Question

Find the area of the described region. Enclosed by one petal of \(r=\sin (3 \theta)\).

Short answer

The area of one petal is \( \frac{\pi}{12} \).

Step-by-step solution

Understand the Polar Equation

The equation given is \( r = \sin(3\theta) \). This is a polar equation representing a rose curve with three petals. However, due to symmetry, each complete petal can be represented by cycling the angle \( \theta \) from 0 to \( \frac{\pi}{3} \).

Setup the Integral for Area

The area \( A \) of one petal in polar coordinates is given by the integral: \[ A = \frac{1}{2} \int_{a}^{b} (r(\theta))^2 \, d\theta \]For our specific petal, substitute \( r(\theta) = \sin(3\theta) \) and the integration limits are \( a = 0 \) to \( b = \frac{\pi}{3} \) because one petal only spans this range of \( \theta \).

Evaluate the Integral

Substitute \( r = \sin(3\theta) \) into the formula:\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \sin^2(3\theta) \, d\theta \]Use the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) to transform the integral:\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \frac{1 - \cos(6\theta)}{2} \, d\theta \]This simplifies to:\[ A = \frac{1}{4} \int_{0}^{\frac{\pi}{3}} (1 - \cos(6\theta)) \, d\theta \]

Integrate and Calculate the Area

Integrate the simplified expression:\[ = \frac{1}{4} \left[ \theta - \frac{1}{6}\sin(6\theta) \right]_{0}^{\frac{\pi}{3}} \]Calculate the definite integral:\[ = \frac{1}{4} \left[ \frac{\pi}{3} - \frac{1}{6} \sin(2\pi) - (0 - 0) \right] \]Since \( \sin(2\pi) = 0 \), this simplifies to:\[ = \frac{1}{4} \cdot \frac{\pi}{3} = \frac{\pi}{12} \]

Conclusion

The area of one petal of the polar plot \( r = \sin(3\theta) \) is \( \frac{\pi}{12} \).

Key concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with the concept of integrals. An integral can be thought of as the reverse operation of a derivative. In simpler terms, while derivatives tell you how a function changes, integrals give you the value of a function. They help in calculating areas and volumes, among many other applications. For example, to find the area under a curve or a polar plot, we use integrals. Here, the problem is to find the area enclosed by one petal of a polar curve. To solve area problems involving polar curves, you need to set up the integral which represents the area you want to find. This involves using formulas specifically derived for polar coordinates, which integrate a radius function— like our example, the sine function in polar coordinates. We solve the integral to obtain the exact area of the region we're interested in.

Area of a Polar Curve

Finding the area of a polar curve is a common exercise in calculus. When dealing with polar coordinates, the method to find an area differs from that of Cartesian coordinates. The formula for finding the area enclosed by a polar curve is \[ A = \frac{1}{2} \int_{a}^{b} (r(\theta))^2 \, d\theta \]Here, \( r(\theta) \) is the radius as a function of the angle \( \theta \). The limits of integration \( a \) and \( b \) are chosen based on the range you wish to examine. In our specific example with \( r = \sin(3\theta) \), one petal is symmetric and spans from \( \theta = 0 \) to \( \theta = \frac{\pi}{3} \). This procedure turns the polar plot into a calculus problem that becomes solvable using integration techniques.

Trigonometric Identities

Trigonometric identities are key tools in calculus to simplify integrals involving trigonometric functions. They help transform complex trigonometric expressions into simpler forms that are easier to integrate.For instance, the identity \[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]was used to transform the integral in our problem from \[ \int \sin^2(3\theta) \, d\theta \]to \[ \int \frac{1 - \cos(6\theta)}{2} \, d\theta \]This simplification allows for straightforward integration. Without such identities, solving the integral would be more complicated and might require more advanced techniques, such as trigonometric substitutions.

Definite Integrals

Definite integrals correspond to the evaluation of an integral over a certain interval. Unlike indefinite integrals, which provide a general formula for a function's antiderivative, definite integrals yield a numerical value that represents the accumulated quantity between two points—in our case, accumulated area.In our example, after simplifying the integrand, we evaluate the definite integral:\[ \int_{0}^{\frac{\pi}{3}} (1 - \cos(6\theta)) \, d\theta \]Through this calculation, we obtain a final area value of \( \frac{\pi}{12} \) for one petal of \( r = \sin(3\theta) \). To compute a definite integral:

• First, find the antiderivative.
• Then, evaluate this antiderivative at the upper and lower limits of integration.
• Subtract these results to find the total value.