Learning Objectives

  • Derive the formula for area of a region in polar coordinates.
  • Determine the arc length of a polar curve.

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 \ds y=f\left(x\right) defined from \ds x=a to \ds x=b where \ds f\left(x\right)\symbol{"3E}0 on this interval, the area between the curve and the x-axis is given by \ds A=\int\limits_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\,dx . We can also find the arc length of this curve using the formula \ds L=\int\limits_{a}^{b}\sqrt{1+{\left({f}^{\prime }\left(x\right)\right)}^{2}}\,dx . In this section, we study formulas for area and arc length in the polar coordinate system.

Areas of Regions Bounded by Polar Curves

Consider a polar curve defined by the function \ds r=f\left(\theta \right), where \ds \alpha \le \theta \le \beta . We want to derive a formula for the area of the region bounded by the curve and between the radial lines \ds \theta =\alpha and \ds \theta =\beta, see Figure 1 below. When defining areas in rectangular coordinates, we approximated the regions with the union of rectangles, and here we are going to use sectors of a circle. Our first step is to partition the interval \ds \left[\alpha ,\beta \right] into n equal-width subintervals. The width of each subinterval is \ds \Delta \theta =\frac{\left(\beta -\alpha \right)}n, and the ith partition point \ds {\theta }_{i} is given by the formula \ds {\theta }_{i}=\alpha +i\Delta \theta . Each partition point \ds \theta ={\theta }_{i} defines a line with slope \ds \text{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.
                                          Figure 1. A partition of a region in polar coordinates.

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 obtain 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.
Figure 2. The area of a sector of a circle is given by \ds A=\frac{1}{2}\theta {r}^{2}.

Recall that the area of a circle is \ds A=\pi {r}^{2}. A fraction of a circle can be measured by the central angle \ds \theta . The full angle is 2\pi, and so the fraction of the circle is given by \ds \frac{\theta }{2\pi }. The area of the sector is this fraction multiplied by the total area:

\ds A=\left(\frac{\theta }{2\pi }\right)\phantom{\rule{0.2em}{0ex}}\pi {r}^{2}=\frac{1}{2}\theta {r}^{2}.

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

\ds A_i=\frac{1}{2}\left(\Delta \theta \right){\left(f\left({\theta }_{i}\right)\right)}^{2}=\frac12(f(\theta_i))^2\Delta\theta.

Summing the areas of sectors for 1\le i\le n, we obtain a Riemann sum that approximates the polar area:

\ds A\approx\sum_{i=1}^n A_n= \sum _{i=1}^{n}\frac12(f(\theta_i))^2\Delta\theta.

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

\ds A=\lim\limits_{n\to\infty} \sum _{i=1}^{n}\frac12(f(\theta_i))^2\Delta\theta=\frac{1}{2}\int\limits_{\alpha }^{\beta }{\left(f\left(\theta \right)\right)}^{2}d\theta .

This gives the following theorem.

Area of a Region Bounded by a Polar Curve

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

\ds (*)\quad A=\frac{1}{2}\int\limits_{\alpha }^{\beta }{\left[f\left(\theta \right)\right]}^{2}d\theta =\frac{1}{2}\int\limits_{\alpha }^{\beta }{r}^{2}d\theta .

Finding an Area of a Polar Region

Find the area of one petal of the rose defined by the equation \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right).

Solution

The graph of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right) is shown below.

A four-petaled rose with furthest extent 3 from the origin at π/4, 3π/4, 5π/4, and 7π/4.
Figure 2. The graph of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right).

