Calculating area bounded by polar functions
Area can be bounded by a polar 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:
If the measure of the angle of the sector is in radians, we can equate the ratio of the area of the sector and the area of the circle to the ratio of the sector angle and the radian measure of the complete circle, $2\pi$:
$$\frac{A}{\pi r^2} = \frac{\theta}{2 \pi}$$
Now if we multiply both sides by $\pi r^2$, we get
$$A = \frac{\theta \pi r^2}{2 \pi} \; \longrightarrow \; A = \frac{\theta r^2}{2}$$
That's the area of a sector of a perfect circle. Now we can use this idea to calculate the area of a non-circular polar-defined area, much as we integrated rectangular functions by adding up rectangles.
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
$$A = \frac{1}{2} \int r^2 \, d\theta = \frac{1}{2} \int [f(\theta)]^2 \, d\theta$$
Now we just need to do some examples . . .
Area bounded by a polar curve
The area bounded by a polar curve and the rays that intercept it is
$$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta = \frac{1}{2} \int_{\theta_1}^{\theta_2} [f(\theta)]^2 d\theta$$
Example 1
Find the area of the function $f(\theta) = 2 cos(4\theta)$ between $\theta = \frac{\pi}{6}$ and $\theta = \frac{\pi}{3}.$
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):
$$ \begin{align} A &= \frac{1}{2} \int [f(\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int [2 cos(4\theta)]^2 \, d\theta \\ \\ &= 2 \int cos^2 (4 \theta) \, d\theta \end{align}$$
Now if we make a substitution, $u = 4\theta$, then $du = 4 \, dx$, and we have
$$= \frac{2}{4} \int cos^2(u) \, du$$
Now for this integral we need the power reduction identity (see trigonometric integrals),
$$cos^2 (x) = \frac{1 + cos(2x)}{2}$$
The resulting integral is
$$= \frac{1}{2} \int \frac{1 + cos(8\theta)}{2} \, d\theta$$
We can pull out the factor of 2 in the denominator (a constant),
$$= \frac{1}{4} \int 1 + cos(8\theta) \, d\theta$$
and separate the integrals (the integral of a sum is a sum of integrals), and substitute u (another u) for 8θ to get:
$$= \frac{1}{4} \int d\theta + \frac{1}{32} \int cos(u) \, du$$
Now we can integrate and evaluate the limits:
$$= \frac{\theta}{4} + \frac{1}{32} sin(8\theta) \bigg|_{\pi/6}^{\pi/3}$$
The limits give the expression
$$\frac{\pi}{12} + \frac{1}{32}sin\left( \frac{8\pi}{3} \right) - \frac{\pi}{24} - \frac{1}{32} sin \left( \frac{4\pi}{3} \right)$$
which reduces to
$$\frac{\pi}{24} + \frac{\sqrt{3}}{64} + \frac{\sqrt{3}}{64}$$
and finally:
$$= \frac{\pi}{24} + \frac{\sqrt{3}}{32} = 0.185$$
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 $f(\theta) = 2 \cdot cos(4 \theta)$ and $f(\theta) = 2 \cdot cos(5 \theta)$. 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 $[0, \theta]$, 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.
Now the graph of $f(\theta) = 2 \cdot cos(5 \theta)$ is different. Here is the graph drawn by sweeping the angle between 0 and $\pi$. In this case, the full figure is drawn over that interval.
In this graph we sweep over the angle interval $[0, 2\pi]$, and you can see it's the same. We observe that for the graphs of $r = sin(n \theta)$, when $n$ is odd there are $n$ lobes of the "flower," and the full figure is traced over the interval $[0, \pi]$, and when $n$ is even, there are $2n$ lobes, which require the interval $[0, 2\pi]$ to be fully drawn.
Why this matters
If you'd like to calculate the area enclosed by the full graph of $f(\theta) = 2 \cdot cos(4 \theta)$, the limits of integration will be 0 and $2\pi$, but using those same limits on $f(\theta) = 2 \cdot cos(5 \theta)$ would give you twice the enclosed area because you would have traced the figure twice. Be careful.
