Introduction

Welcome to this comprehensive guide on finding areas bounded by two polar curves! This topic builds directly on your knowledge of polar area calculations for single curves, extending the concept to regions enclosed between two different polar curves. While finding the area inside a single polar curve uses one integral, finding the area between two curves requires careful analysis of which curve is the outer boundary and which is the inner boundary at each angle.

The essential knowledge tells us that areas of regions bounded by polar curves can be calculated with definite integrals. This means we can find areas between circles and cardioids, between different rose petals, or between any two polar curves by setting up appropriate definite integrals. The key insight is that we integrate the difference between the outer and inner radii squared, accounting for the sector-based geometry of polar coordinates.

Understanding polar area calculations between curves is essential for solving complex geometric problems involving rotational systems, electromagnetic field regions, and engineering applications where multiple boundaries define regions of interest. This topic combines your knowledge of polar coordinates, definite integrals, and curve analysis to solve area problems that would be extremely challenging in rectangular coordinates.

Understanding Area Between Polar Curves

When finding the area between two polar curves $r=f(\theta )$ (outer) and $r=g(\theta )$ (inner), we subtract the area of the inner region from the area of the outer region:

$A={\int }_{\alpha }^{\beta }\frac{1}{2}[f(\theta ){]}^{2}d\theta -{\int }_{\alpha }^{\beta }\frac{1}{2}[g(\theta ){]}^{2}d\theta$

This simplifies to the fundamental formula:

$A={\int }_{\alpha }^{\beta }\frac{1}{2}\left([f(\theta ){]}^2-[g(\theta ){]}^2\right)d\theta$

The geometric interpretation is that we're finding the area of sectors with outer radius $f(\theta )$ and subtracting the area of sectors with inner radius $g(\theta )$, creating a "washer" or "annulus" shape in polar coordinates.

Critical Point: Determining which curve is outer and which is inner may change throughout the interval $[\alpha ,\beta ]$. We must analyze the curves carefully and potentially split the integral at intersection points.

Finding Intersection Points

Before calculating areas, we must find where the curves intersect by solving: $f(\theta )=g(\theta )$

Example 1: Intersection of Circle and Cardioid

Find the intersection points of $r=2$ and $r=2(1+\cos \theta )$.

Setting the equations equal: $2=2(1+\cos \theta )$

$1=1+\cos \theta$

$\cos \theta =0$

This gives us $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$.

At these angles, both curves have $r=2$, confirming our intersection points are $\left(2,\frac{\pi }{2}\right)$ and $\left(2,\frac{3\pi }{2}\right)$ in polar coordinates.

Area Of Limacon Within a Circle

Example 2: Area Between Polar Curves

Find the area inside $r=2$ but outside $r=1+\cos \theta$.

This requires finding intersection points and setting up appropriate integrals. The cardioid $r=1+\cos \theta$ intersects the circle $r=2$ when: $1+\cos \theta =2⟹\cos \theta =1⟹\theta =0,2\pi$

Since the cardioid is entirely inside the circle, the area is: $A={\int }_0^{2\pi }\frac{1}{2}(2^2)d\theta -{\int }_0^{2\pi }\frac{1}{2}(1+\cos \theta {)}^{2}d\theta$

$=2\pi \cdot 2-6\pi =4\pi -\frac{3\pi }{2}=\frac{5\pi }{2}$

Area Between Circle and Cardioid

Example 3: Area Inside Circle but Outside Cardioid

Find the area inside $r=2$ but outside $r=2(1+\cos \theta )$.

From our intersection analysis, we need to determine which curve is outer in different intervals:

The area inside the circle but outside the cardioid exists only where the circle is the outer boundary:

$A={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\frac{1}{2}\left(2^2-[2(1+\cos \theta ){]}^2\right)d\theta$

$={\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\frac{1}{2}\left(4-4(1+\cos \theta {)}^2\right)d\theta$

$=2{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left(1-(1+\cos \theta {)}^2\right)d\theta$

$=2{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left(1-(1+2\cos \theta +{\cos }^2\theta )\right)d\theta$

$=2{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left(-2\cos \theta -{\cos }^2\theta \right)d\theta$

Using ${\cos }^2\theta =\frac{1+\cos (2\theta )}{2}$:

$=2{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left(-2\cos \theta -\frac{1+\cos (2\theta )}{2}\right)d\theta$

$=2{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left(-2\cos \theta -\frac{1}{2}-\frac{\cos (2\theta )}{2}\right)d\theta$

$=2{\left[-2\sin \theta -\frac{\theta }{2}-\frac{\sin (2\theta )}{4}\right]}_{\frac{\pi }{2}}^{\frac{3\pi }{2}}$

Evaluating at the limits:

$=2\left[\left(-2(-1)-\frac{3\pi }{4}-0\right)-\left(-2(1)-\frac{\pi }{4}-0\right)\right]$

$=2\left[2-\frac{3\pi }{4}+2+\frac{\pi }{4}\right]=2\left[4-\frac{\pi }{2}\right]=8-\pi$

Area Between Rose Curves

Example 4: Area Between Rose Petals

Find the area inside $r=4\cos (2\theta )$ but outside $r=2$.

First, find intersections:

$4\cos (2\theta )=2$

$\cos (2\theta )=\frac{1}{2}$

$2\theta =\frac{\pi }{3},\frac{5\pi }{3}$

$\theta =\frac{\pi }{6},\frac{5\pi }{6}$

The rose curve $r=4\cos (2\theta )$ has petals for $\theta \in \left[0,\frac{\pi }{4}\right]$ and $\theta \in \left[\frac{3\pi }{4},\pi \right]$.

