Geometric Calculations of Polar Curves

Learning Outcomes

Apply the formula for area of a region in polar coordinates
Determine the arc length of a polar curve

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.

Consider a curve defined by the function $r=f\left(\theta \right)$ , where $\alpha \le \theta \le \beta$ . Our first step is to partition the interval $\left[\alpha ,\beta \right]$ into $n$ equal-width subintervals. The width of each subinterval is given by the formula $\Delta \theta =\frac{\left(\beta -\alpha \right)}{n}$ , and the $i$ th partition point ${\theta }_{i}$ is given by the formula ${\theta }_{i}=\alpha +i\Delta \theta$ . Each partition point $\theta ={\theta }_{i}$ defines a line with slope $\tan{\theta }_{i}$ passing through the pole as shown in the following graph.

On the polar coordinate plane, a curve is drawn in the first quadrant, and there are rays from the origin that intersect this curve at a regular interval. Every time one of these rays intersects the curve, a perpendicular line is made from the ray to the next ray. The first instance of a ray-curve intersection is labeled θ = α; the last instance is labeled θ = β. The intervening ones are marked θ1, θ2, …, θn−1.

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.

A circle is drawn with radius r and a sector of angle θ. It is noted that A = (1/2) θ r2.

Recall that the area of a circle is $A=\pi {r}^{2}$ . When measuring angles in radians, 360 degrees is equal to $2\pi$ radians. Therefore a fraction of a circle can be measured by the central angle $\theta$ . The fraction of the circle is given by $\frac{\theta }{2\pi }$ , so the area of the sector is this fraction multiplied by the total area:

A = (\frac{θ}{2 π}) π r^{2} = \frac{1}{2} θ r^{2}

$A=\left(\frac{\theta }{2\pi }\right)\pi {r}^{2}=\frac{1}{2}\theta {r}^{2}$ .

Since the radius of a typical sector in Figure 1 is given by ${r}_{i}=f\left({\theta }_{i}\right)$ , the area of the ith sector is given by

A_{i} = \frac{1}{2} (Δ θ) {(f (θ_{i}))}^{2}

${A}_{i}=\frac{1}{2}\left(\Delta \theta \right){\left(f\left({\theta }_{i}\right)\right)}^{2}$ .

Therefore a Riemann sum that approximates the area is given by

A_{n} = \sum_{i = 1}^{n} A_{i} \approx \sum_{i = 1}^{n} \frac{1}{2} (Δ θ) {(f (θ_{i}))}^{2}

${A}_{n}=\displaystyle\sum _{i=1}^{n}{A}_{i}\approx \displaystyle\sum _{i=1}^{n}\frac{1}{2}\left(\Delta \theta \right){\left(f\left({\theta }_{i}\right)\right)}^{2}$ .

We take the limit as $n\to \infty$ to get the exact area:

A = lim_{n \to \infty} A_{n} = \frac{1}{2} \int_{α}^{β} {(f (θ))}^{2} d θ

$A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\frac{1}{2}{\displaystyle\int }_{\alpha }^{\beta }{\left(f\left(\theta \right)\right)}^{2}d\theta$ .

This gives the following theorem.

theorem: Area of a Region Bounded by a Polar Curve

Suppose $f$ is continuous and nonnegative on the interval $\alpha \le \theta \le \beta$ with $0<\beta -\alpha \le 2\pi$ . The area of the region bounded by the graph of $r=f\left(\theta \right)$ between the radial lines $\theta =\alpha$ and $\theta =\beta$ is

A = \frac{1}{2} \int_{α}^{β} {[f (θ)]}^{2} d θ = \frac{1}{2} \int_{α}^{β} r^{2} d θ

$A=\frac{1}{2}{\displaystyle\int }_{\alpha }^{\beta }{\left[f\left(\theta \right)\right]}^{2}d\theta =\frac{1}{2}{\displaystyle\int }_{\alpha }^{\beta }{r}^{2}d\theta$ .

Recall: Solving Trigonometric Equations

Follow the process below to solve trigonometric equations.

