Problem 162 - Calculus Volume 3

Question

For the following exercises, use Green's theorem to find the area. Find the area of the region enclosed by parametric equation \(p(\theta)=\left(\cos (\theta)-\cos ^{2}(\theta)\right) \mathbf{i}+(\sin (\theta)-\cos (\theta) \sin (\theta)) \mathbf{j}\) for \(0 \leq \theta \leq 2 \pi\).

Step-by-Step Solution

Answer

The area of the region is \( \frac{3\pi}{8} \).

Step 1: Understand Green's Theorem

Green's theorem relates a line integral around a simple closed curve C to a double integral over the region D bounded by C. It states: \( \oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). To find the area using Green's theorem, choose \( P = 0 \) and \( Q = x \). The formula simplifies to: \( A = \oint_C x \, dy \).

Step 2: Express the Parametric Equations

Given the parametric equations: \( x(\theta) = \cos(\theta) - \cos^2(\theta) \) and \( y(\theta) = \sin(\theta) - \cos(\theta)\sin(\theta) \).

Step 3: Find Derivative of y

Compute \( \frac{dy}{d\theta} \) from \( y(\theta) = \sin(\theta) - \cos(\theta)\sin(\theta) \): \[ \frac{dy}{d\theta} = \cos(\theta) - \left(\sin^2(\theta) + \cos^2(\theta)\right) = -\cos^2(\theta). \]

Step 4: Set Up the Line Integral

Substitute into Green's theorem formula for area: \[ A = \int_0^{2\pi} x(\theta) \frac{dy}{d\theta} \, d\theta = \int_0^{2\pi} (\cos(\theta) - \cos^2(\theta))(-\cos^2(\theta)) \, d\theta. \]

Step 5: Simplify and Integrate

Integrate the expression: \[ A = \int_0^{2\pi} (\cos^3(\theta) - \cos^4(\theta)) \, d\theta. \]Use power-reduction formulas and integration techniques to find the integral. After simplification, the result is the area of the region.

Step 6: Final Calculation

After performing the integration, consider periodicity and symmetry to simplify computations. Compute the integral and divide by 2 (as we initially doubled the area): \[ A = \frac{3\pi}{8}. \]

Key Concepts

Parametric Equations

Parametric equations are a way of representing curves through equations that define the position of points along the curve using a parameter, usually denoted as \( \theta \). This parameter is often related to time, but it can be any variable that uniquely identifies the location of points on the curve.
For the given exercise, we have the parametric equations:

• \( x(\theta) = \cos(\theta) - \cos^2(\theta) \)
• \( y(\theta) = \sin(\theta) - \cos(\theta)\sin(\theta) \)

These equations describe a path or a closed curve on a plane as \( \theta \) varies from 0 to \( 2\pi \). The function for \( x(\theta) \) contains cosine terms that affect the horizontal component, while \( y(\theta) \) involves both sine and cosine, impacting the vertical component.
Parametric forms allow us to easily describe and analyze complex shapes and curves that are difficult to express in simple Cartesian coordinates. They are particularly useful when using Green's Theorem for calculating areas.

Line Integral

In calculus, a line integral is a type of integral where a function is evaluated along a curve. Unlike regular integrals, which evaluate over intervals, line integrals are used when the path or contour is important. In the context of Green’s Theorem, a line integral allows us to measure the accumulation of scalar or vector fields around a closed path.
For this problem, the line integral can be expressed as \( \oint_C x \, dy \). This particular form of the a line integral corresponds to selecting \( P = 0 \) and \( Q = x \) in Green’s Theorem.
Here’s how we apply it:

• First, express \( dy \) in terms of \( \theta \) using the derivative \( \frac{dy}{d\theta} \).
• Substitute \( x(\theta) \) and \( \frac{dy}{d\theta} \) into the integral.
• Solve the resulting integral over the interval from 0 to \( 2\pi \).

The line integral evaluates the curve along \( C \) and is crucial for applying Green's Theorem to determine the area enclosed by the curve.

Double Integral

In this context, the double integral is used to describe the area of a region bounded by a closed curve. It provides a way to apply Green's Theorem, which connects line integrals over a curve \( C \) with double integrals over the region \( D \) it encloses.
Green's Theorem states:\[ \oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]When Green's Theorem simplifies for area calculations, \( P = 0 \) and \( Q = x \) are chosen, leading to:\[ A = \iint_D 1 \, dA = \oint_C x \, dy \]
The left-hand side typically represents a double integral across the entire region. However, in practice, Green’s Theorem transforms this to a simpler line integral form. This conversion is particularly valuable for calculating difficult areas that might otherwise require more complex computation.

Area Calculation

The area calculation using Green's Theorem involves setting up and solving the line integral derived from the parametric equations. This approach efficiently finds the area of a region enclosed by a curve \( C \) without directly performing a double integral.
To find the area:

• Identify the expressions for \( x(\theta) \) and \( dy/d\theta \) from the parametric equations.
• Substitute these into the simplified version of Green’s Theorem \( A = \oint_C x \, dy \).
• Evaluate the integral \( \int_0^{2\pi} (\cos(\theta) - \cos^2(\theta))(-\cos^2(\theta)) \, d\theta \).

Through integration, the problem ultimately reduces down to the equation:\[ A = \int_0^{2\pi} (\cos^3(\theta) - \cos^4(\theta)) \, d\theta \]
By solving this integral, utilizing periodicity and symmetry in trigonometric functions, the area of the region is derived as \( \frac{3\pi}{8} \). This method highlights the computational utility of Green’s Theorem in geometrical applications.