For the first petal, the rose is outside the circle from $\theta =0$ to $\theta =\frac{\pi }{6}$, and from $\theta =\frac{\pi }{6}$ to $\theta =\frac{\pi }{4}$:

$A_{petal}={\int }_0^{\frac{\pi }{6}}\frac{1}{2}\left([4\cos (2\theta ){]}^2-2^2\right)d\theta +{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{1}{2}[4\cos (2\theta ){]}^{2}d\theta$

$=8{\int }_0^{\frac{\pi }{6}}\left({\cos }^2(2\theta )-\frac{1}{4}\right)d\theta +8{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}{\cos }^2(2\theta )d\theta$

Using ${\cos }^2(2\theta )=\frac{1+\cos (4\theta )}{2}$:

$=8{\int }_0^{\frac{\pi }{6}}\left(\frac{1+\cos (4\theta )}{2}-\frac{1}{4}\right)d\theta +4{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}(1+\cos (4\theta ))d\theta$

$=8{\int }_0^{\frac{\pi }{6}}\left(\frac{1}{4}+\frac{\cos (4\theta )}{2}\right)d\theta +4{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}(1+\cos (4\theta ))d\theta$

$=2{\int }_0^{\frac{\pi }{6}}(1+2\cos (4\theta ))d\theta +4{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}(1+\cos (4\theta ))d\theta$

$=2\left[\theta +\frac{\sin (4\theta )}{2}\right]0^{\frac{\pi }{6}}+4\left[\theta +\frac{\sin (4\theta )}{4}\right]{\frac{\pi }{6}}^{\frac{\pi }{4}}$

$=2\left[\frac{\pi }{6}+\frac{\sin (\frac{2\pi }{3})}{2}\right]+4\left[\frac{\pi }{4}+0-\frac{\pi }{6}-\frac{\sin (\frac{2\pi }{3})}{4}\right]$

$=2\left[\frac{\pi }{6}+\frac{\sqrt{3}}{4}\right]+4\left[\frac{\pi }{12}-\frac{\sqrt{3}}{16}\right]$

$=\frac{\pi }{3}+\frac{\sqrt{3}}{2}+\frac{\pi }{3}-\frac{\sqrt{3}}{4}=\frac{2\pi }{3}+\frac{\sqrt{3}}{4}$

Area Between Limaçons

Example 5: Area Between Two Limaçons

Find the area inside $r=3+2\cos \theta$ but outside $r=1+2\cos \theta$.

Since both curves have the same $\cos \theta$ term, the outer curve is always $r=3+2\cos \theta$ because it's consistently 2 units farther from the origin.

$A={\int }_0^{2\pi }\frac{1}{2}\left[(3+2\cos \theta {)}^2-(1+2\cos \theta {)}^2\right]d\theta$

Expanding the squares:

$(3+2\cos \theta {)}^2=9+12\cos \theta +4{\cos }^2\theta$

$(1+2\cos \theta {)}^2=1+4\cos \theta +4{\cos }^2\theta$

$A=\frac{1}{2}{\int }_0^{2\pi }(8+8\cos \theta )d\theta$

$=4{\int }_0^{2\pi }(1+\cos \theta )d\theta$

$=4[\theta +\sin \theta {]}_0^{2\pi }=4(2\pi +0-0-0)=8\pi$

Complex Intersection Analysis

Example 6: Area Inside Both Curves

Find the area that lies inside both $r=2+2\sin \theta$ and $r=4\sin \theta$.

First, find intersections:

$2+2\sin \theta =4\sin \theta$

$2=2\sin \theta$

$\sin \theta =1$

$\theta =\frac{\pi }{2}$

We need to determine which curve is inner in the region where both exist. The circle $r=4\sin \theta$ exists for $\theta \in [0,\pi ]$, while the cardioid exists for all $\theta$.

For $\theta \in \left[0,\frac{\pi }{2}\right]$: $4\sin \theta <2+2\sin \theta$

For $\theta \in \left[\frac{\pi }{2},\pi \right]$: $4\sin \theta >2+2\sin \theta$

The area inside both curves is:

$A={\int }_0^{\frac{\pi }{2}}\frac{1}{2}[4\sin \theta {]}^{2}d\theta +{\int }_{\frac{\pi }{2}}^{\pi }\frac{1}{2}[2+2\sin \theta {]}^{2}d\theta$

$=8{\int }_0^{\frac{\pi }{2}}{\sin }^2\theta d\theta +2{\int }_{\frac{\pi }{2}}^{\pi }(1+\sin \theta {)}^{2}d\theta$

$=8{\int }_0^{\frac{\pi }{2}}\frac{1-\cos (2\theta )}{2}d\theta +2{\int }_{\frac{\pi }{2}}^{\pi }(1+2\sin \theta +{\sin }^2\theta )d\theta$

$=4{\int }_0^{\frac{\pi }{2}}(1-\cos (2\theta ))d\theta +2{\int }_{\frac{\pi }{2}}^{\pi }\left(\frac{3}{2}+2\sin \theta -\frac{\cos (2\theta )}{2}\right)d\theta$

$=4\left[\theta -\frac{\sin (2\theta )}{2}\right]0^{\frac{\pi }{2}}+2\left[\frac{3\theta }{2}-2\cos \theta -\frac{\sin (2\theta )}{4}\right]{\frac{\pi }{2}}^{\pi }$

$=4\left[\frac{\pi }{2}-0\right]+2\left[\frac{3\pi }{2}-2(-1)-0-\frac{3\pi }{4}+2(0)+0\right]$

$=2\pi +2\left[\frac{3\pi }{4}+2\right]=2\pi +\frac{3\pi }{2}+4=\frac{7\pi }{2}+4$

Practice Section

Test your understanding with these multiple-choice questions: