Calculating area bounded by polar functions

Area can be bounded by apolar function, and we can use the definite integral to calculate it. Here is a typical polar area problem. The function $r = f(\theta)$ is intercepted by two rays making angles $\theta_a$ and $\theta_b$ with the axis system, as shown.

We integrate by "sweeping" a ray through the area from $\theta_a$ to $\theta_b$, adding up the area of infinitessimally small sectors. The way we do this is completely analogous to Riemann integration of a rectangular function using rectangles. For polar functions, the rectangles just become sectors of circles, which approximate sectors of any polar function if they're narrow enough.

Reminder: Area of a sector of a circle (angle in radians)

Before we derive the formula for integrating area bounded by polar functions, we should review the concept of determining the area of a sector of a circle. Here's a sector cut out of a circle:

Putting it together to calculate areas

Here's a sector of a non-circular polar function, the area of which we might want to calculate. Imagine marking out small sectors of "width" $d\theta$. Now the circular ends of those sectors don't exacly match the curvature of the polar function $f(\theta)$, but as we reduce the angle of those sectors to a limit of zero, the fit becomes perfect, and we can add up the exact area.

So the integrated area of a polar sector (a region bounded by a polar function and the rays that intercept it) is

$$A=\frac{1}{2} \int r^2 d \theta = \frac{1}{2} \int f(\theta)]^2 d\theta$$

Example 1

Find the area of the function $f(\theta) = 2 \text{cos}(4 \theta)$ between $\theta=\pi/6$ and $\theta=\pi/3$.

The polar graph of this kind of function looks like a flower. The multiplier (4) of $\theta$ is half of the number of "petals," and the multiplier of the cosine function is just a scaling factor, which causes each petal to be 2 units long. It looks like this:

Now the area we're determining is just that of the one petal that lies between $\theta=\pi/6$ and $\theta=\pi/2$. Remember that if you wanted to calculate the area of the whole figure, you could just multiply by 8. Make sure to use symmetry to your advantage.

The integral is set up like this (I'll leave the limits off for now):

$$ \begin{align} A &= \frac{1}{2} \int [f(\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int [2 \, \text{cos}(4\theta)]^2 \, d\theta \\[5pt] &= 2 \int \text{cos}^2(4 \theta) \, d\theta \end{align}$$

Now if we make a substitution, $u=4\theta$, then $du=4 \, dx$, and we have

A caution: Polar functions are quirky

Be cautious about expecting polar functions and graphs to behave like rectangular functions and graphs. There are some big differences.

One major difference is illustrated in the graphs below. Consider the similar functions $f(\theta)=2 \cdot \text{cos}(4\theta)$ and $f(\theta)=2 \cdot \text{cos}(5\theta)$. They differ only in whether the multiplier of $\theta$ is even or odd. The graph of the first is shown twice below. In the first plot, it's been rendered by rotating $r$ through the angle interval $[0, \; \pi]$, and you can see that only half of the graph is drawn.

To fully draw the graph, we need to sweep the angle through the interval $[0, \; 2\pi]$. The full graph is below. You can see how it's traced by following the arrows in the plot above.

Example 2

Find the area of the function $f(\theta)=\text{cos}(\theta)$ between $\theta=-\frac{\pi}{2}$ and $\theta=\frac{\pi}{2}$.

The graph of this function is a circle of radius $r = 0.5$ centered on the $x$-axis and tangent to the origin. The limits of integration tell us that we should arrive at the area of a circle of radius one, $A = \pi$.

Note: The graphs of $r=A \, \text{cos}(\theta)$ and $r=A \, \text{sin}(\theta)$ are circles of radius $A/2$. It's a common mistake to think that $A$ is the radius of such a circle. The normal rules that apply in Cartesian coordinates don't apply in polar coordinates.

We set up the integral like this:

$$ \begin{align} A &= \frac{1}{2} \int [f(\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \text{ cos}^2(\theta) \, d\theta \end{align}$$

Example 3

Find the area inside of the function $f(\theta)=1$ and outside of the cardioid $g(\theta)=1-\text{cos}(\theta)$

This region is shown in the graph below. It lies within the circle of radius $r=1$ centered at the origin, and outside of the "cardioid" figure $r=1-\text{cos}(\theta)$.

The idea, completely analogous to finding the area between Cartesian curves, is to find the area inside the circle, from one angle-endpoint to the other (the points of intersection), and to subtract the corresponding area of the cardioid, so that the remaining area is what we seek.

The first job is to find the endpoints. The functions are

$$r=1-\text{cos}(\theta) \phantom{00000} r=1$$

and equating them by eliminating r (the transitive property, remember?), we get

$$ \begin{align} 1-\text{cos}(\theta) &= 1 \\[5pt] \text{cos}(\theta) &= 0 \end{align}$$

In the domain of interest, the two solutions are

$$\theta = \pm \frac{\pi}{2}$$

Practice problems

1. Calculate the area enclosed by the polar function   $r = 2·\text{sin}(\theta).$

   Solution

   
   

   First graph this function on a calculator or some other device (or by hand if you have the patience). It traces a complete circle for θ between 0 and π Integrating between 0 and 2π would give twice the actual area, so be careful.

   
   

   

   
   

   To do this integral, you'll need the power reduction formula for sin². Review that here.

   $$ \begin{align} A &= \frac{1}{2} \int_0^{\pi} [f(\theta)]^2 \, d\theta = \frac{1}{2} \int_0^{\pi} [2 \text{sin}(\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int_0^{2\pi} 4 \text{sin}^2 (\theta) \, d\theta \\[5pt] &= 2 \int_0^{2\pi} \frac{1 - \text{cos}(2\theta)}{2} \, d\theta \\[5pt] &= \int_0^{\pi} 1 - \text{cos}(2\theta) \, d\theta \\[5pt] &= \left[ \theta + \frac{1}{2} \text{sin}(2\theta) \right]_0^{\pi} \\[5pt] &= \pi + \frac{1}{2} \text{sin}(2\pi) - 0 - \frac{1}{2} \text{sin}(0) \\[5pt] &= \pi \; units^2 \end{align}$$




2. Calculate the area enclosed by the polar function   $r = 1 - \text{sin}(\theta).$

   Solution

   
   

   First graph this function on a calculator or some other device (or by hand if you have the patience). It traces a cardioid (heart-shaped figure) when graphed over θ = 0 to 2π.

   
   

   

   
   

   To do this integral, you'll need the power-reduction formula for sin². Review that here.

   $$ \begin{align} A &= \frac{1}{2} \int_0^{2\pi} [f(\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int_0^{2\pi} [1 - \text{sin}(\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int_0^{2\pi} 1 - 2 \text{sin}(\theta) + \text{sin}^2(\theta) \, d\theta \\[5pt] &= \frac{1}{2} \int_0^{2\pi} 1 - \text{sin}(\theta) + \frac{1 - \text{cos}(2\theta)}{2} d\theta \\[5pt] &= \frac{1}{2}\left[ \theta + 2 \text{cos}(\theta) + \frac{1}{2}\theta + \frac{1}{4} \text{sin}(2\theta) \right]_0^{2\pi} \\[5pt] &= \frac{1}{2} \left[ \frac{3}{2} \theta + 2 \text{cos}(\theta) + \frac{1}{4} \text{sin}(2\theta) \right]_0^{2\pi} \\[5pt] &= \frac{1}{2} \left[ \frac{3}{2}(2\pi) + 2 \text{cos}(2\pi) - 0 - 2 \text{cos}(0) \right] \\[5pt] &= \frac{1}{2} [3\pi + 2 - 2] = \frac{3\pi}{2} \; units^2 \end{align}$$




3. Calculate the area enclosed by one loop of the polar function   $r = 2·\text{sin}(5\theta).$

   Solution

   
   

   First graph this function on a calculator or some other device (or by hand if you have the patience). It traces a five-petaled flower. We need to find the limits of integration to find the area of one of them. The vertical one is the easiest. It is centered at π/2 and extends right and left by one half of one-fifth of the domain:

   $$\frac{\pi}{2} ± \left( \frac{2\pi}{5} \right) \frac{1}{2} = \frac{\pi}{2} ±\frac{\pi}{5} = \frac{7\pi}{10}, \frac{3\pi}{10}$$

   
   

   

   
   

   To do this integral, you'll need the power-reduction formula for sin². Review that here.

   $$ \begin{align} A &= \frac{1}{2} \int_{3\pi/10}^{7\pi/10} [f(\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int_{3\pi/10}^{7\pi/10} [2\text{sin}(5\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int_{3\pi/10}^{7\pi/10} 4 \text{sin}^2(5\theta) \, d\theta \\[5pt] &= 2 \int_{3\pi/10}^{7\pi/10} \frac{1 - \text{cos}(10\theta)}{2} \, d\theta \\[5pt] &= \left[ \theta - \frac{1}{10} \text{sin}(10\theta) \right]_{3\pi/10}^{7\pi/10} \\[5pt] &= \frac{7\pi}{10} - \frac{1}{10} \text{sin}(7\pi) - \frac{3}{10} + \frac{1}{10}\text{sin}(3\pi) \\[5pt] &= \frac{4\pi}{10} = \frac{2\pi}{5} \; units^2 \\[5pt] \end{align}$$

   Of course, you could have also done this problem by integrating between 0 and 2π and dividing by 5 to get the same result.




