Problem 18 - Calculus Early Transcendentals

Question

\(17-21\) Find the area of the region enclosed by one loop of the curve. $$ r=4 \sin 3 \theta $$

Step-by-Step Solution

Answer

The area of one loop is \(\frac{4\pi}{3}\).

Step 1: Understanding the Curve

The given polar equation is \(r = 4 \sin 3\theta\). This equation represents a limaçon with three loops. We are interested in the area of one loop.

Step 2: Identify the Range for One Loop

Each loop of the curve corresponds to a complete cycle of the sine function from 0 to \(\pi\). Therefore, the range of \(\theta\) for one loop is from \(\theta = 0\) to \(\theta = \frac{\pi}{3}\), since \(3\cdot0 = 0\) and \(3\cdot\frac{\pi}{3} = \pi\).

Step 3: Formula for the Area in Polar Coordinates

For a curve described by \(r = f(\theta)\), the area \(A\) of the region enclosed by the curve from \(\theta = a\) to \(\theta = b\) is given by \[ A = \frac{1}{2} \int_a^b [f(\theta)]^2 \, d\theta \]

Step 4: Substitute and Integrate

Substitute \(f(\theta) = 4 \sin 3\theta\) and the limits from 0 to \(\frac{\pi}{3}\) into the area formula:\[ A = \frac{1}{2} \int_0^{\frac{\pi}{3}} (4 \sin 3\theta)^2 \, d\theta \]Simplifying, \[ A = 8 \int_0^{\frac{\pi}{3}} \sin^2 3\theta \, d\theta \]Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to simplify further:\[ A = 8 \int_0^{\frac{\pi}{3}} \frac{1 - \cos 6\theta}{2} \, d\theta \]\[ = 4 \int_0^{\frac{\pi}{3}} (1 - \cos 6\theta) \, d\theta \]

Step 5: Calculate the Integral

Calculate the integral:\[ A = 4 \left[ \theta - \frac{\sin 6\theta}{6} \right]_0^{\frac{\pi}{3}} \]Evaluate at the bounds:\[ A = 4 \left( \frac{\pi}{3} - \frac{\sin 2\pi}{6} \right) - 4 \left( 0 - \frac{\sin 0}{6} \right) \]Since \(\sin 2\pi = 0\) and \(\sin 0 = 0\), the expression simplifies to:\[ A = 4 \times \frac{\pi}{3} = \frac{4\pi}{3} \]

Step 6: Conclusion

The area of one loop of the cardioid-like, three-looped limaçon described by \(r = 4 \sin 3\theta\) is \(\frac{4\pi}{3}\).

Key Concepts

limaçon

A limaçon is a fascinating curve that finds its basis in polar coordinates. The word "limaçon" comes from the French word for snail because of the curve's shape, which can sometimes resemble a snail's shell. Limaçons are special types of polar graphs defined by equations of the form:

• \( r = a + b ext{cos} heta \)
• \( r = a + b ext{sin} heta \)

In these equations, \( a \) and \( b \) are constants that determine the shape and size of the limaçon.
A key feature of limaçons is the loop, which can vary depending on the values of \( a \) and \( b \). Some limaçons have an inner loop, while others may not, and if \( a = b \), they form a cardioid.
In our exercise, the equation is \( r = 4 \sin 3\theta \), having three loops, typical of certain limaçons where the number of loops can increase with the frequency inside the trigonometric function as it is influenced by the multiplier of \( \theta \). Understanding the

• core equation structure,
• relationship between \( a \) and \( b \),
• and how these parameters influence the appearance and shape

of the limaçon is essential for mastering problems in polar coordinates.

area calculation

Calculating the area of a region in polar coordinates involves integrating, which may seem challenging at first. However, it's quite straightforward once you break it down. For any polar curve described by \( r = f(\theta) \), the formula for the area \( A \) it encloses between two angular limits \( \theta = a \) and \( \theta = b \) is:\[ A = \frac{1}{2} \int_a^b [f(\theta)]^2 \, d\theta \]Using this formula, we can find the area enclosed by a limaçon or any polar graph.
In the given problem, the polar equation is \( r = 4 \sin 3\theta \), and we focused on one loop from \( \theta = 0 \) to \( \theta = \frac{\pi}{3} \). This involved plugging the specific \( f(\theta) = 4 \sin 3\theta \) into the formula and calculating:\[ A = 8 \int_0^{\frac{\pi}{3}} \sin^2 3\theta \, d\theta \]The integral simplifies using specific trigonometric identities and the integration of trigonometric functions. After evaluating it, the final result for the area is:\[ A = \frac{4\pi}{3} \]It's important to:

• Understand the polar area formula,
• Correctly determine the limits for integration,
• and Apply trigonometric identities to simplify the integrals.

These steps ensure the correct calculation of areas enclosed by polar curves.

trigonometric identities

Trigonometric identities are tools that simplify and transform trigonometric expressions, which are essential counterparts in problems involving polar coordinates.Working on polar curves like the limaçon requires familiarity with sine and cosine functions and how they can be modified or simplified.A particularly useful identity in this exercise is the power-reducing identity:\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]This identity was key in transforming \( \sin^2 3\theta \) into a manageable form for integration:\[ \int_0^{\frac{\pi}{3}} \sin^2 3\theta \, d\theta \rightarrow \int_0^{\frac{\pi}{3}} \frac{1 - \cos 6\theta}{2} \, d\theta \]Key trigonometric identities like the above are:

• Relevant for simplifying quadratic trigonometric expressions,
• Vital for both integration and differentiation tasks in polar coordinates,
• and Allow us to perform calculations otherwise complex without them.

Understanding and using these identities can streamline problem-solving strategies in diverse applications, making them indispensable for students grappling with polar curves and related topics.