Problem 4 - The Calculus with Analytic Geometry

Question

Find the area of the region enclosed by the graph of the given equation.\(r=4 \sin ^{2} \frac{1}{2} \theta\)

Step-by-Step Solution

Answer

The area is \( 6 \pi \).

Step 1: Understanding the Given Equation

The equation given is in polar form: \( r = 4 \sin^2 \left( \frac{1}{2} \theta \right) \). This represents a polar graph and the goal is to find the area enclosed by this graph.

Step 2: Setting Up the Area Integral

The general formula for the area enclosed by a polar curve \( r = f(\theta) \) is:\[ A = \frac{1}{2} \int_{a}^{b} [f(\theta)]^2 \, d\theta \]

Step 3: Determine the Limits for Integration

Observe the function \( 4 \sin^2 \left( \frac{1}{2} \theta \right) \). This function is periodic with a period of \( 2\pi \). Therefore, the limits of integration are from \( 0 \) to \( 2\pi \).

Step 4: Plugging in the Function

Substitute \( r = 4 \sin^2 \left( \frac{1}{2} \theta \right) \) into the area formula:\[ A = \frac{1}{2} \int_{0}^{2\pi} [4 \sin^2 \left( \frac{1}{2} \theta \right)]^2 \, d\theta \]

Step 5: Simplifying the Integrand

Simplify the integrand:\[ [4 \sin^2 \left( \frac{1}{2} \theta \right)]^2 = 16 \sin^4 \left( \frac{1}{2} \theta \right) \]Thus, the area becomes:\[ A = 8 \int_{0}^{2\pi} \sin^4 \left( \frac{1}{2} \theta \right) \, d\theta \]

Step 6: Using Trigonometric Identity

Use the identity for \( \sin^4(x) \):\[ \sin^4(x) = \frac{3}{8} - \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x) \]Let \( x = \frac{1}{2} \theta \), so the integral becomes:\[ A = 8 \int_{0}^{2\pi} \left( \frac{3}{8} - \frac{1}{2} \cos \theta + \frac{1}{8} \cos 2\theta \right) \, d\theta \]

Step 7: Evaluating the Integral

Integrate each term separately:\[ A = 8 \left[ \frac{3}{8} \theta \Bigg|_{0}^{2\pi} - \frac{1}{2} \sin \theta \Bigg|_{0}^{2\pi} + \frac{1}{8} \cdot \frac{1}{2} \sin(2\theta) \Bigg|_{0}^{2\pi} \right] \]Since \( \sin \theta \) and \( \sin(2\theta) \) are periodic functions with zero-average over their periods:\[ A = 8 \left[ \frac{3}{8} (2\pi) \right] = 8 \cdot \frac{3}{8} \cdot 2\pi = 6\pi \]

Key Concepts

polar coordinates

Polar coordinates offer a different way to describe the position of a point in a plane. Instead of using the Cartesian coordinate system's x and y coordinates, polar coordinates use a radius and an angle. The radial coordinate (r) represents the distance from the origin to the point, and the angular coordinate (θ) represents the angle between the positive x-axis and the line connecting the origin to the point. This is particularly useful when dealing with curves that are naturally circular or spiral in shape, such as our given equation: \( r = 4 \sin^2 \left( \frac{1}{2} \theta \right) \).
Using polar coordinates simplifies finding areas and can help in visualizing problems better. The formula for the area of a region enclosed by a polar graph is quite elegant and different from Cartesian methods.

area integrals

Finding the area of a region enclosed by a polar curve involves integrating the square of the radius function over the given range of angles. The general formula for the area \(A\) enclosed by a polar curve \( r = f(\theta) \) is:
\[ A = \frac{1}{2} \, \int_{a}^{b} [f(\theta)]^2 \, d\theta \]
This formula accounts for the circular nature of the polar coordinate system. In our exercise, we apply this formula by first determining the limits of integration, which for our function are from \(0\) to \(2π\), because the function is periodic with this period.
We then substitute our function \( r = 4 \sin^2 \left( \frac{1}{2} \theta \right) \) into the formula:
\[ A = \frac{1}{2} \, \int_{0}^{2π} [4 \sin^2 \left( \frac{1}{2} \theta \right)]^2 \, d\theta \]
Understanding how to set up the area integral is crucial for solving problems involving polar coordinates and ensures we accurately capture the area enclosed by these curves.

trigonometric identities

Trigonometric identities are essential in simplifying the integrand in polar area integrals. In our problem, the integrand was simplified using the identity for \(\sin^4(x)\). These identities transform complex trigonometric expressions into more manageable forms. We used:
\[ \sin^4(x) = \frac{3}{8} - \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x) \]
Substituting \( x = \frac{1}{2} \theta \) allows us to transform \( 16 \sin^4 \left( \frac{1}{2} \theta \right) \) into a sum of simpler trigonometric functions. This makes it easier to integrate:
\[ A = 8 \, \int_{0}^{2π} \left( \frac{3}{8} - \frac{1}{2} \cos \theta + \frac{1}{8} \cos 2\theta \right) \, d\theta \]
Using trigonometric identities helps break down the problem into parts that can be solved with basic integration techniques.

integration

Integration is a fundamental concept in calculus used to find areas under curves. In our problem, integration helps us determine the area enclosed by the polar curve. After simplifying our integrand using a trigonometric identity, we perform the following integration:
\[ A = 8 \, \left[ \frac{3}{8} \theta \Bigg|_{0}^{2π} - \frac{1}{2} \sin \theta \Bigg|_{0}^{2π} + \frac{1}{8} \, \frac{1}{2} \sin(2\theta) \Bigg|_{0}^{2π} \right] \]
Each term is integrated separately. Since \( \sin \theta \) and \( \sin(2 \theta) \) are periodic functions with zero-average over their periods, their integrals evaluate to zero over one period. Thus, our area simplifies to:
\[ A = 8 \, \left[ \frac{3}{8} \, (2π) \right] = 8 \, \cdot \frac{3}{8} \cdot 2π = 6π \]
By understanding integration, we transform our setup into a numeric value that represents the enclosed area.