Each petal starts and ends at the pole, so to find the endpoints of the \theta interval that corresponds to one petal, we need to solve the equation r=0. We have \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\left(\theta\right)\right)=0, and the solutions are \ds 0,\pm\frac{\pi}2,\pm\pi,\pm\frac{3\pi}2,\ldots since the zeros of the sine function are of the form k\pi, where k is integer. It follows that the petal in the first quadrant corresponds to \theta\ds\in\left[0,\frac{\pi}2\right]. To find the area inside this petal, use (*) from the above theorem with \ds f\left(\theta \right)=3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right), \ds \alpha =0, and \ds \beta =\frac{\pi}2\text{:}

\ds \begin{array}{cc}\ds \hfill A&\ds =\frac{1}{2}\int\limits_{\alpha }^{\beta }{\left[f\left(\theta \right)\right]}^{2}d\theta \hfill \\[5mm]\ds &\ds =\frac{1}{2}\int\limits_{0}^{\pi \text{/}2}{\left[3\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right)\right]}^{2}d\theta \hfill \\[5mm]\ds &\ds =\frac{1}{2}\int\limits_{0}^{\pi \text{/}2}9\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\left(2\theta \right)\phantom{\rule{0.2em}{0ex}}d\theta .\hfill \end{array}

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

\ds \begin{array}{cc}\ds \hfill A&\ds =\frac{1}{2}\int\limits_{0}^{\pi \text{/}2}9\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\left(2\theta \right)\phantom{\rule{0.2em}{0ex}}d\theta \hfill \\[5mm]\ds &\ds =\frac{9}{2}\int\limits_{0}^{\pi \text{/}2}\frac{\left(1-\text{cos}\left(4\theta \right)\right)}{2}d\theta \hfill \\[5mm]\ds &\ds =\frac{9}{4}\int\limits_{0}^{\pi \text{/}2}\left(1-\text{cos}\left(4\theta \right)\right)d\theta \hfill \\[5mm]\ds &\ds =\frac{9}{4}{\left(\theta -\frac{\text{sin}\left(4\theta \right)}{4}\right)}\Big|_{0}^{\pi \text{/}2}\hfill \\[5mm]\ds &\ds =\frac{9}{4}\left(\frac{\pi }{2}-\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}(2\pi) }{4}\right)-\frac{9}{4}\left(0-\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\left(0\right)}{4}\right)\hfill \\[5mm]\ds &\ds =\frac{9\pi }{8}.\hfill \end{array}

Find the area inside the cardioid defined by the equation \ds r=1-\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) .

Answer

\ds A=3\pi \text{/}2

Hint

Use (*). Be sure to determine the correct limits of integration before evaluating.

In the previous example, we were looking for the area inside one curve. We can also use (*) to find the area between two polar curves. For this, we need to find the points of intersection of the curves, determine which function defines the outer curve and which defines the inner curve, and then subtract the corresponding polar areas. This process is illustrated in the example below.

Finding the Area between Two Polar Curves

Find the area outside the cardioid \ds r=2+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) and inside the circle \ds r=6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) .

Solution

First draw a graph containing both curves as shown below.

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.
Figure 3. The region between the curves \ds r=2+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) and \ds r=6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) .

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for \ds \theta \text{:}

\ds \begin{array}{ccc}\hfill 6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds 2+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds \hfill 4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds 2\hfill \\[5mm]\ds \hfill \text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds \frac{1}{2}.\hfill \end{array}

This gives the solutions \ds \theta =\frac{\pi }{6} and \ds \theta =\frac{5\pi }{6} in the interval (-\pi,\pi], which are the limits of integration since from the picture we see that 6\sin(\theta)\ge2+2\sin(\theta) on \ds\left[\frac{\pi}6,\frac{5\pi}6\right]. The circle \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) is the red graph, which is the outer function, and the cardioid \ds r=2+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\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 \ds \theta =\frac{\pi }{6} and \ds \theta =\frac{5\pi }{6}, then subtract the area inside the cardioid between \ds \theta =\frac{\pi }{6} and \ds \theta =\frac{5\pi }{6}\text{:}

