Areas Enclosed by Polar Curves

Areas Enclosed by Polar Curves

Sometimes we are interested in determining an area enclosed by a polar curve $r = f(\theta)$. First, recall that a sector is essentially a slice of a circle, and has an area $A = \frac{1}{2} r^2 \theta$ as shown:


Now suppose that we wanted to find the area of the region enclosed by $r = f(\theta)$, $\theta = a$, and $\theta = b$ as shown in the diagram below:


What we can do is essentially subdivide this interval [a, b] into n sections with endpoints $\theta_0, \theta_1, ..., \theta_n$, with width $\Delta \theta$. Note that the rays extending from the origin to these endpoints create the subdivisions of the area:


We can then take these area subdivisions an approximate the areas of these sectors. That is, $\Delta A_i \approx \frac{1}{2} [ f(\theta_i^*)]^2 \Delta \theta$. If we sum up all of these smaller areas, we will get an approximation to the total area A, that is:

\begin{align} A \approx \sum_{i = 1}^n \frac{1}{2} [ f(\theta_i^*)]^2 \Delta \theta \end{align}

And if we take the limit of this as n goes to infinity, we get the exact area of this region, that is:

\begin{align} \quad A = \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{2} [ f(\theta_i^*)]^2 \Delta \theta = \int_a^b \frac{1}{2} [f(\theta)]^2 \: d \theta = \int_a^b \frac{1}{2} r^2 \: d \theta \end{align}

Example 1

Determine the area enclosed in the inner loop of the polar curve $r = 4\cos \theta + 2$.

We note that the area we are trying to evaluate is highlighted in green in the following graph:


To determine this area, we need to find limits of integration. In this case, the limits of integration will be the points which start and terminate the loop, or more precisely, the points where the curve intersects the origin and r = 0. Setting our polar curve to 0, we get that:

\begin{align} 0 = 4 \cos \theta + 2 \\ -\frac{1}{2} = \cos \theta \\ \theta = \frac{2 \pi}{3}, \frac{4 \pi}{3} \end{align}

We are now ready to set up our integral and evaluate it as follows (using the trigonometric substitution that $\cos ^2 \theta = \frac{1}{2}(1 + \cos {2 \theta})$):

\begin{align} A = \int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} \frac{1}{2} (4\cos \theta + 2)^2 \: d \theta \\ \quad A = \int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} \frac{1}{2} \cdot [16\cos ^2 \theta + 16 \cos \theta + 4 ] \: d \theta \\ A = \int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} [8\cos ^2 \theta + 8 \cos \theta + 2 ] \: d \theta \\ \quad A = \int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} [4(1 + \cos {2\theta}) + 8 \cos \theta + 2 ] \: d \theta \\ A = \left ( 4 \theta + 2 \sin {2 \theta}) + 8 \sin \theta + 2 \theta \right ) \bigg |_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} \\ A = \left ( 6 \theta + 2 \sin {2 \theta} + 8 \sin \theta \right ) \bigg |_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} \\ A = \frac{8 \pi}{3} \\ A = 4\pi - 6\sqrt{3} \end{align}
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