Area Computations in Polar Coordinates

The graph of the polar-coordinate equations $r=f(\theta)$ may bound an area, as does the cardioid $r=2(1+\cos\theta)$. To calculate the area of this region, we may find it convenient to work directly with polar coordinates rather than to change to rectangular coordinates.

To see how to set up an area integral using polar coordinates, we consider the region $R$ bounded by the two radial lines $\theta=\alpha$ and $\theta=\beta$ and by the curve $r=f(\theta)$, $\alpha\leq \theta\leq \beta$. To approximate the area $A$ of $R$, we begin with the partition $$ \alpha=\theta_0\lt \theta_1\lt \theta_2\lt . . . \lt \theta_n=\beta. $$ of the interval $[\alpha,\beta]$ into $n$ subintervals, all with the same length $\Delta\theta=(\beta-\alpha)/n$. We select a point $\theta_i^*$ in the $i$th subinterval $[\theta_{i-1},\theta_i]$ for $i=1$, 2, 3, $. . .$, $n$.

Let $\Delta A_i$ denote the area of the sector bounded by the lines $\theta=\theta_{i-1}$ and $\theta=\theta_i$ and by the curve $y=f(\theta)$. We see that for small values of $\Delta \theta$, $\Delta A_i$ is approximately equal to the area of the circular sector that has radius $r_i^*=f(\theta_i^*)$ and is bounded by the same lines. That is, $$ \Delta A_i \approx \frac12 (r_i^*)^2\Delta \theta=\frac12 [f(\theta_i^*)]^2 \Delta \theta. $$ We add the areas of these sectors for $i=1$, 2, 3, $. . .$, $n$ and thereby find that $$ A=\sum_{i=1}^n \Delta A_i\approx \frac12 \left[f(\theta_i^*\right]^2\, \Delta\theta. $$ The right-hand sum is a Riemann sum for the integral $$ \int_\alpha^\beta \frac12\left[f(\theta)\right]^2\, d\theta. $$ hence, if $f$ is continuous, the value of this integral is the limit, as $\Delta \theta\to 0$, of the preceding sum. We therefore conclude that the area $A$ of the region $R$ bounded by the lines $\theta=\alpha$ and $\theta=\beta$ and the curve $r=f(\theta)$ is \begin{equation}\label{eqn1} A=\int_\alpha^\beta \frac12\left[f(\theta)\right]^2\, d\theta. \end{equation} The infinitesimal sector with radius $r$ and central angle $d\theta$, and area $dA=\frac12 r^2d\theta$, serves as a useful device for remembering Eq. (\ref{eqn1}) in the abbreviated form \begin{equation}\label{eqn2} A=\int_\alpha^\beta \frac12 r^2\, d\theta. \end{equation}

