Problem 31 - Thomas Calculus

Question

Green's Theorem Area Formula Area of \(R=\frac{1}{2} \oint_{C} x d y-y d x\) Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves. The circle \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)

Step-by-Step Solution

Answer

The area is \( \pi a^2 \).

Step 1: Understand the Parametrization

Given the circle parametrization \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \) with \( 0 \leq t \leq 2 \pi \), identify that \( x = a \cos t \) and \( y = a \sin t \).

Step 2: Determine Derivatives

Find the derivatives \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = a \cos t \).

Step 3: Setup Green's Theorem Area Formula

Using the formula \[ Area \, of \ R = \frac{1}{2} \oint_C (x \, dy - y \, dx) \], substitute \( x, y, dx, dy \) using the parametrization: \( x \, dy = (a \cos t)(a \cos t \, dt) = a^2 \cos^2 t \, dt \) and \( y \, dx = (a \sin t)(-a \sin t \, dt) = -a^2 \sin^2 t \, dt \).

Step 4: Substitute in Area Formula

The area is given by \[ Area = \frac{1}{2} \int_{0}^{2\pi} (a^2 \cos^2 t - (-a^2 \sin^2 t)) \, dt = \frac{1}{2} \int_{0}^{2\pi} a^2 (\cos^2 t + \sin^2 t) \, dt \].

Step 5: Simplify Trigonometric Expression

Utilize the identity \( \cos^2 t + \sin^2 t = 1 \): \[ Area = \frac{1}{2} \int_{0}^{2\pi} a^2 \, dt = \frac{a^2}{2} [t]_{0}^{2\pi} \].

Step 6: Evaluate the Integral

Calculate the integral: \( Area = \frac{a^2}{2} (2\pi - 0) = \pi a^2 \).

Key Concepts

area calculation

Calculating the area of a region, particularly one enclosed by a curve like a circle, involves using mathematical theorems and formulas. In this exercise, we're exploring how Green's Theorem provides a powerful method to find areas. For the circle parametrized by \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \), Green's Theorem offers an elegant solution. The theorem relates a line integral around a simple closed curve to a double integral over the region it encloses. By setting up the area formula, you integrate along the boundary of the region.
This exercise uses the formula \[ Area = \frac{1}{2} \oint_C (x \, dy - y \, dx) \]which simplifies the calculation for regions like circles. This approach is not just about finding a number but understanding the relationship between the geometry and calculus involved.

parametrization

Parametrization involves expressing a geometric object such as a curve in terms of parameters, usually denoted by \( t \). This concept simplifies calculations, especially in a scenario where the curve is complex. In this exercise, the circle is parametrized by \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \).
This gives us:

• \( x = a \cos t \)
• \( y = a \sin t \)

Understanding this parametrization allows us to calculate derivatives, which are crucial in finding the necessary elements for Green's Theorem. This technique translates geometric curves into algebraic expressions, making integration manageable.

integrals

Integrals are fundamental in calculus, serving as a tool to calculate areas, volumes, and other quantities. In the context of Green's Theorem, you perform a line integral to find an area. Here, the focus is on setting up an integral related to the circle's boundary:

• Define the function components by separating \( x \, dy \) and \( y \, dx \)
• Substitute the derivatives: \( x \, dy = a^2 \cos^2 t \cdot dt \) and \( y \, dx = -a^2 \sin^2 t \cdot dt \)
• Set up the integral expression: \[ \int_{0}^{2\pi} (a^2 \cos^2 t - (-a^2 \sin^2 t)) \, dt \]

Calculating this integral provides the area enclosed by the circle. It's essential to understand how to manipulate and evaluate integrals to effectively solve problems, like this, involving area.

trigonometric identities

Trigonometric identities simplify complex expressions, making them easier to integrate or differentiate. In this exercise, they play a pivotal role in resolving the equation derived from Green's Theorem. Specifically, the identity used is:

• \( \cos^2 t + \sin^2 t = 1 \)

This identity allows the simplification of the integral expression by combining \( a^2 \cos^2 t \) and \( -a^2 \sin^2 t \) into a single term:\[ Area = \frac{1}{2} \int_{0}^{2\pi} a^2 \, dt \].
By using this identity, we can easily integrate the expression to find the circle's area \( \pi a^2 \). Understanding and effectively applying trigonometric identities is crucial in simplifying mathematical problems.