Question

Determine a definite integral that represents the area.Region enclosed by one petal of \(r=8 \sin (2 \theta)\)

Short answer

The area of one petal is \(8\pi\).

Step-by-step solution

Identify the Curve

The curve given by the polar equation is a "four-petaled" rose, represented by the equation \( r = 8 \sin(2\theta) \). This equation is called a rose curve, and it has petals because of the periodic sinusoidal function. Each petal represents the symmetrical properties of the circle.

Determine the Range for One Petal

For polar curves of form \( r = a \sin(n\theta) \), petals form when \( n \) is odd, covering \( 0 \leq \theta \leq \pi \), and when \( n \) is even, covering \( 0 \leq \theta \leq \pi/n \). Since \( n=2 \) here, one petal is formed between \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \).

Set Up the Integral for the Area

The formula for the area enclosed by a polar curve \( r(\theta) \) is \( A = \frac{1}{2} \int (r(\theta))^2 \, d\theta \). For this petal, substitute \( r = 8 \sin(2\theta) \) to get the integral for the area of one petal: \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (8 \sin(2\theta))^2 \, d\theta \].

Simplify the Integrand

First, simplify \((8 \sin(2\theta))^2\) to \(64 \sin^2(2\theta)\). Now, \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 64 \sin^2(2\theta) \, d\theta \]. This further simplifies to \[ A = 32 \int_{0}^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta \].

Use Trigonometric Identity

Utilize the identity \( \sin^2(u) = \frac{1 - \cos(2u)}{2} \) to simplify the integrand. Substitute \( u = 2\theta \), so \( \sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2} \). Therefore, \[ A = 32 \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(4\theta)}{2} \, d\theta \].

Evaluate the Integral

Split the integral, simplify and evaluate: \[ A = 32 \cdot \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \int_{0}^{\frac{\pi}{2}} \cos(4\theta) \, d\theta \right) = 16 \cdot \left( \frac{\pi}{2} - 0 \right) \].The integral of \(1\) is simply \(\theta\), evaluated over the interval \([0, \frac{\pi}{2}]\) becomes \(\frac{\pi}{2}\). The integral of \(\cos(4\theta)\) over \([0, \frac{\pi}{2}]\) cancels to zero since it is a complete sinusoidal wave over a half-cycle.

Compute the Area

Substituting and calculating gives \[ A = 16 \cdot \frac{\pi}{2} = 8\pi \].This is the area of one petal of the rose curve \( r = 8 \sin(2\theta) \).

Key concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction. Polar coordinates are presented in the form \((r, \theta)\), where:

• \(r\) is the radius or distance from the origin.
• \(\theta\) is the angle measured from the positive x-axis.

This system is particularly useful for problems involving circular or rotational symmetry, such as those involving rose curves.
While Cartesian coordinates use x and y to define a point in a plane, polar coordinates rely on the radial distance and angular displacement. This method can simplify the equations of various curves, such as spirals and circles.

Rose Curve

A rose curve is a type of polar graph represented by equations like \(r = a \sin(n\theta)\) or \(r = a \cos(n\theta)\). They are called "rose" curves because the resulting plot resembles the shape of a flower with petals.

• If \(n\) is odd, the rose will feature \(n\) petals.
• If \(n\) is even, the curve will have \(2n\) petals.

In our exercise, the equation \(r = 8 \sin(2\theta)\) gives us a four-petaled rose, describing a symmetry with the number of petals dependent on the argument of the sine function. These petals are distributed evenly around the circle, allowing one to systematically calculate the area of a single petal.
When analyzing rose curves, understanding the symmetrical nature can help simplify calculations, such as when determining the area enclosed by one or multiple petals.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. These are vital in simplifying expressions and solving equations in calculus and trigonometry. A key identity used in this exercise is the Pythagorean identity:

• \(\sin^2(u) + \cos^2(u) = 1\).

Another useful identity is the double angle formula, such as:

• \(\sin^2(u) = \frac{1 - \cos(2u)}{2}\).

In this context, converting \(\sin^2(2\theta)\) using the identity \(\sin^2(u) = \frac{1 - \cos(2u)}{2}\) simplifies the integral in the area calculation. Recognizing these identities and applying them correctly streamlines solving complex integral problems in polar coordinates.

Area Calculation

Calculating the area enclosed by polar curves involves integrating with respect to \(\theta\) from the polar area formula. The general formula for the area \(A\) is:\[ A = \frac{1}{2} \int (r(\theta))^2 \, d\theta \]For our rose curve \(r = 8 \sin(2\theta)\), we calculate the area of a single petal formed between \(\theta = 0\) and \(\theta = \frac{\pi}{2}\).
Substituting the equation into the formula and simplifying involves:

• Implementing the trigonometric identity to simplify \(\sin^2(2\theta)\).
• Evaluating the definite integral over the range \([0, \frac{\pi}{2}]\).

After simplifying, this results in a straightforward computation of the areas represented by basic trigonometric integrals. The final calculation yields the area of one petal as \(8\pi\). Understanding these steps is critical for similar problems involving symmetric shapes in polar coordinates.