Question

Determine a definite integral that represents the area.Region enclosed by \(r=3 \sin \theta\)

Short answer

The area is \( \frac{9\pi}{4} \).

Step-by-step solution

Understand the Polar Equation

The region we need to consider is defined by the polar equation \( r = 3 \sin \theta \). This represents a circle in polar coordinates with a radius of 3 and its center at \( (0, \frac{3}{2}) \) in Cartesian coordinates.

Identify the Limits of Integration

For the polar equation \( r = 3 \sin \theta \), the full circle is traced as \( \theta \) varies from \( 0 \) to \( \pi \). These will be our limits of integration for the integral that calculates the area.

Set Up the Integral for Area in Polar Coordinates

The formula for the area \( A \) of a region enclosed by a polar curve \( r(\theta) \) from angle \( \theta = a \) to \( \theta = b \) is given by \( A = \frac{1}{2} \int_{a}^{b} [r(\theta)]^2 \, d\theta \). Here, \( a = 0 \) and \( b = \pi \), and \( r(\theta) = 3 \sin \theta \).

Write the Definite Integral

Using the formula from the previous step, the definite integral for the area is: \[ A = \frac{1}{2} \int_{0}^{\pi} (3 \sin \theta)^2 \, d\theta \].

Simplify and Integrate

Simplify the integrand to get \[ A = \frac{1}{2} \int_{0}^{\pi} 9 \sin^2 \theta \, d\theta \]. Use the trigonometric identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \) to rewrite the integral as: \[ A = \frac{9}{2} \int_{0}^{\pi} \frac{1 - \cos(2\theta)}{2} \, d\theta \].Simplify further:\[ A = \frac{9}{4} \int_{0}^{\pi} (1 - \cos(2\theta)) \, d\theta \].

Evaluate the Integral

Divide the integral and evaluate separately:\[ A = \frac{9}{4} \left[ \int_{0}^{\pi} 1 \, d\theta - \int_{0}^{\pi} \cos(2\theta) \, d\theta \right] \].First Integral: \( \int_{0}^{\pi} 1 \, d\theta = \theta \bigg|_0^{\pi} = \pi \).Second Integral: \( \int_{0}^{\pi} \cos(2\theta) \, d\theta = \frac{1}{2} \sin(2\theta) \bigg|_0^{\pi} = 0 \).Combine results:\[ A = \frac{9}{4} \left( \pi - 0 \right) = \frac{9\pi}{4} \].

Key concepts

Definite Integral

Understanding definite integrals is essential for finding the area under a curve in calculus. A definite integral is an integral with limits of integration, which specifies the range over which we calculate the area. It is denoted as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits, and \( f(x) \) is the integrand, representing the function whose area we want to find.

In polar coordinates, the function is expressed in terms of \( \theta \) (angle) and \( r(\theta) \) (radius). For the exercise given, the polar function \( r = 3 \sin \theta \) describes a circle. The definite integral has the form:\[ A = \frac{1}{2} \int_{a}^{b} [r(\theta)]^2 \, d\theta \]where \( a = 0 \) and \( b = \pi \). This integral helps calculate the total enclosure area of the circle in terms of theta.

When solving this definite integral, each step involves breaking down the polar function into integral terms, combining them using trigonometric identities, and finally calculating the enclosed area.

Area Under a Curve

Finding the area under a curve is a classic problem in calculus. It is crucial for determining the space contained within boundaries defined by functions like those in polar coordinates. The beauty of polar coordinates lies in their ability to handle cyclic and radial functions succinctly.

In the given exercise, you use the definite integral to compute the total area under the curve described by the polar equation \( r = 3 \sin \theta \) from \( \theta = 0 \) to \( \theta = \pi \). Since this curve forms a circle, traditional Cartesian methods aren't as efficient.

The integral approach, particularly in polar form, can directly sum up infinitesimal wedges of the circle, achieving an accurate measurement of area. The formula \( A = \frac{1}{2} \int_{0}^{\pi} (3 \sin \theta)^2 \, d\theta \) elegantly gives us that total area.

Trigonometric Identities

Using trigonometric identities is a powerful method for simplifying complex integrals. In calculus, especially when dealing with integrals of trigonometric functions, identities can transform difficult tasks into manageable steps.

The exercise deals with \( \sin^2 \theta \), a trigonometric expression. To integrate this efficiently, you employ the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \). This transformation simplifies the integral into a sum of terms that are straightforward to compute:

• \( \int_{0}^{\pi} 1 \, d\theta \)
• \( \int_{0}^{\pi} \cos(2\theta) \, d\theta \)

Each sub-integral has a known solution, enabling you to find the enclosed area: \( A = \frac{9\pi}{4} \).

Identities like \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \) make complex integrations accessible, by breaking them into smaller, solvable pieces.