Question

Finding an Area of a Polar Region Find the area of one petal of the rose defined by the equation \(r=3 \sin (2 \theta)\).

Short answer

The area of one petal is \(\frac{9\pi}{8}\).

Step-by-step solution

Identify Number of Petals

The polar equation given is \(r=3\sin(2\theta)\). In general, for equations of the form \(r=a\sin(n\theta)\), there are \(n\) petals if \(n\) is odd and \(2n\) petals if \(n\) is even. Since \(n=2\) here, the rose has 4 petals.

Determine the Bounds for One Petal

To find the area of one petal, we need to determine the values of \(\theta\) for which the petal is traced. A single petal of \(r=3\sin(2\theta)\) is traced when \(\sin(2\theta)=0\). The bounds for one petal occur from \(\theta=0\) to \(\theta=\frac{\pi}{2}\). Since the sine function completes one cycle from 0 to \(\pi\) (not \(2\pi\) since \(\sin(2\theta)\)), a single petal is traced from \(0\) to \(\frac{\pi}{2}\).

Set Up the Integral for Area

The formula to find the area \(A\) of one petal of a polar curve \(r=f(\theta)\) is \(A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 \, d\theta\). Substitute \(f(\theta) = 3\sin(2\theta)\) and the limits from Step 2: \(A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [3\sin(2\theta)]^2 \, d\theta\).

Compute the Integral

Simplify the integrand: \([3\sin(2\theta)]^2 = 9\sin^2(2\theta)\).Area then becomes:\[A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 9\sin^2(2\theta) \, d\theta.\]Use the identity \(\sin^2(2\theta) = \frac{1-\cos(4\theta)}{2}\) to simplify:\[A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 9 \times \frac{1-\cos(4\theta)}{2} \, d\theta = \frac{9}{4} \int_{0}^{\frac{\pi}{2}} (1-\cos(4\theta)) \, d\theta.\]

Evaluate the Integral

We now need to calculate the integral:\[\frac{9}{4} \int_{0}^{\frac{\pi}{2}} (1-\cos(4\theta)) \, d\theta = \frac{9}{4} \left[ \theta - \frac{1}{4} \sin(4\theta) \right]_{0}^{\frac{\pi}{2}}\]Computing from 0 to \(\frac{\pi}{2}\),\[\frac{9}{4} \left[ \frac{\pi}{2} - 0 \right] = \frac{9\pi}{8} - 0 \]

Finalize the Area of One Petal

The computed area for one petal is given by \(\frac{9\pi}{8}\). This is the final answer for the area enclosed by one petal.

Key concepts

Area of Polar Region

To find the area of a polar region, we often deal with the intriguing and unique shapes formed by polar curves. Polar equations like roses, circles, and spirals are represented in a polar coordinate system using a radius and an angle. To find the area of one petal of a polar region, like the rose curve given by the equation \( r=3\sin(2\theta) \), we have to follow a specific method.
First, you need to determine how many repetitions or 'petals' these curves make. For example, the rose curve \( r=a\sin(n\theta) \) or \( r=a\cos(n\theta) \), will have \( n \) petals if \( n \) is odd and \( 2n \) petals if \( n \) is even. In our exercise, where \( n=2 \), the rose has 4 petals.
Next, you need to find the section of the graph that displays one petal. This involves determining the angular bounds (\( \theta \) values) over which a single petal is traced. Identifying these bounds is critical, as they will be used as the limits in our integral to calculate the area of one petal.

Polar Equation

Polar equations play a key role in defining the shape of a polar region. They differ from Cartesian equations as they express a relationship between the radius \( r \) and the angle \( \theta \) directly. A polar equation such as \( r=3\sin(2\theta) \) dictates the radial distance from the origin to any point on the curve, based on the angle \( \theta \).
In polar coordinates, every point in the plane is determined by two values: the distance from a fixed reference point (often called the pole, similar to the origin in the Cartesian system), and the angle from a fixed direction (usually the positive x-axis), known as the polar axis. This system excels at representing spirals and periodic structures, with equations often characterized by trigonometric functions.
Understanding the role of \( n \) in the equations \( r=a\sin(n\theta) \) and \( r=a\cos(n\theta) \) is crucial, as it influences the symmetry and number of petals in flower-like curves. Our exercise used \( n = 2 \), resulting in additional petals due to the even nature of \( n \), doubling the anticipated outcome.

Integrals in Polar Coordinates

Integrals in polar coordinates offer a way to calculate areas for complex polar shapes that don't conform to simple geometric formulas. Unlike Cartesian coordinates, polar integrals involve integrating with respect to the angle \( \theta \).
For the rose curve example \( r=3\sin(2\theta) \), the area \( A \) of one petal can be determined using the strategy:

• Identify the bounds for one petal, typically from \( \alpha \) to \( \beta \).
• Apply the integral formula for polar areas: \( A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 \, d\theta \), where \( f(\theta) = r \).
• Simplify and solve the integral to find the area.

In this specific situation, \( f(\theta) \) becomes the rose equation, leading to \( A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [3\sin(2\theta)]^2 \, d\theta \). Simplifying using trigonometric identities is often required, like noting \( \sin^2(2\theta) = \frac{1-\cos(4\theta)}{2} \).
Finally, perform the integration over the specified bounds to calculate the enclosed area. This method highlights how polar integrals are instrumental in calculating areas that are not easily addressed using traditional Cartesian methods.