Example 2
Find the area of the function $f(\theta) = cos(\theta)$ 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 $r = A \cdot cos(\theta)$ and $r = A\cdot sin(\theta)$ 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:
$$ \begin{align} A &= \frac{1}{2} \int [f(\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int_{-\pi/2}^{\pi/2} \, cos^2(\theta) \, d\theta \end{align}$$
The limits of integration are from -π/2 to π/2. I'll leave those off now until the end, just for clarity. We again use the power reduction formula (very handy in polar integrals) to reduce the integral to the integral of a sum:
$$= \frac{1}{2} \int \frac{1 + cos(2\theta)}{2} \, d\theta$$
The 2 in the denominator can come out of the integral:
$$= \frac{1}{4} \int 1 + cos(2 \theta) \, d\theta$$
and now with a little bit of substitution, the integral is:
$$= \frac{\theta}{4} + \frac{1}{8} \, sin(2\theta) \bigg|_{-\pi/2}^{\pi/2}$$
Evaluating the limits is pretty easy because the sine function is zero at multiples of π:
$$= \frac{\pi}{8} + \frac{1}{8} sin(\pi) - \frac{-\pi}{8} - \frac{1}{8} sin(3 \theta)$$
and the result is
$$= \frac{\pi}{4}$$
This is just what we'd expect from a circle of radius $\frac{1}{2}A = \pi r^2 = \pi/4$
Example 3
Find the area inside of the function $f(\theta) = 1$ and outside of the cardioid $g(\theta) = 1 - cos(\theta).$
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 $r = 1 - cos(\theta)$.
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
$$r = 1 - cos(\theta) \; \; \; r = 1$$
and equating them by eliminating r (the transitive property, remember?), we get
$$ \begin{align} 1 - cos(\theta) &= 1 \\ cos(\theta) &= 0 \end{align}$$
In the domain of interest, the two solutions are
$$\theta = ± \frac{\pi}{2}$$
So the integral will be the square of the outer function (the circle) minus the square of the inner one, the cardioid:
$$A = \frac{1}{2} \int_{-\pi/2}^{\pi/2} \, 1^2 - (1 - cos(\theta))^2 \, d\theta $$
We can take advantage of symmetry here. We just cut the interval in half and multiply the whole thing by 2:
$$= \int_0^{\pi/2} \, 1^2 - (1 - cos(\theta))^2 \, d\theta$$
Now expand that squared term,
$$= \int_0^{\pi/2} \, 1 - (1 - 2 cos(\theta) + cos^2(\theta)) \, d\theta$$
... and do the subtractions to get
$$= \int_0^{\pi/2} \, 2 cos(\theta) - cos^2(\theta) \, d\theta$$
To address the $cos^2(\theta)$ term, we'll employ the same power-reduction identity we did in the examples above:
$$= \int_0^{\pi/2} \, 2 cos(\theta) - \frac{1 + cos(2\theta)}{2} \, d\theta$$
And the solution pops right out:
$$= \int_0^{\pi/2} \, 2 cos(\theta) - \frac{1 + cos(2 \theta)}{2} \, d\theta$$
Evaluating the limits gives us the area:
$A = 2 - \frac{\pi}{4}$ units2
Practice problems
| 1. | Calculate the area enclosed by the polar function $r = 2·sin(\theta).$ | |
| 2. | Calculate the area enclosed by the polar function $r = 1 - sin(\theta).$ | |
| 3. | Calculate the area enclosed by one loop of the polar function $r = 2·sin(5\theta).$ | |
| 4. | Calculate the area that lies inside of the curve $r = 2 + sin(\theta)$ and outside of the curve $r = 3·sin(\theta).$ | |
| 5. | Find the area between one large loop of the curve, $1 + 2·cos(3\theta),$ and its enclosed small loop. |
Problem 1 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.
$$ \begin{align} A &= \frac{1}{2} \int_0^{\pi} [f(\theta)]^2 \, d\theta = \frac{1}{2} \int_0^{\pi} [2 sin(\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int_0^{2\pi} 4 sin^2 (\theta) \, d\theta \\ \\ &= 2 \int_0^{2\pi} \frac{1 - cos(2\theta)}{2} \, d\theta \\ \\ &= \int_0^{\pi} 1 - cos(2\theta) \, d\theta \\ \\ &= \left[ \theta + \frac{1}{2} sin(2\theta) \right]_0^{\pi} \\ \\ &= \pi + \frac{1}{2} sin(2\pi) - 0 - \frac{1}{2} sin(0) \\ &= \pi \; units^2 \end{align}$$
Problem 2 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.