\begin{example} Find the area of the region bounded by the limaçon with equation $r=3+2\cos\theta$, $0\leq \theta\leq 2\pi$. \textbf{Solution: } We could apply Eq. (\ref{eqn2}) with $\alpha=0$ and $\beta=2\pi$. Here, instead, we will make use of symmetry. We will calculate the area of the upper half of the region and then double the result. Note that the infinitesimal sector sweeps out the upper half of the limaçon as $\theta$ increases from $0$ to $\pi$. Hence \begin{eqnarray*} A &=& 2\int_\alpha^\beta \frac12 r^2\, d\theta =\int_0^\pi (3+2\cos\theta)^2\, d\theta\\ &=& \int_0^\pi (9+12\cos\theta+4\cos^2\theta)\, d\theta. \end{eqnarray*} Because $$ 4\cos^2\theta=4\cdot \frac{1+\cos 2\theta}{2}=2+2\cos 2\theta, $$ we now get \begin{eqnarray*} A &=& (11+12\cos\theta+2\cos 2\theta)\, d\theta \\ &=& \left[11\theta+12\sin\theta+\sin 2\theta\right]_0^\pi=11\pi. \end{eqnarray*} \end{example} \begin{example} Find the area bounded by each loop of the limaçon with equation $r=1+2\cos\theta$. \textbf{Solution: } The equation $1+2\cos\theta=0$ has two solutions for $\theta$ in the interval $[0,2\pi]$: $\theta=2\pi/3$ and $\theta=4\pi/3$. The upper half of the outer loop of the limaçon corresponds to values of $\theta$ between 0 and $2\pi/3$, where $r$ is positive. Because the curve is symmetric around the $x$-axis, we can find the total area $A_1$ bounded by the outer loop by integrating from 0 to $2\pi/3$ and then doubling. Thus \begin{eqnarray*} A_1 &=& 2\int_0^{2\pi/3}\frac12(1+2\cos\theta)^2\, d\theta = \int_0^{2\pi/3} (1+4\cos\theta+4\cos^2\theta)\, d\theta \\ &=& \int_0^{2\pi/3} (3+4\cos\theta+2\cos^2\theta)\, d\theta \\ &=& \left[ 3\theta+4\sin\theta+\sin 2\theta \right]_0^{2\pi/3} =2\pi+\frac32\sqrt{3}. \end{eqnarray*} The inner loop of the limaçon corresponds to values of $\theta$ between $2\pi/3$ and $4\pi/3$ where $r$ is negative. Hence the area bounded by the inner loop is \begin{eqnarray*} A_2 &=& \int_{2\pi/3}^{4\pi/3}\frac12(1+2\cos\theta)^2\, d\theta \\ &=& \left[ 3\theta+4\sin\theta+\sin 2\theta \right]_{2\pi/3}^{4\pi/3} =\pi-\frac32\sqrt{3}. \end{eqnarray*} The area of the region lying between the two loops of the limaçon is then $$ A=A_1-A_2=2\pi+\frac32\sqrt{3}-(\pi-\frac32\sqrt{3})=\pi+3\sqrt{3}. $$ \end{example}

The Area Between Two Polar Curves

Now consider two curves $r=f(\theta)$ and $r=g(\theta)$, with $f(\theta)\geq g(\theta)\geq 0$ for $\alpha\leq \theta\leq \beta$. Then we can find the area of the region bounded by these curves and the rays (radial lines) $\theta=\alpha$ and $\theta=\beta$ by subtracting the area bounded by the inner curve from that bounded by the outer curve. That is, the area $A$ between the two curves is given by $$ A=\int_\alpha^\beta \frac12[f(\theta)]^2\, d\theta-\int_\alpha^\beta \frac12[g(\theta)]^2\, d\theta, $$ so that \begin{equation}\label{eqn3} A=\frac12\int_\alpha^\beta \left\{[f(\theta)]^2-[g(\theta)]^2\right\}\, d\theta. \end{equation} \begin{example} Find the area $A$ of the region that lies within the limaçon $r=1+2\cos\theta$ and outside the circle $r=2$. \textbf{Solution: } The circle and limaçon are shown in the figure below:
Figure 1. The Region of Example 3.

The points of intersection of the circle and limaçon are given by $$ 1+2\cos\theta=2,\quad\mbox{so}\quad\cos\theta=\frac12, $$ and the figure shows that we should choose the solutions $\theta=\pm\pi/3$. These two values of $\theta$ are the needed limits of integration. When we use Eq. (\ref{eqn3}), we find that \begin{eqnarray*} A &=& \frac12 \int_{-\pi/3}^{\pi/3} [(1+2\cos\theta)^2-2^2]\, d\theta \\ &=& \int_0^{\pi/3} (4\cos\theta+4\cos^2\theta-3)\, d\theta \\ &=& \int_0^{\pi/3} (4\cos\theta+2\cos 2\theta-1)\, d\theta \\ &=& \left[ 4\sin\theta+\sin 2\theta-\theta\right]_0^{\pi/3}=\frac{15\sqrt{3}-2\pi}{6}. \end{eqnarray*} \end{example}