Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the given region \(S\) by calculating \(\iint_{S} r d r d \theta .\) Be sure to make a sketch of the region first. \(S\) is the region outside the circle \(r=2\) and inside the lemniscate \(r^{2}=9 \cos 2 \theta\).

Step-by-Step Solution

Answer

The area of region S is \( \frac{9}{2} - \pi \)."

Step 1: Understand the region S

The region S is defined as being outside the circle given by \( r = 2 \) and inside the lemniscate given by \( r^2 = 9 \cos(2\theta) \). The circle has a radius of 2, and the lemniscate is a polar curve that is shaped like a figure eight.

Step 2: Identify the bounds

For the circle \( r = 2 \), the area that lies outside implies \( r > 2 \). For the lemniscate \( r^2 = 9 \cos(2\theta) \), solve for \( r \) to obtain \( r = \sqrt{9\cos(2\theta)} = 3\sqrt{\cos(2\theta)} \), meaning \( r \leq 3\sqrt{\cos(2\theta)} \). Therefore, the bounds are \( 2 < r < 3\sqrt{\cos(2\theta)} \).

Step 3: Set the angular bounds

To find the angular bounds, check when \( r^2 = 9\cos(2\theta) \) is defined. The lemniscate is plotted between \( -\frac{\pi}{4} \) and \( \frac{\pi}{4} \). Therefore, the angular bounds for theta will be \( -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \).

Step 4: Formulate the double integral

The area \( A \) of the region can be expressed using a double integral in polar coordinates: \[ A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{2}^{3\sqrt{\cos(2\theta)}} r \, dr \, d\theta \]

Step 5: Compute the inner integral

The integral with respect to \( r \) is \[ \int_{2}^{3\sqrt{\cos(2\theta)}} r \, dr = \left[ \frac{r^2}{2} \right]_{2}^{3\sqrt{\cos(2\theta)}} = \frac{(3\sqrt{\cos(2\theta)})^2}{2} - \frac{2^2}{2} \] This simplifies to \[ \frac{9\cos(2\theta)}{2} - 2 \]

Step 6: Compute the outer integral

Now, integrate with respect to \( \theta \):\[ A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{9\cos(2\theta)}{2} - 2 \right) d\theta \] Separate the integrals: \[ = \frac{9}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2\theta) \, d\theta - 2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \theta \, d\theta \]

Step 7: Integrate trigonometric function

For the integral \( \int \cos(2\theta) \; d\theta \), use the identity: \[ \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C \] Therefore, \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2\theta) \, d\theta = \frac{1}{2} \left[ \sin(2\theta) \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{2} (1 - (-1)) = 1 \]

Step 8: Integrate constant function

The definite integral of a constant:\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, d\theta = \left[ \theta \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2} \]

Step 9: Final calculations

Substitute both results into the area expression besides their integral factor: \[ A = \frac{9}{2} \times 1 - 2 \times \frac{\pi}{2} = \frac{9}{2} - \pi \] Therefore, the area of region S is \( \frac{9}{2} - \pi \).

Key Concepts

Double Integral

A double integral is a way to integrate over a two-dimensional area. It's the extension of a single integral, which deals with functions of one variable, to functions of two variables. In this exercise, we use a double integral to calculate the area of a specific region in polar coordinates. Instead of the usual Cartesian coordinates, polar coordinates are used, which specify locations with a radius and an angle.
The double integral is represented in the form:

• \( \iint_{D} f(x,y) \, dx \, dy \) for Cartesian coordinates.
• \( \iint_{S} f(r,\theta) \, r \, dr \, d\theta \) for polar coordinates.

In our problem, the region \( S \) is the difference between two curves described in polar coordinates. This means our integral involves both the radius \( r \) and the angle \( \theta \), considering the area outside one shape and inside another.” Calculating the double integral lets us find the total area for this specific polar region.

Lemniscate

A lemniscate is a beautiful and unique shape often likened to a figure-eight or the symbol for infinity. In polar coordinates, a lemniscate can be described mathematically in various ways. For this exercise, the lemniscate is given by the equation \( r^2 = 9 \cos(2\theta) \).

• Its appearance and size are affected by the coefficient in front of the trigonometric function.
• It creates a symmetric curve about the origin that loops back on itself.

Understanding the lemniscate in this exercise is crucial as it defines one of the boundaries for calculating the area. The integral's bounds are set using this shape to construct the area correctly, considering the region's section inside the lemniscate.

Area Calculation

Calculating an area using polar coordinates involves integrating over a region defined by bounds on \( r \), the radius, and on \( \theta \), the angle. The area in polar coordinates is not a simple \( dx \times dy \) since it captures circular segments; hence each infinitesimal area element is \( r \, dr \, d\theta \).

• The outer integral defines the range for \( \theta \), the angular boundary.
• The inner integral describes the range for \( r \), the radial boundary.

To find the area inside the lemniscate but outside the circle, calculate the double integral\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{2}^{3\sqrt{\cos(2\theta)}} r \, dr \, d\theta \] This formula ensures all required bounds are respected. The computation involves separating the double integral into parts that can be solved independently, focusing on the inner and then the outer integrals. This step-by-step approach allows determining the net area accurately after eliminating unwanted sections from other shapes, resulting in a final expression that finely describes the area in question.