Use trigonometric identities to rewrite the expression in terms of a single trigonometric function, if necessary.
Use algebra to isolate the trigonometric expression.
Identify all angles on the unit circle that satisfy the equation.
Note the period of the function to state all possible solutions for the angle.
Solve for the variable in the angle expression, and identify the angles that lie within the desired interval.

For example, to solve the equation $1 + 3\cos 2\theta = 5 \cos 2\theta$ on the interval $\left[0, 2\pi \right)$

Subtract the $3\cos 2\theta$ term and then divide both sides of the equation by $2$ :

$\cos 2\theta = \frac{1}{2}$

There are two angles on the unit circle where cosine is $\frac{1}{2}$ :

$2\theta = \frac{\pi}{3}$ and $2\theta = \frac{5\pi}{3}$

Since $\cos x$ has a period of $2\pi$ , all possible solutions are given by:

$2\theta = \frac{\pi}{3} + 2\pi k, 2\theta = \frac{5\pi}{3} + 2\pi k$ for any integer $k$ .

Divide by $2$ to isolate $\theta$ :

$\theta = \frac{\pi}{6} + \pi k, \theta = \frac{5\pi}{6} + \pi k$ for any integer $k$

The possible values for $k$ that result in angles within the interval $\left[0, 2\pi \right)$ are $k = 0 \: \text{and} \: 1$ , yielding the solution:

$\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

Example: Finding an Area of a Polar Region

Find the area of one petal of the rose defined by the equation $r=3\sin\left(2\theta \right)$ .

Show Solution

The graph of $r=3\sin\left(2\theta \right)$ follows.

A four-petaled rose with furthest extent 3 from the origin at π/4, 3π/4, 5π/4, and 7π/4.

When $\theta =0$ we have $r=3\sin\left(2\left(0\right)\right)=0$ . The next value for which $r=0$ is $\theta =\frac{\pi}{2}$ . This can be seen by solving the equation $3\sin\left(2\theta \right)=0$ for $\theta$ . Therefore the values $\theta =0$ to $\theta =\frac{\pi}{2}$ trace out the first petal of the rose. To find the area inside this petal, use the theorem with $f\left(\theta \right)=3\sin\left(2\theta \right)$ , $\alpha =0$ , and $\beta =\frac{\pi}{2}\text{:}$

\begin{array}{cc} A & = \frac{1}{2} \int_{α}^{β} {[f (θ)]}^{2} d θ \\ = \frac{1}{2} \int_{0}^{\frac{π}{2}} {[3 \sin (2 θ)]}^{2} d θ \\ = \frac{1}{2} \int_{0}^{\frac{π}{2}} 9 \sin^{2} (2 θ) d θ . \end{array}

$\begin{array}{cc}\hfill A& =\frac{1}{2}{\displaystyle\int }_{\alpha }^{\beta }{\left[f\left(\theta \right)\right]}^{2}d\theta \hfill \\ & =\frac{1}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}{\left[3\sin\left(2\theta \right)\right]}^{2}d\theta \hfill \\ & =\frac{1}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}9{\sin}^{2}\left(2\theta \right)d\theta .\hfill \end{array}$

To evaluate this integral, use the formula ${\sin}^{2}\alpha =\frac{\left(1-\cos\left(2\alpha \right)\right)}{2}$ with $\alpha =2\theta \text{:}$

\begin{array}{cc} A & = \frac{1}{2} \int_{0}^{\frac{π}{2}} 9 \sin^{2} (2 θ) d θ \\ = \frac{9}{2} \int_{0}^{\frac{π}{2}} \frac{(1 - \cos (4 θ))}{2} d θ \\ = \frac{9}{4} (\int_{0}^{\frac{π}{2}} 1 - \cos (4 θ) d θ) \\ = \frac{9}{4} {(θ - \frac{\sin (4 θ)}{4})}_{0}^{\frac{π}{2}} \\ = \frac{9}{4} (\frac{π}{2} - \frac{\sin 2 π}{4}) - \frac{9}{4} (0 - \frac{\sin 4 (0)}{4}) \\ = \frac{9 π}{8} . \end{array}

