Problem 103 - Calculus Volume 3

Question

Find the area of the region bounded by \(x=2 \sin ^{2} \theta, y=2 \sin ^{2} \theta \tan \theta,\) for \(0 \leq \theta \leq \frac{\pi}{2}\).

Step-by-Step Solution

Answer

The area of the region is 1 square unit.

Step 1: Parametric Equations Identification

The given problem provides parametric equations: \( x = 2 \sin^2 \theta \) and \( y = 2 \sin^2 \theta \tan \theta \). We are asked to find the area of the region bounded by these equations over the interval \( 0 \leq \theta \leq \frac{\pi}{2} \).

Step 2: Area Element in Parametric Form

The area \( A \) under a parametric curve from \( \theta = \alpha \) to \( \theta = \beta \) given by \( x = f(\theta) \) and \( y = g(\theta) \) is \( \int_{\alpha}^{\beta} y \frac{dx}{d\theta} \, d\theta \). Here, \( \alpha = 0 \) and \( \beta = \frac{\pi}{2} \).

Step 3: Compute dx/dθ

First, find \( \frac{dx}{d\theta} \) for \( x = 2 \sin^2 \theta \): \( \frac{dx}{d\theta} = 4 \sin \theta \cos \theta = 2 \sin 2\theta \) using the trigonometric identity \( \cos 2\theta = 2 \sin \theta \cos \theta \).

Step 4: Set Up Integral for the Area

Using the expression for \( \frac{dx}{d\theta} \), the area becomes \[ \int_{0}^{\frac{\pi}{2}} (2 \sin^2 \theta \tan \theta) (2 \sin 2\theta) \, d\theta \]. Simplify to get \[ 4 \int_{0}^{\frac{\pi}{2}} \sin^3 \theta \cos \theta \, d\theta \].

Step 5: Simplify Integral

The integral \( \int \sin^3 \theta \cos \theta \, d\theta \) can be simplified using the substitution \( u = \sin \theta \), \( du = \cos \theta \, d\theta \). The integral becomes \( \int u^3 \, du \), which is \( \frac{u^4}{4} + C \).

Step 6: Evaluate the Integral

Substitute back \( u = \sin \theta \) and evaluate from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \): \[ \frac{4}{4} [\sin^4 \theta]_{0}^{\frac{\pi}{2}} = [\sin^4 \theta]_{0}^{\frac{\pi}{2}} \]. Since \( \sin \frac{\pi}{2} = 1 \) and \( \sin 0 = 0 \), the final result is 1.

Key Concepts

Parametric equations

Parametric equations are a way to represent a curve by defining both the x and y coordinates as functions of a third variable, often denoted as \( \theta \), which is called a parameter. This allows us to describe more complex curves that may not be easily defined by a single function in terms of x and y alone. In the given problem, the parametric equations are \( x = 2 \sin^2 \theta \) and \( y = 2 \sin^2 \theta \tan \theta \). This means for each value of \( \theta \) within a specified range, you can calculate corresponding x and y coordinates.
Parametric equations are particularly useful in describing curves that loop back on themselves or curves that aren't functions in the traditional sense since every x value is associated with exactly one y value. They provide the flexibility to represent a variety of shapes and trajectories in a simple way.

Trigonometric identities

Trigonometric identities are mathematical equations that involve trigonometric functions and are true for all values of the variables involved. These identities are very useful in simplifying and solving trigonometric expressions. For example, a key identity used in this problem is the double-angle formula for sine: \( \sin 2\theta = 2 \sin \theta \cos \theta \). This helps us to transform and simplify expressions involving trigonometric functions.
In the area calculation problem, understanding and applying these identities lets us move from the expression \( 2 \sin \theta \cos \theta \) to \( \sin 2\theta \), which produces a simpler equivalent expression. Knowing these identities by heart or understanding how they are derived will significantly enhance your problem-solving skills.

Calculus integration

Integration is a fundamental concept in calculus that allows us to find monumental values such as areas under curves, volumes, and other related measures. It is the process of finding a function from its derivative — essentially 'summarizing' the accumulation of quantity.
In this problem, integration is used to calculate the total area under the curve defined by our parametric equations, which involves setting up an integral that refines and simplifies the functions of \( \theta \). Here, we used the integral form \( A = \int_{\alpha}^{\beta} y \frac{dx}{d\theta} \, d\theta \), to determine the area under the curve over the interval \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \). Understanding integration allows one to calculate not only areas, but also other quantitative properties of curves.

Parametric form area calculation

To find the area under a parametric curve, it involves integrating with respect to the parameter rather than x or y directly. In our given exercise, we use the formula for the area \( A \) under a parametric curve: \[ A = \int_{\alpha}^{\beta} y \frac{dx}{d\theta} \, d\theta \]. This method essentially "pieces together" infinitesimally small sections of area to find the total area under the curve.
We found \( \frac{dx}{d\theta} \) from the parametric equation \( x = 2 \sin^2 \theta \), which was found to be \( 2 \sin 2\theta \) through differentiation. With \( y = 2 \sin^2 \theta \tan \theta \), the area was calculated by evaluating the integral \[4 \int_{0}^{\frac{\pi}{2}} \sin^3 \theta \cos \theta \, d\theta \]. The final area was found to be 1, which shows the complete understanding and application of parametric integration to solve real mathematical problems.