Areas of Regions Bounded by Polar Curves

In rectangular coordinates, you calculate the area under a curve [latex]y = f(x)[/latex] from [latex]x = a[/latex] to [latex]x = b[/latex] using the definite integral [latex]A = \int_a^b f(x) dx[/latex]. You also found arc length using [latex]L = \int_a^b \sqrt{1 + (f'(x))^2} dx[/latex].

The key insight is adapting the Riemann sum approach. For rectangular coordinates, we used rectangles to approximate area. For polar curves, we use circular sectors instead.

Consider a polar curve [latex]r = f(\theta)[/latex] where [latex]\alpha \leq \theta \leq \beta[/latex]. Here’s how we build the area formula:

Divide [latex][\alpha, \beta][/latex] into [latex]n[/latex] equal subintervals, each with width: [latex]\Delta\theta = \frac{\beta - \alpha}{n}[/latex]

The partition points are: [latex]\theta_i = \alpha + i\Delta\theta[/latex]

Each [latex]\theta_i[/latex] creates a line through the origin, dividing our region into sectors. The radius of each sector is [latex]r_i = f(\theta_i)[/latex].

The line segments are connected by arcs of constant radius, creating sectors whose areas we can calculate using a geometric formula. We’ll use these sector areas to approximate the area between successive line segments.

A full circle has area [latex]\pi r^2[/latex] and central angle [latex]2\pi[/latex]. A sector represents the fraction [latex]\frac{\theta}{2\pi}[/latex] of the full circle, so:

Since the radius of each sector is [latex]r_i = f(\theta_i)[/latex], the area of the [latex]i[/latex]th sector becomes:

Taking the limit as [latex]n \to \infty[/latex], we get the exact area:

This derivation leads us to our main theorem.

Find the area of one petal of the rose defined by the equation [latex]r=3\sin\left(2\theta \right)[/latex].

Show Solution

The graph of [latex]r=3\sin\left(2\theta \right)[/latex] follows.

When [latex]\theta =0[/latex] we have [latex]r=3\sin\left(2\left(0\right)\right)=0[/latex]. The next value for which [latex]r=0[/latex] is [latex]\theta =\frac{\pi}{2}[/latex]. This can be seen by solving the equation [latex]3\sin\left(2\theta \right)=0[/latex] for [latex]\theta[/latex]. Therefore the values [latex]\theta =0[/latex] to [latex]\theta =\frac{\pi}{2}[/latex] trace out the first petal of the rose. To find the area inside this petal, use the theorem with [latex]f\left(\theta \right)=3\sin\left(2\theta \right)[/latex], [latex]\alpha =0[/latex], and [latex]\beta =\frac{\pi}{2}\text{:}[/latex]

[latex]\begin{array}{cc}\hfill A& =\frac{1}{2}{\displaystyle\int }_{\alpha }^{\beta }{\left[f\left(\theta \right)\right]}^{2}d\theta \hfill \\ & =\frac{1}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}{\left[3\sin\left(2\theta \right)\right]}^{2}d\theta \hfill \\ & =\frac{1}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}9{\sin}^{2}\left(2\theta \right)d\theta .\hfill \end{array}[/latex]

 

To evaluate this integral, use the formula [latex]{\sin}^{2}\alpha =\frac{\left(1-\cos\left(2\alpha \right)\right)}{2}[/latex] with [latex]\alpha =2\theta \text{:}[/latex]

[latex]\begin{array}{cc}\hfill A& =\frac{1}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}9{\sin}^{2}\left(2\theta \right)d\theta \hfill \\ & =\frac{9}{2}{\displaystyle\int }_{0}^{\frac{\pi}{2}}\frac{\left(1-\cos\left(4\theta \right)\right)}{2}d\theta \hfill \\ & =\frac{9}{4}\left({\displaystyle\int }_{0}^{\frac{\pi}{2}}1-\cos\left(4\theta \right)d\theta \right)\hfill \\ & =\frac{9}{4}{\left(\theta -\frac{\sin\left(4\theta \right)}{4}\right)}_{0}^{\frac{\pi}{2}}\hfill \\ & =\frac{9}{4}\left(\frac{\pi }{2}-\frac{\sin2\pi }{4}\right)-\frac{9}{4}\left(0-\frac{\sin4\left(0\right)}{4}\right)\hfill \\ & =\frac{9\pi }{8}.\hfill \end{array}[/latex]

 

The previous example showed how to find area inside one curve. You can also find the area between two polar curves using the same formula, but this requires some additional steps.