4. Calculate the area that lies inside of the curve   $r = 2 + \text{sin}(\theta)$   and outside of the curve   $r = 3·\text{sin}(\theta).$

   Solution

   
   

   First graph these two functions. They trace two ircles of different radius and center location. We're after the area between the blue and red circles. (Note: these don't look like circles on the calculator screen because of the aspect ratio (ratio of number horizontal to vertical pixels, but they are.)

   
   

   

   
   

   To do this integral, you'll need the power-reduction formula for sin². Review that here.

   $$ \begin{align} A &= \frac{1}{2} \int_{0}^{\pi} [(2 + \text{sin}(\theta))^2 - (3 \text{sin}(\theta))^2] \, d\theta \\[5pt] &= \frac{1}{2} \int_{0}^{\pi} [4 + 2 \text{sin}(\theta) + \text{sin}^2(\theta) - 9 \text{sin}^2(\theta)] \, d\theta \\[5pt] &= \frac{1}{2} \int_{0}^{\pi} [4 + 2 \text{sin}(\theta) - 8 \text{sin}^2(\theta)] \, d\theta \\[5pt] &= \frac{1}{2} \int_{0}^{\pi} \left[ 4 + 2 \text{sin}(\theta) - \frac{8(1 - \text{cos}(2\theta))}{2}\right] \, d\theta \\[5pt] &= \int_{0}^{\pi} 2 + \text{sin}(\theta) - 2 - 2 \text{cos}(2\theta) \, d\theta \\[5pt] &= \int_{0}^{\pi} \text{sin}(\theta) - 2 \text{cos}(2\theta) \, d\theta \\[5pt] &= -\text{cos}(\theta) - \text{sin}(2\theta) \bigg|_0^{\pi} \\[5pt] &= -\text{cos}(\pi) - \text{sin}(2\pi) + \text{cos}(0) + \text{sin}(0) \\[5pt] &= 1 + 1 = 2 \; units^2 \end{align}$$




5. Find the area between one large loop of the curve,   $1 + 2·\text{cos}(3\theta),$   and its enclosed small loop.

   Solution

   
   

   When this figure is drawn, half of it is rendered between θ = 0 - π, and the other half between π and 2π. For the loops on the right, the big loop is first then the smaller. That gives us our integration strategy.

   
   

   

   
   

   To do this integral, you'll need the power-reduction formula for sin². Review that here.

   Outer loop:

   $$ \begin{align} A &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} [1 + 2 \text{cos}(3\theta)]^2 \, d\theta \\[5pt] &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} [1 + 4 \text{cos}(3\theta) + 4 \text{cos}^2(3\theta)] \, d\theta \\[5pt] &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} \left[1 + 4 \text{cos}(3\theta) \frac{4(1 - \text{cos}(6\theta))}{2}\right] \, d\theta \\[5pt] &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} \left[ \frac{1}{2} + 2 \text{cos}(3\theta) + 2 - 2\text{cos}(6\theta) \right] \, d\theta \\[5pt] &= \int_{-\pi/6}^{\pi/6} \left[ \frac{5}{2} + 2 \text{cos}(3\theta) - 2 \text{cos}{6\theta)} \right] \, d\theta \\[5pt] &= \left[ \frac{5}{2} \theta + \frac{2}{3} \text{sin}(3\theta) - \frac{1}{3} \text{sin}(6\theta) \right]_{-\pi/6}^{\pi/6} \\[5pt] &= 2\left[ \frac{5}{2} \theta + \frac{2}{3} \text{sin}(3\theta) - \frac{1}{3} \text{sin}(6\theta) \right]_0^{\pi/6} \\[5pt] &= 2\left[ \frac{5}{2} \frac{\pi}{6} + \frac{2}{3} \text{sin} \left( \frac{\pi}{2} \right) - \frac{1}{3} \text{sin}(\pi) \right] \\[5pt] &= 2 \left[ \frac{5\pi}{12} + \frac{2}{3} - 0 \right] \\[5pt] &= 2 \left[ \frac{15\pi + 24}{36}\right] = \frac{15\pi + 24}{18} \\[5pt] &= \frac{5\pi + 8}{6} \; units^2 \end{align}$$

   Now to find the area in the small lobe, evaluate the same integral between the limits of 2π ± π/6, and subtract. I'll leave that part for you.