Problem 25 - Thomas Calculus

Question

The reason is that by Equation \((4),\) run backward, $$\begin{aligned} \text { Area of } R &=\iint_{R} d y d x=\iint_{K}\left(\frac{1}{2}+\frac{1}{2}\right) d y d x \\ &=\oint_{C} \frac{1}{2} x d y-\frac{1}{2} y d x \end{aligned}$$ Use the Green's Theorem area formula given above to find the areas bof the regions enclosed by the curves in Exercises \(25-28 .\) $$\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi$$

Step-by-Step Solution

Answer

The area of the circle is \( \pi a^2 \).

Step 1: Identify the Region and Curve

The problem involves finding the area enclosed by the parametric curve \( \mathbf{r}(t)=(a \cos t) \, \mathbf{i} + (a \sin t) \, \mathbf{j} \) for \( 0 \leq t \leq 2\pi \). This represents a circle with radius \( a \) centered at the origin.

Step 2: Understand Green's Theorem Application

Green's Theorem relates the line integral around a simple, closed curve \( C \) to a double integral over the plane region \( R \) bounded by \( C \). The formula provided can be simplified to yield the area \( A \) as follows: \( A = \oint_{C} \frac{1}{2} x \, dy - \frac{1}{2} y \, dx \).

Step 3: Compute the Line Integrals

Substitute the parameterization into the expression: \( x = a \cos t \) and \( y = a \sin t \). Compute \( dx = -a \sin t \, dt \) and \( dy = a \cos t \, dt \). Substitute these into Green's Theorem formula: \[ A = \oint_{0}^{2\pi} \left( \frac{1}{2} a \cos t \cdot a \cos t \, dt - \frac{1}{2} a \sin t \cdot (-a \sin t) \, dt \right) \].

Step 4: Simplify the Integrals

Simplify the resulting expression: \[ A = \frac{a^2}{2} \int_{0}^{2\pi} (\cos^2 t + \sin^2 t) \, dt \]. Since \( \cos^2 t + \sin^2 t = 1 \), the equation becomes \( A = \frac{a^2}{2} \int_{0}^{2\pi} 1 \, dt = \frac{a^2}{2} [t]_{0}^{2\pi}\).

Step 5: Evaluate the Integral

Evaluate the integral: \( A = \frac{a^2}{2} \times (2\pi - 0) = \frac{a^2}{2} \times 2\pi = \pi a^2 \). This result confirms that the area of the region enclosed by the curve is \( \pi a^2 \), which is consistent with the area of a circle of radius \( a \).

Key Concepts

Parametric Equations

A parametric equation defines a group of related quantities as functions of an independent variable, often "time". This approach is essential in describing specific paths or curves in space. In the context of the exercise, the parametric equation given is \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \), where \( t \) is the parameter that varies from 0 to \( 2\pi \). This particular equation describes a circle centered at the origin with a radius \( a \).

The reason parametric equations are useful lies in their ability to represent curves more conveniently, without the direct need for a single function form \( y = f(x) \). Instead, both \( x \) and \( y \) are specified independently through another variable \( t \).

In many mathematical problems, such as finding areas or solving the motions of objects, parametric equations provide a flexible and powerful tool. They allow for the description of more complex curves, as they are not subjected to the vertical line test that specific function-based forms must adhere to.

Line Integrals

A line integral generalizes the concept of integration to functions that are integrated along a curve. Imagine summing up values not along straight distances, but over a path. In Green's Theorem, the line integral is crucial because it connects the area calculation to boundary traversal.

The line integral around a closed curve \( C \) is represented by \( \oint_{C} \). In our problem, evaluating the line integral involves integrating around the circumference of a circle represented by parametric equations.

What distinguishes line integrals from regular integrals is their ability to handle vector fields. For example, the integral \( \oint_{C} \frac{1}{2}x \, dy - \frac{1}{2}y \, dx \) doesn't sum values over a straight line but along a circle's edge in the plane. By applying these techniques, areas enclosed by curves can be found in an elegant manner, often employing vector calculus concepts for seamless calculations.

Double Integrals

Double integrals are used to integrate functions over two-dimensional regions. This allows for the calculation of volume under a surface, or, as in our case, areas on a plane.

In this problem, Green's Theorem helps transform a line integral into a double integral over the region \( R \) bounded by the curve \( C \). It effectively links the boundary behavior of a function to its behavior over the entire region. Originally, the exercise shows that by applying Green’s Theorem, the double integrals over the region \( R \) yield the same result as the line integral around its boundary.

This transformation *streamlines the calculation of areas in many cases,*brings versatility, especially when dealing with complex curves,*links the local properties (within \( R \)) to global properties (along \( C \)).
Double integrals also introduce varieties of coordinate systems like polar (which coincidentally matches the nature of this circular problem), simplifying calculations further.

Circle Area Calculation

Calculating the area of a circle is a classic problem in geometry, formulaically given by \( \pi a^2 \), where \( a \) is the radius. This formula fundamentally arises from integrating over the circle's plane region.

Using Green’s Theorem offers an alternative perspective to arrive at this conclusion. By converting the line integral around a circle into a double integral over the circle's interior, we find the area using advanced calculus techniques rather than relying solely on memorized geometric formulas.
The exercise illustrates this process:

• Begin by setting up parametric equations for a circle.
• Apply the Green's Theorem to transform the problem into line and double integrals.
• Simplify these integrals.
• Evaluate to derive \( \pi a^2 \), reaffirming the circle's area formula.

This approach not only proves consistent with the standard area formula but also provides insight into how calculus can solve geometry problems.