$\begin{array}{cc}\hfill A& =\frac{1}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}9{\sin}^{2}\left(2\theta \right)d\theta \hfill \\ & =\frac{9}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}\frac{\left(1-\cos\left(4\theta \right)\right)}{2}d\theta \hfill \\ & =\frac{9}{4}\left({\displaystyle\int }_{0}^{\frac{\pi}{2}}1-\cos\left(4\theta \right)d\theta \right)\hfill \\ & =\frac{9}{4}{\left(\theta -\frac{\sin\left(4\theta \right)}{4}\right)}_{0}^{\frac{\pi}{2}}\hfill \\ & =\frac{9}{4}\left(\frac{\pi }{2}-\frac{\sin2\pi }{4}\right)-\frac{9}{4}\left(0-\frac{\sin4\left(0\right)}{4}\right)\hfill \\ & =\frac{9\pi }{8}.\hfill \end{array}$

Watch the following video to see the worked solution to Example: Finding an Area of a Polar Region.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “7.4 Area and Arc Length in Polar Coordinates” here (opens in new window).

Try It

try it

Find the area inside the cardioid defined by the equation $r=1-\cos\theta$ .

Hint

Show Solution

$A=\frac{3\pi}{2}$

The previous example involved finding the area inside one curve. We can also use Area of a Region Bounded by a Polar Curve 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.

Example: Finding the Area between Two Polar Curves

Find the area outside the cardioid $r=2+2\sin\theta$ and inside the circle $r=6\sin\theta$ .

Show Solution

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 $\theta \text{:}$

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

$\begin{array}{ccc}\hfill 6\sin\theta & =\hfill & 2+2\sin\theta \hfill \\ \hfill 4\sin\theta & =\hfill & 2\hfill \\ \hfill \sin\theta & =\hfill & \frac{1}{2}.\hfill \end{array}$

This gives the solutions $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}$ , which are the limits of integration. The circle $r=3\sin\theta$ is the red graph, which is the outer function, and the cardioid $r=2+2\sin\theta$ is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}$ , then subtract the area inside the cardioid between $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}\text{:}$

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

$\begin{array}{cc}\hfill A& =\text{circle}-\text{cardioid}\hfill \\ & =\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}{\left[6\sin\theta \right]}^{2}d\theta -\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}{\left[2+2\sin\theta \right]}^{2}d\theta \hfill \\ & =\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}36{\sin}^{2}\theta d\theta -\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}4+8\sin\theta +4{\sin}^{2}\theta d\theta \hfill \\ & =18{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\frac{1-\cos\left(2\theta \right)}{2}d\theta -2{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}1+2\sin\theta +\frac{1-\cos\left(2\theta \right)}{2}d\theta \hfill \\ & =9{\left[\theta -\frac{\sin\left(2\theta \right)}{2}\right]}_{\frac{\pi}{6}}^{\frac{5\pi}{6}}-2{\left[\frac{3\theta }{2}-2\cos\theta -\frac{\sin\left(2\theta \right)}{4}\right]}_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\hfill \\ & =9\left(\frac{5\pi }{6}-\frac{\sin2\left(\frac{5\pi}{6}\right)}{2}\right)-9\left(\frac{\pi }{6}-\frac{\sin2\left(\frac{\pi}{6}\right)}{2}\right)\hfill \\ & \text{-}\left(3\left(\frac{5\pi }{6}\right)-4\cos\frac{5\pi }{6}-\frac{\sin2\left(\frac{5\pi}{6}\right)}{2}\right)+\left(3\left(\frac{\pi }{6}\right)-4\cos\frac{\pi }{6}-\frac{\sin2\left(\frac{\pi}{6}\right)}{2}\right)\hfill \\ & =4\pi .\hfill \end{array}$

Watch the following video to see the worked solution to Example: Finding the Area between Two Polar Curves.