When finding the area between two polar curves, you need to:

• Find the intersection points of the curves
• Determine which curve is the outer curve and which is the inner curve in each region
• Set up the integral as: [latex]A = \frac{1}{2}\int_{\alpha}^{\beta} [r_{\text{outer}}^2 - r_{\text{inner}}^2] d\theta[/latex]

Find the area outside the cardioid [latex]r=2+2\sin\theta[/latex] and inside the circle [latex]r=6\sin\theta[/latex].

Show Solution

First draw a graph containing both curves as shown.

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for [latex]\theta \text{:}[/latex]

[latex]\begin{array}{ccc}\hfill 6\sin\theta & =\hfill & 2+2\sin\theta \hfill \\ \hfill 4\sin\theta & =\hfill & 2\hfill \\ \hfill \sin\theta & =\hfill & \frac{1}{2}.\hfill \end{array}[/latex]

 

This gives the solutions [latex]\theta =\frac{\pi }{6}[/latex] and [latex]\theta =\frac{5\pi }{6}[/latex], which are the limits of integration. The circle [latex]r=3\sin\theta[/latex] is the red graph, which is the outer function, and the cardioid [latex]r=2+2\sin\theta[/latex] is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between [latex]\theta =\frac{\pi }{6}[/latex] and [latex]\theta =\frac{5\pi }{6}[/latex], then subtract the area inside the cardioid between [latex]\theta =\frac{\pi }{6}[/latex] and [latex]\theta =\frac{5\pi }{6}\text{:}[/latex]

[latex]\begin{array}{cc}\hfill A& =\text{circle}-\text{cardioid}\hfill \\ & =\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}{\left[6\sin\theta \right]}^{2}d\theta -\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}{\left[2+2\sin\theta \right]}^{2}d\theta \hfill \\ & =\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}36{\sin}^{2}\theta d\theta -\frac{1}{2}{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}4+8\sin\theta +4{\sin}^{2}\theta d\theta \hfill \\ & =18{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\frac{1-\cos\left(2\theta \right)}{2}d\theta -2{\displaystyle\int }_{\frac{\pi}{6}}^{\frac{5\pi}{6}}1+2\sin\theta +\frac{1-\cos\left(2\theta \right)}{2}d\theta \hfill \\ & =9{\left[\theta -\frac{\sin\left(2\theta \right)}{2}\right]}_{\frac{\pi}{6}}^{\frac{5\pi}{6}}-2{\left[\frac{3\theta }{2}-2\cos\theta -\frac{\sin\left(2\theta \right)}{4}\right]}_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\hfill \\ & =9\left(\frac{5\pi }{6}-\frac{\sin2\left(\frac{5\pi}{6}\right)}{2}\right)-9\left(\frac{\pi }{6}-\frac{\sin2\left(\frac{\pi}{6}\right)}{2}\right)\hfill \\ & \text{-}\left(3\left(\frac{5\pi }{6}\right)-4\cos\frac{5\pi }{6}-\frac{\sin2\left(\frac{5\pi}{6}\right)}{2}\right)+\left(3\left(\frac{\pi }{6}\right)-4\cos\frac{\pi }{6}-\frac{\sin2\left(\frac{\pi}{6}\right)}{2}\right)\hfill \\ & =4\pi .\hfill \end{array}[/latex]

 

In the previous example, we found the area inside the circle and outside the cardioid by first finding their intersection points. When we solved [latex]2 + 2\sin\theta = 6\sin\theta[/latex] directly, we got two solutions: [latex]\theta = \frac{\pi}{6}[/latex] and [latex]\theta = \frac{5\pi}{6}[/latex].

However, looking at the graph reveals three intersection points. The third intersection point is the origin.

The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of [latex]\theta[/latex].