Finding the Area of the Region Bounded by Two Polar Curves - AP Calculus BC Study Guide

The Core Idea: Finding the Area of the Region Bounded by Two Polar Curves

This topic extends the concept of finding the area of a region defined by a single polar curve to finding the area of a region situated between two distinct polar curves. The fundamental approach is analogous to finding the area between two functions in Cartesian coordinates, which involves subtracting the area of the inner region from the area of the outer region. Instead of summing the areas of infinitesimally thin rectangles, in the polar system, we sum the areas of infinitesimally thin sectors.

The core problem is to calculate the area of the region bounded by an "outer" curve, $r=f(\theta )$, and an "inner" curve, $r=g(\theta )$, over a specific interval of angles from $\theta =\alpha$ to $\theta =\beta$. This is achieved by integrating the difference between the area element of the outer curve and the area element of the inner curve across the specified angular range.

Key Formulas

The area of the region bounded by two polar curves, $r=f(\theta )$ and $r=g(\theta )$, for the interval $\alpha \le \theta \le \beta$ is given by a single definite integral.

Area Between Two Polar Curves:

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

Conditions for the Formula:

For this formula to be valid, the following conditions must be met over the interval $[\alpha ,\beta ]$:

1. $f(\theta )$ represents the outer curve (farther from the origin).

2. $g(\theta )$ represents the inner curve (closer to the origin).

3. The radii must satisfy $f(\theta )\ge g(\theta )\ge 0$.

Understanding the Setup

The primary challenge in finding the area between two polar curves lies in correctly setting up the definite integral. This process involves more than just substituting functions into the formula; it requires a careful analysis of the geometry of the region.

The formula $A=\frac{1}{2}{\int }_{\alpha }^{\beta }\left((f(\theta ){)}^2-(g(\theta ){)}^2\right)d\theta$ is a direct consequence of the principle of subtracting the inner area from the outer area. It can be expanded conceptually as:

$$A=\left(\text{Area of outer region}\right)-\left(\text{Area of inner region}\right)$$

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

The key steps to correctly set up the integral are:

1. Identify the Curves: Determine which function, $f(\theta )$ or $g(\theta )$, corresponds to the outer boundary and which corresponds to the inner boundary for the region of interest. A sketch of the curves is often essential.

2. Find the Bounds of Integration ($\alpha$ and $\beta$): The bounds are the starting and ending angles that trace the desired region. These are typically found by setting the two polar equations equal to each other, $f(\theta )=g(\theta )$, and solving for $\theta$. These solutions represent the angles at which the curves intersect.

3. Construct the Integral: Substitute the outer function for $f(\theta )$, the inner function for $g(\theta )$, and the intersection angles for $\alpha$ and $\beta$ into the area formula.

Core Concepts & Rules

• The area of a region bounded by two polar curves is found by subtracting the area of the inner region from the area of the outer region.

• The definitive formula for the area between $r=f(\theta )$ (outer) and $r=g(\theta )$ (inner) from $\theta =\alpha$ to $\theta =\beta$ is $A=\frac{1}{2}{\int }_{\alpha }^{\beta }\left((f(\theta ){)}^2-(g(\theta ){)}^2\right)d\theta$.

• A critical first step is to determine the bounds of integration, $\alpha$ and $\beta$. This is typically done by finding the intersection points of the two curves by solving $f(\theta )=g(\theta )$.

• It is essential to correctly identify which function represents the outer radius ($f(\theta )$) and which represents the inner radius ($g(\theta )$) over the integration interval. Visualizing the graphs is the most reliable method.

• The formula requires squaring each radius function individually before finding the difference. The expression is $(f(\theta ){)}^2-(g(\theta ){)}^2$, not $(f(\theta )-g(\theta ){)}^2$.

Step-by-Step Example 1: Basic Application

Problem: Find the area of the region that lies inside the cardioid $r=4+4\cos \theta$ and outside the circle $r=6$.

Step 1: Find the points of intersection to determine the bounds.

Set the two equations equal to each other:

$$4+4\cos \theta =6$$

$$4\cos \theta =2$$

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

The principal solutions for $\theta$ in the interval $[-\pi ,\pi ]$ are $\theta =-\frac{\pi }{3}$ and $\theta =\frac{\pi }{3}$. These will be our bounds of integration, so $\alpha =-\frac{\pi }{3}$ and $\beta =\frac{\pi }{3}$.

Step 2: Identify the outer and inner curves.

The problem asks for the area inside the cardioid and outside the circle. This means the cardioid is the outer boundary and the circle is the inner boundary.

• Outer curve: $f(\theta )=4+4\cos \theta$

• Inner curve: $g(\theta )=6$

We can verify this by testing a point within the interval, such as $\theta =0$.

• $f(0)=4+4\cos (0)=4+4=8$

• $g(0)=6$

Since $8>6$, the cardioid is indeed the outer curve in this region.

Step 3: Set up the definite integral.

Using the area formula $A=\frac{1}{2}{\int }_{\alpha }^{\beta }\left((f(\theta ){)}^2-(g(\theta ){)}^2\right)d\theta$:

$$A=\frac{1}{2}{\int }_{-\pi /3}^{\pi /3}\left((4+4\cos \theta {)}^2-(6{)}^2\right)d\theta$$

Step 4: Simplify the integrand.

$$A=\frac{1}{2}{\int }_{-\pi /3}^{\pi /3}\left((16+32\cos \theta +16{\cos }^2\theta )-36\right)d\theta$$

$$A=\frac{1}{2}{\int }_{-\pi /3}^{\pi /3}\left(16{\cos }^2\theta +32\cos \theta -20\right)d\theta$$

Use the power-reducing identity ${\cos }^2\theta =\frac{1+\cos (2\theta )}{2}$:

$$A=\frac{1}{2}{\int }_{-\pi /3}^{\pi /3}\left(16\left(\frac{1+\cos (2\theta )}{2}\right)+32\cos \theta -20\right)d\theta$$

$$A=\frac{1}{2}{\int }_{-\pi /3}^{\pi /3}\left(8+8\cos (2\theta )+32\cos \theta -20\right)d\theta$$

$$A=\frac{1}{2}{\int }_{-\pi /3}^{\pi /3}\left(8\cos (2\theta )+32\cos \theta -12\right)d\theta$$

$$A={\int }_{-\pi /3}^{\pi /3}\left(4\cos (2\theta )+16\cos \theta -6\right)d\theta$$

Step 5: Evaluate the integral.

Find the antiderivative:

$$A={\left[2\sin (2\theta )+16\sin \theta -6\theta \right]}_{-\pi /3}^{\pi /3}$$

Evaluate at the upper and lower bounds:

$$A=\left(2\sin \left(\frac{2\pi }{3}\right)+16\sin \left(\frac{\pi }{3}\right)-6\left(\frac{\pi }{3}\right)\right)-\left(2\sin \left(-\frac{2\pi }{3}\right)+16\sin \left(-\frac{\pi }{3}\right)-6\left(-\frac{\pi }{3}\right)\right)$$

$$A=\left(2\left(\frac{\sqrt{3}}{2}\right)+16\left(\frac{\sqrt{3}}{2}\right)-2\pi \right)-\left(2\left(-\frac{\sqrt{3}}{2}\right)+16\left(-\frac{\sqrt{3}}{2}\right)+2\pi \right)$$

$$A=\left(\sqrt{3}+8\sqrt{3}-2\pi \right)-\left(-\sqrt{3}-8\sqrt{3}+2\pi \right)$$

$$A=(9\sqrt{3}-2\pi )-(-9\sqrt{3}+2\pi )$$

$$A=18\sqrt{3}-4\pi$$

Step-by-Step Example 2: Exam-Style Application

Problem: Let $R$ be the region in the first quadrant that is bounded by the polar curves $r=2\theta$ and $r=4\sin \theta$. Find the area of $R$.

Step 1: Find the points of intersection.

The region is in the first quadrant, so we are interested in $\theta \in [0,\pi /2]$. Set the equations equal:

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

This equation, $\theta =2\sin \theta$, cannot be solved algebraically. This is a strong indicator that this problem is intended for a calculator. However, we can observe by inspection that $\theta =0$ is one intersection point (at the origin). We must use a calculator to find the other non-zero intersection point. Graphing $y=2\theta$ and $y=4\sin \theta$ and finding their intersection gives $\theta \approx 1.89549$. Let's call this value $k$.

So, the bounds of integration are $\alpha =0$ and $\beta =k\approx 1.89549$.

Step 2: Identify the outer and inner curves.

We need to determine which function has a larger radius on the interval $(0,k)$. Let's test a value in this interval, for example, $\theta =\pi /4\approx 0.785$.

• $r=2\theta ⟹r=2(\pi /4)=\pi /2\approx 1.57$

• $r=4\sin \theta ⟹r=4\sin (\pi /4)=4(\sqrt{2}/2)=2\sqrt{2}\approx 2.828$

Since $2.828>1.57$, the curve $r=4\sin \theta$ is the outer curve and $r=2\theta$ is the inner curve on this interval.

• Outer curve: $f(\theta )=4\sin \theta$

• Inner curve: $g(\theta )=2\theta$

Step 3: Set up the definite integral.

Using the area formula:

$$A=\frac{1}{2}{\int }_0^k\left((4\sin \theta {)}^2-(2\theta {)}^2\right)d\theta$$

$$A=\frac{1}{2}{\int }_0^{1.89549}\left(16{\sin }^2\theta -4{\theta }^2\right)d\theta$$

Step 4: Evaluate the integral using a calculator.

This integral is best evaluated numerically.

$$A=\frac{1}{2}{\int }_0^{1.89549}(16{\sin }^2\theta -4{\theta }^2)d\theta \approx 3.835$$

The area of region $R$ is approximately 3.835.

(Note: If this were a non-calculator problem, the intersection point would be solvable by hand, e.g., an intersection between $r=2\cos \theta$ and $r=1$.)

Using Your Calculator

A graphing calculator is an indispensable tool for finding the area between two polar curves, especially when finding intersection points or evaluating the final integral is complex.

Typical Workflow (TI-84 Style):

1. Set Mode: Press [MODE] and change the graphing mode to POLAR.

2. Enter Equations: Press [Y=] and enter the two polar functions. For example, enter $r_1=f(\theta )$ into $r1$ and $r_2=g(\theta )$ into $r2$.

3. Graph and Visualize: Press [GRAPH]$.Adjustthe$[WINDOW]$settings(especially$\theta_{min}$,$\theta_{max}$,and$\theta_{step}) to get a clear view of the region. This step is crucial for identifying the outer and inner curves and estimating the intersection points. 4. **Find Intersection Points (Bounds):** * **Graphically:** Use the `[2nd]` `[TRACE]` (CALC) menu and select $5: intersect. The calculator will prompt you for the first curve, second curve, and a guess. Move the cursor near an intersection point and press [ENTER]` three times. Note the $\theta$ value. Repeat for other intersection points.

   • Analytically: On the home screen, use the numerical solver or graph the two functions in function mode (e.g., $Y_1=f(X)$ and $Y_2=g(X)$) to find where they intersect. Store the intersection value in a variable (e.g., STO-> A).

4. Calculate the Area:

   • On the home screen, use the numerical integration function (fnInt is under [MATH] $9:`).

   • The syntax is: (1/2) * fnInt( (r1(\theta))^2 - (r2(\theta))^2, \theta, α, β )

   • To access $r1$ and $r2$, press [VARS]`, go to the `Y-VARS` menu, select $3: Polar, and choose $r1$ or $r2$.

   • $\theta$ is the $[X,T,\theta ,n]$ key.

   • $\alpha$ and $\beta$ are the lower and upper bounds you found in Step 4.

Example Command: To solve the problem from Example 2 with $r_1=4\sin \theta$, $r_2=2\theta$, and the upper bound stored as $A$:

0.5 * fnInt( (r1(\theta))^2 - (r2(\theta))^2, \theta, 0, A )

AP Exam Quick Hit

Common Question Types

• Calculator-Active Area Calculation: You will be given two polar functions, $r=f(\theta )$ and $r=g(\theta )$, and asked to find the area of the region inside one curve and outside the other. The primary tasks are to find the intersection points using your calculator, correctly identify the outer and inner curves, and use your calculator's numerical integrator to find the value.

  • Example: "Find the area of the region inside $r=3$ and outside $r=2-2\cos \theta$."

• Integral Setup Only (Non-Calculator): You will be shown a graph of two polar curves or given their equations and asked to write, but not evaluate, an integral expression for the area of a specified region. This tests your ability to find the bounds (which will be common angles like $\pi /3,\pi /4$, etc.) and correctly construct the integrand.

  • Example: "Write an integral expression for the area of the region common to the curves $r=3\cos \theta$ and $r=\sqrt{3}\sin \theta$."

• Area of a Shared Region: You may be asked to find the area of a region common to two curves. This often requires splitting the total area into two or more integrals, as the "outer" curve may change.

  • Example: "Find the area of the region lying inside both the circle $r=2$ and the cardioid $r=2+2\cos \theta$." (This would require two integrals: one for the cardioid part and one for the circle part, each over different intervals of $\theta$).

Common Mistakes

• Forgetting the $\frac{1}{2}$ Multiplier: The most frequent and simple error is omitting the $\frac{1}{2}$ coefficient in front of the integral.

• Forgetting to Square the Radii: Students may integrate $(f(\theta )-g(\theta ))$ instead of the correct $(f(\theta ){)}^2-(g(\theta ){)}^2$. The area formula is based on $r^2$, not $r$.

• Incorrectly Squaring the Difference: A major algebraic error is writing $(f(\theta )-g(\theta ){)}^2$ instead of $(f(\theta ){)}^2-(g(\theta ){)}^2$. These are not equivalent. You must square each radius function individually before subtracting.

• Incorrect Bounds of Integration: Students often fail to find the correct intersection points or use a default interval like $[0,2\pi ]$ when a smaller, specific interval is required to trace the region only once. Always sketch the curves to understand how the region is formed.

• Reversing Outer and Inner Curves: Mistaking the inner curve for the outer curve (i.e., calculating $\int (g^2-f^2)$ instead of $\int (f^2-g^2)$). This will result in the negative of the correct area. A quick check of the radii at a midpoint angle can prevent this.