You can view the transcript for this segmented clip of “7.4 Area and Arc Length in Polar Coordinates” here (opens in new window).

try it

Find the area inside the circle $r=4\cos\theta$ and outside the circle $r=2$ .

Hint

Show Solution

$A=\frac{4\pi }{3}+4\sqrt{3}$

In the previous example we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for $\theta$ yielded two solutions: $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}$ . 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 $\theta$ . For example, for the cardioid we get

\begin{array}{ccc} 2 + 2 \sin θ & = & 0 \\ \sin θ & = & - 1, \end{array}

$\begin{array}{ccc}\hfill 2+2\sin\theta & =\hfill & 0\hfill \\ \hfill \sin\theta & =\hfill & -1,\hfill \end{array}$

so the values for $\theta$ that solve this equation are $\theta =\frac{3\pi }{2}+2n\pi$ , where n is any integer. For the circle we get

6 \sin θ = 0

$6\sin\theta =0$ .

The solutions to this equation are of the form $\theta =n\pi$ 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 $\left(x\left(t\right),y\left(t\right)\right)$ for $a\le t\le b$ is given by

L = \int_{a}^{b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t

$L={\displaystyle\int }_{a}^{b}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt$ .

In polar coordinates we define the curve by the equation $r=f\left(\theta \right)$ , where $\alpha \le \theta \le \beta$ . In order to adapt the arc length formula for a polar curve, we use the equations

x = r \cos θ = f (θ) \cos θ and y = r \sin θ = f (θ) \sin θ

$x=r\cos\theta =f\left(\theta \right)\cos\theta \:\:\text{and} \:\:y=r\sin\theta =f\left(\theta \right)\sin\theta$ ,

and we replace the parameter t by $\theta$ . Then

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

$\begin{array}{c}\frac{dx}{d\theta }={f}^{\prime }\left(\theta \right)\cos\theta -f\left(\theta \right)\sin\theta \hfill \\ \frac{dy}{d\theta }={f}^{\prime }\left(\theta \right)\sin\theta +f\left(\theta \right)\cos\theta .\hfill \end{array}$

We replace $dt$ by $d\theta$ , and the lower and upper limits of integration are $\alpha$ and $\beta$ , respectively. Then the arc length formula becomes

\begin{array}{cc} L & = \int_{a}^{b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t \\ = \int_{α}^{β} \sqrt{{(\frac{d x}{d θ})}^{2} + {(\frac{d y}{d θ})}^{2}} d θ \\ = \int_{α}^{β} \sqrt{{(f^{'} (θ) \cos θ - f (θ) \sin θ)}^{2} + {(f^{'} (θ) \sin θ + f (θ) \cos θ)}^{2}} d θ \\ = \int_{α}^{β} \sqrt{{(f^{'} (θ))}^{2} (\cos^{2} θ + \sin^{2} θ) + {(f (θ))}^{2} (\cos^{2} θ + \sin^{2} θ)} d θ \\ = \int_{α}^{β} \sqrt{{(f^{'} (θ))}^{2} + {(f (θ))}^{2}} d θ \\ = \int_{α}^{β} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ . \end{array}

$\begin{array}{cc}\hfill L& ={\displaystyle\int }_{a}^{b}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt\hfill \\ & ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{d\theta }\right)}^{2}+{\left(\frac{dy}{d\theta }\right)}^{2}}d\theta \hfill \\ & ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left({f}^{\prime }\left(\theta \right)\cos\theta -f\left(\theta \right)\sin\theta \right)}^{2}+{\left({f}^{\prime }\left(\theta \right)\sin\theta +f\left(\theta \right)\cos\theta \right)}^{2}}d\theta \hfill \\ & ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left({f}^{\prime }\left(\theta \right)\right)}^{2}\left({\cos}^{2}\theta +{\sin}^{2}\theta \right)+{\left(f\left(\theta \right)\right)}^{2}\left({\cos}^{2}\theta +{\sin}^{2}\theta \right)}d\theta \hfill \\ & ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left({f}^{\prime }\left(\theta \right)\right)}^{2}+{\left(f\left(\theta \right)\right)}^{2}}d\theta \hfill \\ & ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta .\hfill \end{array}$

This gives us the following theorem.

theorem: Arc Length of a Curve Defined by a Polar Function

Let $f$ be a function whose derivative is continuous on an interval $\alpha \le \theta \le \beta$ . The length of the graph of $r=f\left(\theta \right)$ from $\theta =\alpha$ to $\theta =\beta$ is

L = \int_{α}^{β} \sqrt{{[f (θ)]}^{2} + {[f^{'} (θ)]}^{2}} d θ = \int_{α}^{β} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ

$L={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left[f\left(\theta \right)\right]}^{2}+{\left[{f}^{\prime }\left(\theta \right)\right]}^{2}}d\theta ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta$ .

Example: finding the arc length of a polar curve

Find the arc length of the cardioid $r=2+2\cos\theta$ .

Show Solution

When $\theta =0,r=2+2\cos0=4$ . Furthermore, as $\theta$ goes from $0$ to $2\pi$ , the cardioid is traced out exactly once. Therefore these are the limits of integration. Using $f\left(\theta \right)=2+2\cos\theta$ , $\alpha =0$ , and $\beta =2\pi$ , the theorem equation becomes

\begin{array}{cc} 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{array}

$\begin{array}{cc}\hfill L& ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left[f\left(\theta \right)\right]}^{2}+{\left[{f}^{\prime }\left(\theta \right)\right]}^{2}}d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\sqrt{{\left[2+2\cos\theta \right]}^{2}+{\left[-2\sin\theta \right]}^{2}}d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi}\sqrt{4+8\cos\theta +4{\cos}^{2}\theta +4{\sin}^{2}\theta }d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\sqrt{4+8\cos\theta +4\left({\cos}^{2}\theta +{\sin}^{2}\theta \right)}d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\sqrt{8+8\cos\theta }d\theta \hfill \\ & =2{\displaystyle\int }_{0}^{2\pi }\sqrt{2+2\cos\theta }d\theta .\hfill \end{array}$

Next, using the identity $\cos\left(2\alpha \right)=2{\cos}^{2}\alpha -1$ , add $1$ to both sides and multiply by $2$ . This gives $2+2\cos\left(2\alpha \right)=4{\cos}^{2}\alpha$ . Substituting $\alpha =\frac{\theta}{2}$ gives $2+2\cos\theta =4{\cos}^{2}\left(\frac{\theta}{2}\right)$ , so the integral becomes

\begin{array}{cc} 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{array}

$\begin{array}{cc}\hfill L& =2{\displaystyle\int }_{0}^{2\pi }\sqrt{2+2\cos\theta }d\theta \hfill \\ & =2{\displaystyle\int }_{0}^{2\pi }\sqrt{4{\cos}^{2}\left(\frac{\theta }{2}\right)}d\theta \hfill \\ & =2{\displaystyle\int }_{0}^{2\pi }2|\cos\left(\frac{\theta }{2}\right)|d\theta .\hfill \end{array}$

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 $\pi$ and double the answer. This strategy works because cosine is positive between $0$ and $\frac{\pi }{2}$ . Thus,

\begin{array}{cc} 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{array}

$\begin{array}{cc}\hfill L& =4{\displaystyle\int }_{0}^{2\pi }|\cos\left(\frac{\theta }{2}\right)|d\theta \hfill \\ & =8{\displaystyle\int }_{0}^{\pi }\cos\left(\frac{\theta }{2}\right)d\theta \hfill \\ & =8{\left(2\sin\left(\frac{\theta }{2}\right)\right)}_{0}^{\pi }\hfill \\ & =16.\hfill \end{array}$

Watch the following video to see the worked solution to Example: finding the arc length of a polar curve.

You can view the transcript for this segmented clip of “7.4 Area and Arc Length in Polar Coordinates” here (opens in new window).

try it

Find the total arc length of $r=3\sin\theta$ .

Hint

Show Solution

$s=3\pi$

Module 1: Parametric Equations and Polar Coordinates