\ds \begin{array}{cc}\ds \hfill A&\ds =\text{circle}-\text{cardioid}\hfill \\[5mm]\ds &\ds =\frac{1}{2}\int\limits_{\pi \text{/}6}^{5\pi \text{/}6}{\left[6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \right]}^{2}d\theta -\frac{1}{2}\int\limits_{\pi \text{/}6}^{5\pi \text{/}6}{\left[2+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \right]}^{2}d\theta \hfill \\[5mm]\ds &\ds =\frac{1}{2}\int\limits_{\pi \text{/}6}^{5\pi \text{/}6}36\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}(\theta) \phantom{\rule{0.2em}{0ex}}d\theta -\frac{1}{2}\int\limits_{\pi \text{/}6}^{5\pi \text{/}6}\left(4+8\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +4\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}(\theta)\right)\,d\theta \hfill \\[5mm]\ds &\ds =18\int\limits_{\pi \text{/}6}^{5\pi \text{/}6}\frac{1-\text{cos}\left(2\theta \right)}{2}d\theta -2\int\limits_{\pi \text{/}6}^{5\pi \text{/}6}\left(1+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +\frac{1-\text{cos}\left(2\theta \right)}{2}\right)d\theta \hfill \\[5mm]\ds &\ds =9{\left(\theta -\frac{\text{sin}\left(2\theta \right)}{2}\right)}\Big|_{\pi \text{/}6}^{5\pi \text{/}6}-2{\left(\frac{3\theta }{2}-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) -\frac{\text{sin}\left(2\theta \right)}{4}\right)}\Big|_{\pi \text{/}6}^{5\pi \text{/}6}\hfill \\[5mm]\ds &\ds =9\left(\frac{5\pi }{6}-\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\left(5\pi \text{/}3\right)}{2}\right)-9\left(\frac{\pi }{6}-\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\left(\pi \text{/}3\right)}{2}\right)\hfill \\[5mm]\ds &\ds \phantom{\rule{0.4em}{0ex}} - \left(3\left(\frac{5\pi }{6}\right)-4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\frac{5\pi }{6}-\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\left(5\pi \text{/}3\right)}{2}\right)+\left(3\left(\frac{\pi }{6}\right)-4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{6}-\frac{\text{sin}\phantom{\rule{0.2em}{0ex}}\left(\pi \text{/}3\right)}{2}\right)\hfill \\[5mm]\ds &\ds =4\pi .\hfill \end{array}

Analysis

Notice that equating the formulas and solving for \ds \theta only yielded two solutions: \ds \theta =\frac{\pi }{6} and \ds \theta =\frac{5\pi }{6}. However, in the graph there are three intersection points. The third intersection point is the pole. The reason why this point did not show up as a solution is because the pole is on both graphs but corresponds to different values of \ds \theta . Indeed, for the cardioid we get

\ds \begin{array}{ccc}\hfill 2+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds 0\hfill \\[5mm]\ds \hfill \text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) &\ds =\hfill &\ds -1,\hfill \end{array}

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

\ds 6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) =0.

The solutions to this equation are of the form \ds \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 pole. This case must always be taken into consideration, although in this partcular example it did not affect the calculations.

Find the area inside the circle \ds r=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) and outside the circle \ds r=2.

Answer

\ds A=\frac{4\pi }{3}+2\sqrt{3}

Hint

Use (*) and take advantage of symmetry.

Arc Length for 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 \ds \left(x\left(t\right),y\left(t\right)\right) for \ds a\le t\le b is given by

\ds L=\int\limits_{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 \ds r=f\left(\theta \right), where \ds \alpha \le \theta \le \beta . In order to adapt the arc length formula for a polar curve, we use the equations

\ds x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) =f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) =f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) .

Differentiating, we obtain

\ds \begin{array}{c}\ds\frac{\,dx }{d\theta }={f}^{\prime }\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) -f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \hfill \\[5mm]\ds \frac{dy}{d\theta }={f}^{\prime }\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) .\hfill \end{array}

