2} + {cos}^{2} θ} d θ

r = 2 sec θ on the interval 0 \leq θ \leq \frac{π}{3}

r = e^{θ} on the interval 0 \leq θ \leq 1

\sqrt{2} \int_{0}^{1} e^{θ} d θ

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

r = 6 on the interval 0 \leq θ \leq \frac{π}{2}

r = e^{3 θ} on the interval 0 \leq θ \leq 2

\frac{\sqrt{10}}{3} (e^{6} - 1)

r = 6 cos θ on the interval 0 \leq θ \leq \frac{π}{2}

r = 8 + 8 cos θ on the interval 0 \leq θ \leq π

r = 1 - sin θ on the interval 0 \leq θ \leq 2 π

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

[T] $r = 3 θ on the interval 0 \leq θ \leq \frac{π}{2}$

6.238

[T] $r = \frac{2}{θ} on the interval π \leq θ \leq 2 π$

[T] $r = {sin}^{2} (\frac{θ}{2}) on the interval 0 \leq θ \leq π$

[T] $r = 2 θ^{2} on the interval 0 \leq θ \leq π$

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

4.39

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 \leq θ \leq π

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

A = π {(\frac{\sqrt{2}}{2})}^{2} = \frac{π}{2} and \frac{1}{2} \int_{0}^{π} (1 + 2 sin θ cos θ) d θ = \frac{π}{2}

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

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 \leq θ \leq π

C = 2 π (\frac{3}{2}) = 3 π and \int_{0}^{π} 3 d θ = 3 π

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

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

C = 2 π (5) = 10 π and \int_{0}^{π} 10 d θ = 10 π

Verify that if $y = r sin θ = f (θ) sin θ$

then $\frac{d y}{d θ} = f' (θ) sin θ + f (θ) cos θ .$

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 $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ}$

and the product rule to derive the derivative of a polar equation.

\frac{d y}{d x} = \frac{f^{'} (θ) sin θ + f (θ) cos θ}{f^{'} (θ) cos θ - f (θ) sin θ}

r = 1 - sin θ;

(\frac{1}{2}, \frac{π}{6})

r = 4 cos θ;

(2, \frac{π}{3})

The slope is $\frac{1}{\sqrt{3}} .$

r = 8 sin θ;

(4, \frac{5 π}{6})

r = 4 + sin θ;

(3, \frac{3 π}{2})

The slope is 0.

r = 6 + 3 cos θ;

(3, π)

r = 4 cos (2 θ);

tips of the leaves

At $(4, 0),$

the slope is undefined. At $(−4, \frac{π}{2}),$

the slope is 0.

r = 2 sin (3 θ);

tips of the leaves

r = 2 θ;

(\frac{π}{2}, \frac{π}{4})

The slope is undefined at $θ = \frac{π}{4} .$

Find the points on the interval $− π \leq θ \leq π$

at which the cardioid $r = 1 - cos θ$

has a vertical or horizontal tangent line.

For the cardioid $r = 1 + sin θ,$

find the slope of the tangent line when $θ = \frac{π}{3} .$

Slope = −1.

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 = \frac{π}{3}

r = θ,

θ = \frac{π}{2}

Slope is $\frac{−2}{π} .$

r = ln θ,

θ = e

[T] Use technology: $r = 2 + 4 cos θ$

at $θ = \frac{π}{6}$

Calculator answer: −0.836.

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

r = 4 cos θ

r^{2} = 4 cos (2 θ)

Horizontal tangent at $(± \sqrt{2}, \frac{π}{6}),$

(± \sqrt{2}, - \frac{π}{6}) .

r = 2 sin (2 θ)

The cardioid $r = 1 + sin θ$

Horizontal tangents at $\frac{π}{2}, \frac{7 π}{6}, \frac{11 π}{6} .$

Vertical tangents at $\frac{π}{6}, \frac{5 π}{6}$

and also at the pole $(0, 0) .$

Show that the curve $r = sin θ tan θ$

(called a cissoid of Diocles) has the line $x = 1$

as a vertical asymptote.