Problem 75 - Essential Calculus

Question

Find the area of the region. $$ \begin{array}{l} x=2 \sin ^{2} \theta \\ y=2 \sin ^{2} \theta \tan \theta \\ 0 \leq \theta<\frac{\pi}{2} \end{array} $$

Step-by-Step Solution

Answer

The area of the region enclosed by the given coordinates is \(\frac{3\pi}{4} - 1\) square units.

Step 1: Convert to polar coordinates

The given system is already in polar form. We have, \(r=2 \sin^{2} \theta\) and \(0 \leq \theta < \frac{\pi}{2}\)

Step 2: Use the formula for area in polar coordinates

The area \(A\) enclosed by a polar curve \(r = f(\theta)\) from \(\theta = a\) to \(\theta = b\) is given by \(A = \frac{1}{2} \int_{a}^{b} [f(\theta)]^2 d\theta\). Substituting \(f(\theta) = 2 \sin^{2} \theta\) and \(a = 0\), \(b = \frac{\pi}{2}\) we get, \(A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [2 \sin^{2} \theta]^2 d\theta = 2\int_{0}^{\frac{\pi}{2}} \sin^{4} \theta d\theta\)

Step 3: Integrate the function

To simplify the integration, use the power reduction identity: \(\sin^{4} \theta = \frac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}\), then integrate: \(2\int_{0}^{\frac{\pi}{2}} \sin^{4} \theta d\theta = \frac{1}{2} \left[ 3\theta - 2\sin(2\theta) + \frac{1}{4}\sin(4\theta) \right]_{0}^{\frac{\pi}{2}} = \frac{3\pi}{4} - 1\)

Step 4: Area of the bound region

The area of the region bound by the curve is thus, \(\frac{3\pi}{4} - 1\) square units

Key Concepts

Integration in Calculus

Integration in calculus is a fundamental concept that involves finding the accumulated value, such as areas under curves. When we talk about integration, we usually refer to an operation that brings us from a "derivative" back to a function. This is known as the antiderivative or the integral of the function.

In this exercise, our task is to find the area of a particular region using integration. Polar coordinates provide a beautiful way to describe equations and their curves in a two-dimensional plane. When equations are expressed in polar forms, integration helps calculate areas between such curves efficiently. To do this, we use methods specific to polar equations, focusing on the concept of accumulation to find total areas.

Key points in integration include:

• Understanding the functions involved and what they represent.
• Applying the right formulas or methods to find the area.
• Identifying the correct limits of integration, which in this problem are from 0 to \(\frac{\pi}{2}\).

Trigonometric Identities

Trigonometric identities are vital in simplifying complex expressions, especially when integrating. They relate the basic trigonometric functions to each other and are used extensively to transform complicated integrals into solvable forms.

In this solution, one of the essential steps was using a power reduction identity to simplify \(\sin^4 \theta\). By expressing \(\sin^4 \theta\) in terms of cosines, we can integrate the function easily. The identity:

• \(\sin^4 \theta = \frac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}\).

This transformation is crucial as it breaks down the power of sine into a combination of terms that are straightforward to integrate. Using identities saves time and reduces errors by decomposing expressions into more manageable parts.

Such identities effectively bridge the gap between complex functions and basic operations, allowing us to solve integration problems that initially seem daunting.

Definite Integrals

Definite integrals provide the exact accumulated quantity between two specific limits. In terms of geometry, they help calculate the total area under a curve between two points along the x-axis or, in this case, along the \(\theta\)-axis in polar coordinates.

Definite integration is distinct from indefinite integration. In definite integration:

• The result is a fixed number representing the area, volume, or other accumulated value.
• The integration is computed between two limits, known as the lower and upper bounds.
• The specific bounds in this problem are \(0\) and \(\frac{\pi}{2}\).

Using definite integrals, we calculated the area enclosed by the curve \(r = 2 \sin^2 \theta\). The integration process involved working through trigonometric transformations to evaluate the integral from the specified limits, yielding an exact area of \(\frac{3\pi}{4} - 1\) square units.

This final numerical result tells us the size of the region enclosed by the polar curve, showcasing the power of definite integrals in solving real-world area problems.