Problem 16 - Thomas Calculus

Question

Find the areas of the regions Inside the lemniscate \(r^{2}=6 \cos 2 \theta\) to the right of the line \(r=(3 / 2) \sec \theta\)

Step-by-Step Solution

Answer

The area is 3 square units.

Step 1: Understand the Lemniscate

The given lemniscate equation is in polar form: \( r^2 = 6 \cos 2\theta \). This is a symmetrical curve about the origin and has a shape similar to a figure-eight. More precisely, the curve lies within the ranges where \( \cos 2\theta \geq 0 \).

Step 2: Understand the Line

The line given is \( r = \frac{3}{2} \sec \theta \) in polar form, which translates to a vertical line in Cartesian coordinates. To simplify, this is equivalent to \( x = \frac{3}{2} \), where the line splits the lemniscate.

Step 3: Find Intersection Points

Find where the lemniscate intersects the line \( x = \frac{3}{2} \). Substitute \( r = \frac{3}{2} \sec \theta \) into the lemniscate equation: \( \left(\frac{3}{2} \sec \theta\right)^2 = 6 \cos 2\theta \). Solving gives \( 9/4 \cdot 1/\cos^2 \theta = 6 (2\cos^2\theta - 1) \), which simplifies to find specific \( \theta \) values.

Step 4: Set Limits of Integration

Using the symmetry and intersections from Step 3, find \( \theta \) intervals where the lemniscate is to the right of the line. As we found intersections correspond to \( \theta = \pi/4, -\pi/4 \) in prior solutions, the area is evaluated for \(-\pi/4 \leq \theta \leq \pi/4 \).

Step 5: Setup Area Integral

The area \( A \) inside the lemniscate to the right of the line can be found using the polar area formula: \( A = \frac{1}{2} \int_{-\pi/4}^{\pi/4} (6 \cos 2\theta) d\theta \).

Step 6: Solve the Integral

Carry out the integration: \( A = \frac{1}{2} \int_{-\pi/4}^{\pi/4} 6 \cos 2\theta \ d\theta \). Use the substitution \( u = 2\theta, \ du = 2d\theta \). This integral evaluates to yield \( A = 3 \left[ \frac{1}{2} \sin 2\theta \right]_{-\pi/4}^{\pi/4} \), giving the area as \( 3 \).

Step 7: Calculate Final Area

The calculated area from Step 6 gives the area of the region inside the lemniscate and to the right of the line as \( 3 \) square units.

Key Concepts

Lemniscate

A lemniscate is a unique and intriguing geometric shape that resembles a figure-eight or an infinity symbol. In polar coordinates, a lemniscate is often defined by equations like:

• \( r^2 = a^2 \cos 2\theta \)

This symmetrical curve is centered around the origin (the pole) and has properties like those observed in the given problem. Here, the equation \( r^{2}=6\cos 2\theta \) represents our lemniscate.
The symmetry about the origin means that the lemniscate will generally look symmetric while crossing both the positive and negative axes.
It is essential to understand these properties to accurately determine areas within the lemniscate in relation to other geometrical structures, such as lines, that may intersect it.

Area of a Region

Calculating the area within certain curves can become quite complex, especially with figures like the lemniscate in polar coordinates. The specific problem requires finding the area inside the lemniscate \( r^2 = 6 \cos 2\theta \) and to the right of the vertical line at \( x = \frac{3}{2} \).
To solve this, we must first substitute and find where the lemniscate intersects the line. This is typically done by solving the polar equations to find where the curves cross each other. Then, using the angles determined by these intersection points, we establish the limits for integration.
Understanding how to set these limits is crucial because they define the specific region's boundaries we are interested in. The intersection points will often guide us because they tell us the range of \( \theta \) that confines the area of interest.

Integration in Polar Coordinates

Integration in polar coordinates is a technique used to calculate areas that shapes encompass or regions within curves that are best represented in polar form. The general formula for area \( A \) in polar coordinates between two angles \( \theta_1 \) and \( \theta_2 \) is:

• \[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \]

For a polar equation like the lemniscate, integration considers the radial value \( r \) as determined by the equation (in this case \( r^2 = 6 \cos 2\theta \)).
Substituting these values along with the appropriate limits of \( \theta \) from intersections, you can solve the integral to find the enclosed area. Simplifications, such as trigonometric identities or substitutions like \( u = 2\theta \), can help evaluate the integral easily.
This process ultimately helps in achieving the desired area within the specified limits efficiently.