$$ \begin{align} A &= \frac{1}{2} \int_0^{2\pi} [f(\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int_0^{2\pi} [1 - sin(\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int_0^{2\pi} 1 - 2 sin(\theta) + sin^2(\theta) \, d\theta \\ \\ &= \frac{1}{2} \int_0^{2\pi} 1 - sin(\theta) + \frac{1 - cos(2\theta)}{2} d\theta \\ \\ &= \frac{1}{2}\left[ \theta + 2 cos(\theta) + \frac{1}{2}\theta + \frac{1}{4} sin(2\theta) \right]_0^{2\pi} \\ \\ &= \frac{1}{2} \left[ \frac{3}{2} \theta + 2 cos(\theta) + \frac{1}{4} sin(2\theta) \right]_0^{2\pi} \\ \\ &= \frac{1}{2} \left[ \frac{3}{2}(2\pi) + 2 cos(2\pi) - 0 - 2 cos(0) \right] \\ \\ &= \frac{1}{2} [3\pi + 2 - 2] = \frac{3\pi}{2} \; units^2 \end{align}$$
Problem 3 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.
$$ \begin{align} A &= \frac{1}{2} \int_{3\pi/10}^{7\pi/10} [f(\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int_{3\pi/10}^{7\pi/10} [2sin(5\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int_{3\pi/10}^{7\pi/10} 4 sin^2(5\theta) \, d\theta \\ \\ &= 2 \int_{3\pi/10}^{7\pi/10} \frac{1 - cos(10\theta)}{2} \, d\theta \\ \\ &= \left[ \theta - \frac{1}{10} sin(10\theta) \right]_{3\pi/10}^{7\pi/10} \\ \\ &= \frac{7\pi}{10} - \frac{1}{10} sin(7\pi) - \frac{3}{10} + \frac{1}{10}sin(3\pi) \\ \\ &= \frac{4\pi}{10} = \frac{2\pi}{5} \; units^2 \\ \\ \end{align}$$
Of course, you could have also done this problem by integrating between 0 and 2π and dividing by 5 to get the same result.
Problem 4 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.
$$ \begin{align} A &= \frac{1}{2} \int_{0}^{\pi} [(2 + sin(\theta))^2 - (3 sin(\theta))^2] \, d\theta \\ \\ &= \frac{1}{2} \int_{0}^{\pi} [4 + 2 sin(\theta) + sin^2(\theta) - 9 sin^2(\theta)] \, d\theta \\ \\ &= \frac{1}{2} \int_{0}^{\pi} [4 + 2 sin(\theta) - 8 sin^2(\theta)] \, d\theta \\ \\ &= \frac{1}{2} \int_{0}^{\pi} \left[ 4 + 2 sin(\theta) - \frac{8(1 - cos(2\theta))}{2}\right] \, d\theta \\ \\ &= \int_{0}^{\pi} 2 + sin(\theta) - 2 - 2 cos(2\theta) \, d\theta \\ \\ &= \int_{0}^{\pi} sin(\theta) - 2 cos(2\theta) \, d\theta \\ \\ &= -cos(\theta) - sin(2\theta) \bigg|_0^{\pi} \\ \\ &= -cos(\pi) - sin(2\pi) + cos(0) + sin(0) \\ \\ &= 1 + 1 = 2 \; units^2 \end{align}$$
Problem 5 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:
$$ \begin{align} A &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} [1 + 2 cos(3\theta)]^2 \, d\theta \\ \\ &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} [1 + 4 cos(3\theta) + 4 cos^2(3\theta)] \, d\theta \\ \\ &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} \left[1 + 4 cos(3\theta) \frac{4(1 - cos(6\theta))}{2}\right] \, d\theta \\ \\ &= \frac{1}{2} \int_{-\pi/6}^{\pi/6} \left[ \frac{1}{2} + 2 cos(3\theta) + 2 - 2cos(6\theta) \right] \, d\theta \\ \\ &= \int_{-\pi/6}^{\pi/6} \left[ \frac{5}{2} + 2 cos(3\theta) - 2 cos{6\theta)} \right] \, d\theta \\ \\ &= \left[ \frac{5}{2} \theta + \frac{2}{3} sin(3\theta) - \frac{1}{3} sin(6\theta) \right]_{-\pi/6}^{\pi/6} \\ \\ &= 2\left[ \frac{5}{2} \theta + \frac{2}{3} sin(3\theta) - \frac{1}{3} sin(6\theta) \right]_0^{\pi/6} \\ \\ &= 2\left[ \frac{5}{2} \frac{\pi}{6} + \frac{2}{3} sin \left( \frac{\pi}{2} \right) - \frac{1}{3} sin(\pi) \right] \\ \\ &= 2 \left[ \frac{5\pi}{12} + \frac{2}{3} - 0 \right] \\ \\ &= 2 \left[ \frac{15\pi + 24}{36}\right] = \frac{15\pi + 24}{18} \\ \\ &= \frac{5\pi + 8}{6} \; units^2 \end{align}$$
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.
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