Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation and find the area of the region bounded by it. $$ r^{2}=9 \sin 2 \theta $$

Step-by-Step Solution

Answer

The bounded region has an area of 9 square units.

Step 1: Understand the Equation Form

The given polar equation is \( r^2 = 9 \sin 2\theta \). This is a polar equation where \( r \) is defined in terms of \( \theta \). It resembles the form of a limaçon but with a sine double angle function, indicating it may represent a rose curve or a lemniscate.

Step 2: Sketch the Graph

To sketch the graph, consider key values of \( \theta \):1. When \( \theta = 0 \), \( r^2 = 9 \sin 0 = 0 \), so \( r = 0 \).2. When \( \theta = \frac{\pi}{4} \), \( 9 \sin \frac{\pi}{2} = 9 \), so \( r^2 = 9 \). Hence, \( r = 3 \).3. When \( \theta = \frac{\pi}{2} \), \( r^2 = 9 \sin \pi = 0 \), so \( r = 0 \).4. This pattern continues every \( \frac{\pi}{2} \) radians indicating a symmetrical figure about the origin.The symmetry and alternating radii indicate a lemniscate shape, symmetric about the origin.

Step 3: Calculate Area of Bounded Region

The area \( A \) of a region enclosed in polar coordinates by \( r^2 = 9 \sin 2\theta \) is given by the integral:\[ A = \frac{1}{2} \int_{0}^{\pi/2} (9 \sin 2\theta) d\theta \] First, compute \( \sin 2\theta \) integral. Let \( u = 2\theta \), then \( du = 2 d\theta \), \( d\theta = \frac{1}{2} du \).\[ A = \frac{1}{2} \int_{0}^{\pi} 9 \sin u \frac{1}{2} du = \frac{9}{4} \left( -\cos u \right) \bigg|_0^{\pi} = \frac{9}{4} (1 - (-1)) = \frac{9}{2} \]

Step 4: Determine Total Area

Since the area was calculated from \(0\) to \(\frac{\pi}{2}\) and similar sectors are formed in other quadrants due to symmetry, the total area of the lemniscate is \( 2 \times \frac{9}{2} = 9 \).

Key Concepts

Lemniscate

Lemniscate is a fascinating shape that appears elegantly in polar coordinates. It is defined by equations like our equation in the exercise: \( r^2 = a^2 \sin(2\theta) \). Lemniscates resemble the infinity symbol (∞) or a figure-eight. They are symmetrical across both the polar axis and the perpendicular line through the origin.
Key characteristics include:

• It is not a circle but has two lobes extending from the origin.
• At specific angles, the radius, \( r \), takes the value zero, causing the graph to pass through the origin.
• Lemniscates can be sketched by understanding their periodic nature and by calculating \( r \) for various significant \( \theta \) values.

The exercise displays a lemniscate due to its symmetry in equation form and the alternate radii, which give it its unique shape.

Area Calculation Using Integrals

Calculating the area for enclosed shapes like the lemniscate involves integrals, which account for the changing radius over an angle. When given the equation \( r^2 = 9\sin(2\theta) \), we can find the area \( A \) by using the integral:
\[ A = \frac{1}{2} \int_0^{\pi/2} (9 \sin(2\theta)) \, d\theta \]
The steps to solve this are:

• Substitution: Let \( u = 2\theta \), giving \( du = 2 \, d\theta \), and adjust bounds accordingly.
• Integration: Solve \[ \int \sin(u) \, du = -\cos(u) \]
• Convert back with original bounds and compute the value: \( \frac{9}{4} \) of the integral.

This method provides an exact area for part of the lemniscate, which can be extended due to symmetry to find the total area.

Graph Sketching

Sketching the graph of a polar equation, like a lemniscate, requires calculating a few points at different values of \( \theta \). Here’s how to approach it:

• Start with key values: Calculate \( r \) at \( \theta = 0, \frac{\pi}{4}, \frac{\pi}{2} \) etc.
• Plot Accuracy: At these \( \theta \), note \( r \)’s value to see how far the graph reaches.

As \( \sin(2\theta) \) varies, these radii repeat periodically, contributing to the symmetrical nature of the lemniscate:
When plotted correctly, the symmetry about the origin is evident, revealing the distinct looped structure of the lemniscate.

Symmetry in Polar Graphs

Symmetry helps simplify the process of understanding and sketching polar graphs. The equation \( r^2 = 9\sin(2\theta) \) has specific symmetries that are intrinsic to its form.
How to identify symmetry:

• Direct Symmetry: The graph repeats its pattern every \( \pi \), showing symmetry in successive quadrants.
• Origin Symmetry: Points that exist on one side of the graph have counterparts directly opposite related by symmetry.

For such polar graphs, symmetry allows one to calculate areas over a simpler section and multiply it accordingly to find the total area, saving computations and easing the sketching process. By recognizing these symmetries, you gain insight into the elegant structure inherent to polar equations.