Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve - AP Calculus BC Study Guide

The Core Idea: Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

In Cartesian coordinates, we find the area under a curve by summing the areas of an infinite number of infinitesimally thin rectangles. In the polar coordinate system, we adapt this concept to find the area of a region defined by a polar function, $r(\theta )$. Instead of rectangles, the fundamental shape is a sector of a circle. The core idea is to find the area "swept out" by the radius vector $r$ as the angle $\theta$ changes from a starting angle $\alpha$ to an ending angle $\beta$.

The process involves summing the areas of infinitely many infinitesimally small circular sectors. The area of a single sector of a circle is given by $\frac{1}{2}{r}^2\theta$. By treating $d\theta$ as an infinitesimally small change in angle, we can express the area of a tiny sector as $\frac{1}{2}(r(\theta ){)}^{2}d\theta$. Integrating this expression between the angular bounds $\alpha$ and $\beta$ gives the total area of the polar region. This method allows us to calculate areas of complex shapes like cardioids, limaçons, and rose curves that would be difficult or impossible to describe with a single Cartesian function.

Key Formulas

The area of a region in the plane bounded by a polar curve $r=r(\theta )$ and the lines $\theta =\alpha$ and $\theta =\beta$ is given by a single, fundamental formula.

Area of a Polar Region:

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

• $A$ is the area of the region.

• $r(\theta )$ is the polar function defining the boundary of the region.

• $\alpha$ and $\beta$ are the starting and ending angles, respectively, that trace the boundary of the desired region. The integration must occur over an interval $[\alpha ,\beta ]$ that traces the region exactly once.

Understanding the Bounds of Integration

The most critical part of finding the area of a polar region is determining the correct bounds of integration, $\alpha$ and $\beta$. These angles define the "start" and "stop" points for sweeping out the area.

• Given Bounds: In some problems, the bounds are explicitly stated. For example, "Find the area of the region bounded by $r=\theta$ in the first quadrant." Here, the bounds are implicitly $\alpha =0$ and $\beta =\frac{\pi }{2}$.

• Finding Bounds for a "Loop" or "Petal": For curves that form loops or petals (like limaçons or rose curves), you often need to find the area of a single loop. A loop typically begins and ends where the curve passes through the pole (the origin). To find these angles, you must set the function $r(\theta )$ equal to zero and solve for $\theta$. Two consecutive values of $\theta$ for which $r(\theta )=0$ often form the bounds for a single loop.

• Finding Bounds for the Entire Curve: To find the total area enclosed by a curve like a cardioid, you need to determine the interval of $\theta$ values that traces the entire curve exactly once. For many common curves, this interval is $[0,2\pi ]$, but for others, like rose curves of the form $r=a\cos (n\theta )$ where $n$ is even, the full curve is traced on $[0,\pi ]$. Graphing the curve (often with a calculator) is an essential tool for visualizing how the curve is traced and confirming the correct bounds.

Core Concepts & Rules

• The area of a region enclosed by a polar curve $r=r(\theta )$ from angle $\theta =\alpha$ to $\theta =\beta$ is calculated with the definite integral $A=\frac{1}{2}{\int }_{\alpha }^{\beta }(r(\theta ){)}^{2}d\theta$.

• This formula is derived from summing the areas of an infinite number of small circular sectors, each with area $dA=\frac{1}{2}{r}^{2}d\theta$.

• Determining the correct bounds of integration, $\alpha$ and $\beta$, is a crucial first step. These bounds must trace the desired region exactly once to avoid calculating double the area or missing a portion of the region.

• The area of more complex regions can be found by strategically adding or subtracting the areas of simpler regions defined by polar curves. (This concept is more fully explored in Topic 9.9).

Step-by-Step Example 1: Area of a Rose Curve Petal

Problem: Find the area of one petal of the rose curve given by $r=3\cos (2\theta )$.

Step 1: Find the Bounds of Integration

A petal starts and ends when the curve is at the pole, which is where $r=0$. We set the equation to zero and solve for $\theta$.

$$3\cos (2\theta )=0$$

$$\cos (2\theta )=0$$

The principal values for which cosine is zero are when its argument is $\frac{\pi }{2},\frac{3\pi }{2},\dots$ or $-\frac{\pi }{2},-\frac{3\pi }{2},\dots$. Let's find two consecutive solutions.

$$2\theta =-\frac{\pi }{2}\text{and}2\theta =\frac{\pi }{2}$$

$$\theta =-\frac{\pi }{4}\text{and}\theta =\frac{\pi }{4}$$

These two angles form the bounds for the petal centered on the positive x-axis. So, $\alpha =-\frac{\pi }{4}$ and $\beta =\frac{\pi }{4}$.

Step 2: Set Up the Area Integral

Use the polar area formula with the function $r(\theta )=3\cos (2\theta )$ and the bounds we just found.

$$A=\frac{1}{2}{\int }_{-\pi /4}^{\pi /4}[3\cos (2\theta ){]}^{2}d\theta$$

$$A=\frac{1}{2}{\int }_{-\pi /4}^{\pi /4}9{\cos }^2(2\theta )d\theta$$

Step 3: Evaluate the Integral

To integrate ${\cos }^2(2\theta )$, we use the power-reducing identity: ${\cos }^2(x)=\frac{1+\cos (2x)}{2}$. In our case, $x=2\theta$.

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

Substitute this back into the integral:

$$A=\frac{9}{2}{\int }_{-\pi /4}^{\pi /4}\left(\frac{1+\cos (4\theta )}{2}\right)d\theta$$

$$A=\frac{9}{4}{\int }_{-\pi /4}^{\pi /4}(1+\cos (4\theta ))d\theta$$

Now, find the antiderivative:

$$A=\frac{9}{4}{\left[\theta +\frac{1}{4}\sin (4\theta )\right]}_{-\pi /4}^{\pi /4}$$

Evaluate at the bounds:

$$A=\frac{9}{4}\left[\left(\frac{\pi }{4}+\frac{1}{4}\sin (\pi )\right)-\left(-\frac{\pi }{4}+\frac{1}{4}\sin (-\pi )\right)\right]$$

Since $\sin (\pi )=0$ and $\sin (-\pi )=0$:

$$A=\frac{9}{4}\left[\left(\frac{\pi }{4}+0\right)-\left(-\frac{\pi }{4}+0\right)\right]$$

$$A=\frac{9}{4}\left[\frac{\pi }{4}+\frac{\pi }{4}\right]=\frac{9}{4}\left[\frac{2\pi }{4}\right]=\frac{9}{4}\left[\frac{\pi }{2}\right]$$

Step 4: Final Answer

$$A=\frac{9\pi }{8}$$

Step-by-Step Example 2: Area of a Cardioid

Problem: Find the total area enclosed by the cardioid $r=2-2\cos (\theta )$.

Step 1: Find the Bounds of Integration

We need to find the interval of $\theta$ that traces the entire cardioid exactly once. Let's test some points:

• At $\theta =0$, $r=2-2(1)=0$. The curve is at the pole.

• At $\theta =\pi /2$, $r=2-2(0)=2$.

• At $\theta =\pi$, $r=2-2(-1)=4$. This is the maximum distance from the pole.

• At $\theta =3\pi /2$, $r=2-2(0)=2$.

• At $\theta =2\pi$, $r=2-2(1)=0$. The curve returns to the pole.

The full curve is traced once as $\theta$ goes from $0$ to $2\pi$. Therefore, $\alpha =0$ and $\beta =2\pi$.

Step 2: Set Up the Area Integral

Use the polar area formula:

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

Step 3: Evaluate the Integral

First, expand the integrand:

$$[2-2\cos (\theta ){]}^2=4-8\cos (\theta )+4{\cos }^2(\theta )$$

The integral becomes:

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

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

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

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

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

$$A={\int }_0^{2\pi }(3-4\cos (\theta )+\cos (2\theta ))d\theta$$

Now, find the antiderivative:

$$A={\left[3\theta -4\sin (\theta )+\frac{1}{2}\sin (2\theta )\right]}_0^{2\pi }$$

Evaluate at the bounds:

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

Since all sine terms are zero at both bounds:

$$A=(6\pi -0+0)-(0-0+0)$$

Step 4: Final Answer

$$A=6\pi$$

Using Your Calculator

A graphing calculator is an invaluable tool for polar area problems, both for visualizing the curve to find the correct bounds and for evaluating the final definite integral on calculator-active sections of the AP Exam.

1. Visualizing and Finding Bounds:

• Set your calculator to Polar mode.

• Enter the function $r(\theta )$ into one of the $r=$ editors.

• Set an appropriate viewing window. A good starting point for $\theta$ is $\text{thetamin}=0$, $\text{thetamax}=2\pi$, and $\text{thetastep}$ around $0.1$. Adjust $Xmin$, $Xmax$, $Ymin$, and $Ymax$ to see the full graph.

• Use the TRACE feature. As you trace, the calculator shows the values of $\theta$, $x$, and $y$. This helps you see exactly which values of $\theta$ correspond to the start and end of a loop or the full curve.

2. Evaluating the Integral:

• Once you have determined your function $r(\theta )$ and your bounds $\alpha$ and $\beta$, you can use the numerical integration feature. It is often easiest to do this in Function mode.

• Use the numerical integrator (e.g., fnInt on a TI-84, found under the MATH menu).

• The syntax will be: $(1/2)\ast fnInt((r(X){)}^2,X,\alpha ,\beta )$

• Example: To calculate the area from Example 1, $A=\frac{1}{2}{\int }_{-\pi /4}^{\pi /4}(3\cos (2\theta ){)}^{2}d\theta$, you would enter:

  0.5 * fnInt((3*cos(2X))^2, X, -π/4, π/4)

• The calculator will return a decimal approximation of the area (e.g., $\mathrm{3.534...}$ which is approximately $\frac{9\pi }{8}$).

AP Exam Quick Hit

Common Question Types

• Area of a Single Petal/Loop: You will be given a function like $r=5\sin (3\theta )$ and asked to find the area of one petal. The key challenge is finding the correct bounds by setting $r=0$.

• Total Area of a Simple Curve: You will be given a function like a cardioid $r=3+3\sin (\theta )$ or a circle $r=4\cos (\theta )$ and asked for the total area enclosed. The challenge is determining the $\theta$ interval that traces the curve exactly once.

• Area over a Specified Interval: You will be given a function and an explicit interval, such as "Find the area of the region bounded by $r=\sqrt{\theta }$ for $0\le \theta \le \pi$." This is the most straightforward type, as the bounds are given directly.

Common Mistakes

• Forgetting the $\frac{1}{2}$: The most frequent error is omitting the $\frac{1}{2}$ coefficient from the area formula $A=\frac{1}{2}{\int }_{\alpha }^{\beta }[r(\theta ){]}^{2}d\theta$.

• Forgetting to Square $r(\theta )$: A close second is failing to square the function, instead integrating $\frac{1}{2}\int r(\theta )d\theta$. Remember, the formula is based on the area of a sector, $\frac{1}{2}{r}^2\theta$.

• Incorrect Bounds of Integration: This is the most significant conceptual error. Students may use $[0,2\pi ]$ for a rose curve like $r=\cos (2\theta )$, which traces the curve twice and results in double the actual area. Always verify your bounds by finding where $r=0$ or by tracing the graph.

• Power-Reduction Errors: When integrating by hand, many mistakes occur during the application of the power-reducing identities for ${\sin }^2(\theta )$ and ${\cos }^2(\theta )$. Be careful with coefficients and the argument of the resulting cosine term.

• Confusing with Volume Formulas: Students sometimes mix up the polar area formula with the disk/washer method for volumes of revolution, attempting to use an integral like $\pi \int (r(\theta ){)}^{2}d\theta$. These are fundamentally different concepts.