Applying the known arc length formula, we get

\ds \begin{array}{cc}\ds \hfill L&\ds =\int\limits_{\alpha }^{\beta }\sqrt{{\left(\frac{\,dx }{d\theta }\right)}^{2}+{\left(\frac{dy}{d\theta }\right)}^{2}}d\theta \hfill \\[5mm]\ds &\ds =\int\limits_{\alpha }^{\beta }\sqrt{{\big({f}^{\prime }\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) -f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \big)}^{2}+{\big({f}^{\prime }\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +f\left(\theta \right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \big)}^{2}}d\theta \hfill \\[5mm]\ds &\ds =\int\limits_{\alpha }^{\beta }\sqrt{{\left({f}^{\prime }\left(\theta \right)\right)}^{2}\left({\text{cos}}^{2}(\theta) +{\text{sin}}^{2}(\theta) \right)+{\left(f\left(\theta \right)\right)}^{2}\left({\text{cos}}^{2}(\theta) +{\text{sin}}^{2}(\theta) \right)}d\theta \hfill \\[5mm]\ds &\ds =\int\limits_{\alpha }^{\beta }\sqrt{{\left({f}^{\prime }\left(\theta \right)\right)}^{2}+{\left(f\left(\theta \right)\right)}^{2}}d\theta \hfill \\[5mm]\ds &\ds =\int\limits_{\alpha }^{\beta }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta .\hfill \end{array}

This gives us the following theorem.

Arc Length of a Curve Defined by a Polar Function

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

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

Finding the Arc Length of a Polar Curve

Find the arc length of the cardioid \ds r=2+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) .

Solution

As \ds \theta goes from \ds -\pi to \ds \pi, the cardioid is traced out exactly once. Therefore we can use those as the limits of integration in the formula from the above theorem to obtain

\ds \begin{array}{cc}\ds \hfill L&\ds =\int\limits_{-\pi}^{\pi }\sqrt{{\left[2+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) \right]}^{2}+{\left[-2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(\theta) \right]}^{2}}\phantom{\rule{0.2em}{0ex}}d\theta \hfill \\[5mm]\ds &\ds =\int\limits_{-\pi}^{\pi }\sqrt{4+8\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) +4\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.1em}{0ex}}(\theta) +4\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{2}\phantom{\rule{0.1em}{0ex}}(\theta) }\phantom{\rule{0.2em}{0ex}}d\theta \hfill \\[5mm]\ds &\ds =\int\limits_{-\pi}^{\pi }\sqrt{4+8\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) +4\left({\text{cos}}^{2}\phantom{\rule{0.1em}{0ex}}(\theta) +{\text{sin}}^{2}\phantom{\rule{0.1em}{0ex}}(\theta) \right)}\phantom{\rule{0.2em}{0ex}}d\theta \hfill \\[5mm]\ds &\ds =\int\limits_{-\pi}^{\pi }\sqrt{8+8\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) }\phantom{\rule{0.2em}{0ex}}d\theta \hfill \\[5mm]\ds &\ds =2\int\limits_{-\pi}^{\pi }\sqrt{2+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) }\phantom{\rule{0.2em}{0ex}}d\theta .\hfill \end{array}

Next, the identity \ds \text{cos}\left(2\alpha \right)=2\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}(\alpha) -1, implies that\ds 2+2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(2\alpha \right)=4\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}(\alpha). Substituting \ds \alpha =\frac{\theta}2 gives \ds 2+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) =4\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}\left(\frac{\theta}2\right), so the integral becomes

\ds \begin{array}{cc}\ds \hfill L&\ds =2\int\limits_{-\pi}^{\pi }\sqrt{2+2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) }d\theta \hfill \\[5mm]\ds &\ds =2\int\limits_{-\pi}^{\pi }\sqrt{4\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}\left(\frac{\theta }{2}\right)}d\theta \hfill \\[5mm]\ds &\ds =2\int\limits_{-\pi}^{\pi }2\left|\text{cos}\left(\frac{\theta }{2}\right)\right|d\theta .\hfill \end{array}

When \theta\in[-\pi,\pi], we have that \ds\frac{\theta}2\in\left[-\frac{\pi}2,\frac{\pi}2\right] and since cosine is non-negative on this interval, the absolute value can be dropped and we obtain

\ds \begin{array}{cc}\ds \hfill L&\ds =4\int\limits_{-\pi}^{\pi }\text{cos}\left(\frac{\theta }{2}\right)\,d\theta \hfill \\[5mm]\ds &\ds =4{\left(2\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{\theta }{2}\right)\right)}\Big|_{-\pi}^{\pi }\hfill \\[5mm]\ds &\ds =8(1-(-1))=16.\hfill \end{array}

Find the total arc length of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) .

Answer

\ds s=3\pi

Hint

To determine the correct limits, make a table of values.

Key Concepts

  • The area of the region bounded by the polar curve \ds r=f\left(\theta \right) and between the radial lines \ds \theta =\alpha and \ds \theta =\beta is given by the integral \ds A=\frac{1}{2}{\int\limits_{\alpha }^{\beta }\left[f\left(\theta \right)\right]}^{2}d\theta .
  • 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 \ds r=f\left(\theta \right) with \ds \alpha \le \theta \le \beta is given by the integral \ds L=\int\limits_{\alpha }^{\beta }\sqrt{{\left[f\left(\theta \right)\right]}^{2}+{\left[{f}^{\prime }\left(\theta \right)\right]}^{2}}d\theta =\int\limits_{\alpha }^{\beta }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta .

Key Equations

  • Area of a region bounded by a polar curve
    \ds A=\frac{1}{2}\int\limits_{\alpha }^{\beta }{\left[f\left(\theta \right)\right]}^{2}d\theta =\frac{1}{2}\int\limits_{\alpha }^{\beta }{r}^{2}d\theta
  • Arc length of a polar curve
    \ds L=\int\limits_{\alpha }^{\beta }\sqrt{{\left[f\left(\theta \right)\right]}^{2}+{\left[{f}^{\prime }\left(\theta \right)\right]}^{2}}d\theta =\int\limits_{\alpha }^{\beta }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta

Exercises

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

1. Region enclosed by \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

Answer

\ds \frac{9}{2}\int\limits_{0}^{\pi }{\text{sin}}^{2}(\theta) \phantom{\rule{0.2em}{0ex}}d\theta

2. Region in the first quadrant within the cardioid \ds r=1+\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

3. Region enclosed by one petal of \ds r=8\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\theta \right).

Answer

\ds 32\int\limits_{-\pi\text{/}4}^{\pi \text{/}4}{\text{cos}}^{2}\left(2\theta \right)d\theta

4. Region enclosed by one petal of \ds r=\text{sin}\left(3\theta \right).

5. Region below the polar axis and enclosed by \ds r=1-\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

Answer

\ds \frac{1}{2}\int\limits_{\pi }^{2\pi }{\left(1-\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \right)}^{2}d\theta

6. Region in the first quadrant enclosed by \ds r=2-\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta).

7. Region enclosed by the inner loop of \ds r=2-3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

(Hint: inner loop corresponds to r\le0.)

Answer

\ds \int\limits_{{\text{sin}}^{-1}\left(2\text{/}3\right)}^{\pi \text{/}2}{\left(2-3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \right)}^{2}d\theta

8. Region enclosed by the inner loop of \ds r=3-4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta).

9. Region enclosed by \ds r=1-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) and outside the inner loop.

Answer

\ds \int\limits_{\pi\text{/}3}^{\pi }{\left(1-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \right)}^{2}d\theta -\int\limits_{0}^{\pi \text{/}3}{\left(1-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \right)}^{2}d\theta.

10. Region common to \ds r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=2-\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

(Hint: divide the region into several parts so that each of the parts is determined by one of the curves.)

11. Region common to \ds r=2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta).

(Hint: divide the region into several parts so that each of the parts is determined by one of the curves.)

Answer

\ds 4\int\limits_{0}^{\pi \text{/}3}d\theta +16\int\limits_{\pi \text{/}3}^{\pi \text{/}2}{\text{cos}}^{2}(\theta)d\theta

12. Region common to \ds r=3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

(Hint: divide the region into several parts so that each of the parts is determined by one of the curves.)

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

13. Region enclosed by \ds r=6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

Answer

\ds 9\pi

14. Region above the polar axis and enclosed by \ds r=2+\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

15. Region below the polar axis and enclosed by \ds r=2-\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta).

Answer

\ds \frac{9\pi }{4}

16. Region enclosed by one petal of \ds r=4\phantom{\rule{0.2em}{0ex}}\text{cos}\left(3\theta \right).

17. Region enclosed by one petal of \ds r=3\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\theta \right).

Answer

\ds \frac{9\pi }{8}

18. Region enclosed by \ds r=1+\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

19. Region enclosed by the inner loop of \ds r=3+6\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta).

Answer

\ds \frac{18\pi -27\sqrt{3}}{2}

20. Region enclosed by \ds r=2+4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) and outside the inner loop.

21. Common interior of \ds r=4\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\theta \right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=2.

Answer

\ds \frac{4}{3}\left(4\pi -3\sqrt{3}\right)

22. Common interior of \ds r=\sqrt3-2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=-\sqrt3+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta).

23. Common interior of \ds r=6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=3.

Answer

\ds \frac{3}{2}\left(4\pi -3\sqrt{3}\right)

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

24. \ds r=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta),\phantom{\rule{0.2em}{0ex}}0\le \theta \le \frac{\pi }{2}

25. \ds r=1+\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta), \ds -\pi\le \theta \le \pi

Answer

\ds \int\limits_{-\pi}^{\pi }\sqrt{{\left(1+\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \right)}^{2}+{\text{cos}}^{2}(\theta) }d\theta

26. \ds r=2\phantom{\rule{0.2em}{0ex}}\text{sec}\phantom{\rule{0.2em}{0ex}}(\theta),\phantom{\rule{0.2em}{0ex}}0\le \theta \le \frac{\pi }{3}

27. \ds r={e}^{\theta },\phantom{\rule{0.2em}{0ex}}0\le \theta \le 1

Answer

\ds \sqrt{2}\int\limits_{0}^{1}{e}^{\theta }d\theta

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

28. \ds r=6,\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}0\le \theta \le \frac{\pi }{2}

29. \ds r={e}^{3\theta },\phantom{\rule{0.2em}{0ex}}0\le \theta \le 2

Answer

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

30. \ds r=6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta),\phantom{\rule{0.2em}{0ex}}0\le \theta \le \frac{\pi }{2}

31. \ds r=8+8\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta),\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

Answer

32

32. \ds r=1-\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta),\phantom{\rule{0.2em}{0ex}}0\le \theta \le 2\pi

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

33. [T]\ds r=3\theta,\phantom{\rule{0.2em}{0ex}}0\le \theta \le \frac{\pi }{2}

Answer

6.238

34. [T]\ds r=\frac{2}{\theta },\phantom{\rule{0.2em}{0ex}}\pi \le \theta \le 2\pi

35. [T]\ds r={\text{sin}}^{2}\left(\frac{\theta }{2}\right),\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

Answer

2

36. [T]\ds r=2{\theta }^{2},\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

37. [T]\ds r=\text{sin}\left(3\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \right),\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

Answer

4.39

For the following exercises, rewrite the equation of the given polar curve in rectangular coordinates and use the familiar formula from geometry to find the area enclosed by the curve. Confirm your answer by using the definite integral.

38. \ds r=6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +8\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{on the interval}\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

39. \ds r=\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{on the interval}\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

Answer

\ds A=\pi {\left(\frac{\sqrt{2}}{2}\right)}^{2}=\frac{\pi }{2}\phantom{\rule{0.2em}{0ex}}\ \text{and}\ \phantom{\rule{0.2em}{0ex}}\frac{1}{2}\int\limits_{0}^{\pi }\left(1+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \right)d\theta =\frac{\pi }{2}

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

40. \ds r=6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +8\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{on the interval}\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

Answer

\ds C=2\pi \left(5\right)=10\pi \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\int\limits_{0}^{\pi }10\phantom{\rule{0.2em}{0ex}}d\theta =10\pi

41. \ds r=\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) +\text{cos}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{on the interval}\phantom{\rule{0.2em}{0ex}}0\le \theta \le \pi

42. Show that the curve \ds r=\text{sin}\phantom{\rule{0.2em}{0ex}}(\theta) \phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}(\theta) (called a cissoid of Diocles) has the line \ds x=1 as a vertical asymptote.

(Hint: the line x=1 has polar equation r=\sec(\theta).)

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Calculus: Volume 2 (Second University of Manitoba Edition) by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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