Introduction
The area of a section of a circle is given by the formula:
A = \frac{1}{2} r^2 \theta
Where
r
is the radius of the circle.
\theta
is the angle of the section. For example, a complete circle has an angle of
\theta=2\pi
, and its area is
A = \frac{1}{2} r^2 2\pi = \pi r^2
To find the area bounded by a polar curve
r = f(\theta)
between angles
\theta = \alpha
and
\theta = \beta
:
A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 d\theta
In the previous module, we found the area of explicit functions by summing the areas of infinitely many, infinitely thin rectangles.
We find the area under polar curves by summing the areas of infinitely many, infinitely thin slices of circles.
Finding the Area of the Region Bounded by Two Polar Curves
To find the area between two polar curves
r = f(\theta)
and
r = g(\theta)
from
\theta = \alpha
to
\theta = \beta
:
A = \frac{1}{2} \int_{\alpha}^{\beta} \left| [f(\theta)]^2 - [g(\theta)]^2 \right| d\theta
This is the same as finding the area of the infinitely many, infinitely thin circle slices comprising the outer curve, and then subtracting the area of the infinitely many, infinitely thin circle slices comprising the inner curve.
Just like in the last module, we must split the integral wherever the functions cross through, exchanging the inner and outer functions in the integral.
Otherwise, we may accidentally include negative areas, which would cancel out the positive ones and give a smaller area than is there.
Examples - Area Under Polar Curves
Example 1: Circle Verification
Find the area enclosed by the polar circle
r = 5
.
\begin{align*}
A &= \frac{1}{2} \int_{0}^{2\pi} 5^2 \, d\theta \\
&= \frac{1}{2} \, 25 \, \theta\Big|_0^{2\pi} \\
&= 25\pi \, \text{units}^2
\end{align*}
This matches the standard circle area formula
\pi r^2 = 25\pi
.
Example 2: One Petal of a Rose
Calculate the area of one petal of
r = 3\sin(2\theta)
.
The function has a radius of
0
at
\theta = k\pi/2
. Therefore, the first petal will be on
\theta=[0,\pi/2]
\begin{align*}
A &= \frac{1}{2} \int_{0}^{\pi/2} (3\sin2\theta)^2 \, d\theta \\
&= \frac{9}{2} \int_{0}^{\pi/2} \sin^2 2\theta \, d\theta \\
&= \frac{9}{4} \int_{0}^{\pi/2} (1 - \cos4\theta) \, d\theta \\
&= \frac{9}{4} \left(\theta - \frac{\sin4\theta}{4}\right)_0^{\pi/2} \\
&= \frac{9}{4} \cdot \frac{\pi}{2} = \frac{9\pi}{8} \, \text{units}^2
\end{align*}
The area of one petal is
9\pi/8 \text{ units}^2
.
Example 3: Limaçon Area
Find the area inside
r = 2 + \cos(\theta)
.
\begin{align*}
A &= \frac{1}{2} \int_{0}^{2\pi} (2 + \cos(\theta))^2 \, d\theta \\
&= \frac{1}{2} \int_{0}^{2\pi} (4 + 4\cos(\theta) + \cos^2\theta) \, d\theta \\
&= \frac{1}{2} \left(4\theta + 4\sin(\theta) + \frac{\theta}{2} + \frac{\sin2\theta}{4}\right)_0^{2\pi} \\
&= \frac{1}{2} \left(8\pi + 0 + \pi + 0\right) = \frac{9\pi}{2} \, \text{units}^2
\end{align*}
The limaçon encloses an area of
9\pi/2 \text{ units}^2
.
Example 4: Area Between Two Intersecting Circles
Find the area of the space enclosed by
r = 4\sin(\theta)
and
r = 2
.
First, find the intersection points:
4\sin(\theta) = 2 \implies \sin(\theta) = \frac{1}{2} \implies \theta = \frac{\pi}{6}, \frac{5\pi}{6}
Between
\pi/6
and
5\pi/6
,
r = 4\sin(\theta)
is larger. The area is:
\begin{align*}
A &= \frac{1}{2} \int_{\pi/6}^{5\pi/6} \left[(4\sin(\theta))^2 - 2^2\right] d\theta \\
&= \frac{1}{2} \int_{\pi/6}^{5\pi/6} (16\sin^2\theta - 4) d\theta \\
&= 8 \int_{\pi/6}^{5\pi/6} (1 - \cos(2\theta)) d\theta - 2 \int_{\pi/6}^{5\pi/6} d\theta \\
&= \frac{8\pi}{3} + 2\sqrt{3} \, \text{units}^2
\end{align*}
The area equals
\frac{8\pi}{3} + 2\sqrt{3}
.