Question

Use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.\(r=3 \sin \theta\) on the interval \(0 \leq \theta \leq \pi\)

Short answer

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

Step-by-step solution

Identify the region

The given equation is \(r = 3 \sin \theta\). This describes a circle with radius 3 centered at \( (0, \frac{3}{2}) \). The region described is a semicircle because the bounds of \(\theta\) are from \(0\) to \(\pi\).

Formula for the area of a circle

To find the area of the region using geometry, recognize that \(r = 3 \sin \theta\) forms a full circle with diameter \(3\). Therefore, the radius of the circle is \(\frac{3}{2}\). The area of a full circle with radius \(\frac{3}{2}\) is \(\pi (\frac{3}{2})^2 = \frac{9\pi}{4}\).

Calculate the area of the semicircle

Since \(\theta\) ranges from \(0\) to \(\pi\), it only describes a semicircle (half of a circle). Thus, the area of the semicircle is \(\frac{1}{2} \times \frac{9\pi}{4} = \frac{9\pi}{8}\).

Set up definite integral

The area in polar coordinates is calculated using the definite integral \(\int_{a}^{b} \frac{1}{2} r^2 \, d\theta\). Here, \(r = 3 \sin \theta\), thus \(r^2 = 9 \sin^2 \theta\). The integral becomes \(\int_{0}^{\pi} \frac{1}{2} 9 \sin^2 \theta \, d\theta\).

Simplify and solve integral

We use the identity \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\) to simplify the integral to \(\frac{9}{2} \int_{0}^{\pi} \frac{1 - \cos(2\theta)}{2} \, d\theta\). This simplifies to \(\frac{9}{4} \int_{0}^{\pi} (1 - \cos(2\theta)) \, d\theta\).

Perform integration

Calculate the integral \(\int_{0}^{\pi} 1 \, d\theta = [\theta]_{0}^{\pi} = \pi\) and \(\int_{0}^{\pi} \cos(2\theta) \, d\theta = [\frac{1}{2} \sin(2\theta)]_{0}^{\pi} = 0\). Adding the results, the integral is \(\pi\).

Confirm calculated area with integral

The final calculated area is \(\frac{9}{4} \times \pi = \frac{9\pi}{4}\), and because we integrated the full semicircle via the transformation (not calculating symmetry), the answer matches the geometric area. Divide by 2 since this is a semicircle: \(\frac{9\pi}{8}\) confirms Step 3.

Key concepts

Definite Integral

The concept of a definite integral is central to calculus, used to calculate the accumulated value over an interval. In this context, it helps find the area of a region in polar coordinates. With polar equations like \(r = 3 \sin \theta\), the definite integral allows us to determine the exact area of a semicircle over the interval \(0 \leq \theta \leq \pi\).
To set up the definite integral for this problem, we use the polar area formula:

• \(\text{Area} = \int_{a}^{b} \frac{1}{2} r^2 \, d\theta\)

This means for our example, \(r = 3 \sin \theta\) leads to \(r^2 = 9 \sin^2 \theta\), and our integral becomes \(\int_{0}^{\pi} \frac{1}{2} 9 \sin^2 \theta \, d\theta\).
Solving this integral involves using trigonometric identities, which transform \(\sin^2 \theta\) into a more manageable form. Thus, the definite integral not only confirms the geometric calculation of the semicircle's area but also showcases the power of calculus in solving real-world problems.

Area of a Semicircle

Understanding the area of a semicircle is integral when dealing with equations in polar coordinates. A semicircle is essentially half of a full circle. For a circle described by \(r = 3 \sin \theta\), it forms a full circle with a center at \((0, \frac{3}{2})\) and a diameter of 3.
To find the area of this semicircle, first calculate the area of the full circle using the formula \(\pi r^2\), considering the radius to be \(\frac{3}{2}\):

• Area of full circle = \(\pi (\frac{3}{2})^2 = \frac{9\pi}{4}\)

Since the interval is from \(0\) to \(\pi\), only half the circle is considered, producing the area of a semicircle:

• Area of semicircle = \(\frac{1}{2} \times \frac{9\pi}{4} = \frac{9\pi}{8}\)

This geometric understanding ties in seamlessly with calculations performed using definite integrals, providing a practical way to check mathematical derivations.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. These identities are crucial in simplifying and solving complex integrals.
In the context of finding the area of a region defined by \(r = 3 \sin \theta\), there is an important identity used: \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\).
This identity transforms the integral \(\int_{0}^{\pi} 9 \sin^2 \theta \, d\theta\) into a simpler form:

• \(\int_{0}^{\pi} \frac{9}{2}(1 - \cos(2\theta)) \, d\theta\)

Breaking down complex expressions into simpler components using trigonometric identities allows for more straightforward integration. It enables the calculation of the semicircle's area in a straightforward manner using a definite integral. Understanding and applying these identities is vital for solving problems in calculus involving trigonometric functions.