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Magazine Chapter 16: Problem 21
Use the Green's Theorem area formula (Equation 13\()\) to find the areas of the regions enclosed by the curves in Exercises \(21-24\) . The circle \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)
Short Answer
Expert verified
The area is \( \pi a^2 \).
Step by step solution
01
Understanding Green's Theorem
Green's Theorem relates a line integral around a simple, closed curve C and a double integral over the plane region D bounded by C. It is given by the formula:\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \). For calculating the area using Green's Theorem, we choose \( M = -y/2 \) and \( N = x/2 \), so that: \[ A = \frac{1}{2} \oint_{C} x \, dy - y \, dx \]
02
Parameterizing the Curve
The curve given is a circle with the parameterization \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \), with \( 0 \leq t \leq 2 \pi \). Here, \( x(t) = a \cos t \) and \( y(t) = a \sin t \).
03
Calculate Derivatives
Compute the derivatives \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = a \cos t \). These are needed to compute the integrals for finding the area.
04
Apply Green's Theorem
Substitute the derivatives into the line integral expression for the area: \[ A = \frac{1}{2} \int_{0}^{2\pi} \left( a \cos t \cdot a \cos t - a \sin t \cdot (-a \sin t) \right) \, dt \]
05
Simplify the Integral
Simplify to get \[ A = \frac{1}{2} \int_{0}^{2\pi} (a^2 \cos^2 t + a^2 \sin^2 t) \, dt = \frac{1}{2} \int_{0}^{2\pi} a^2 (\cos^2 t + \sin^2 t) \, dt \]. Recall the Pythagorean identity \( \cos^2 t + \sin^2 t = 1 \), so \[ A = \frac{1}{2} \int_{0}^{2\pi} a^2 \, dt \].
06
Evaluate the Integral
Evaluate \[ A = \frac{1}{2} \, a^2 \left[ t \right]_{0}^{2\pi} = \frac{1}{2} a^2 (2\pi - 0) = \pi a^2 \].
07
Conclusion: Area of Circle
The area enclosed by the circle is \( \pi a^2 \), which is consistent with the known formula for the area of a circle.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integral
A line integral is a type of integral where a function is evaluated along a curve. This is particularly useful in physics and engineering, as it helps to calculate work done by a force field or the circulation of a field around a curve. When we perform a line integral, we're essentially "summing up" the effects of a function over a path.
In the context of Green's Theorem, a line integral is taken around a closed curve, denoted by \(C\). Specifically, we have:
In the context of Green's Theorem, a line integral is taken around a closed curve, denoted by \(C\). Specifically, we have:
- \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \) represents the line integral over the curve \(C\).
- This calculates the accumulation of the vector field \(\mathbf{F}\) along the path \(C\).
Double Integral
A double integral involves integrating a function over a two-dimensional region. It's akin to finding the volume under a surface. Double integrals are represented as \(\iint_{D} f(x, y) \, dA\), where \(D\) is the region of integration.
In connection with Green's Theorem, double integrals are used to relate a function's behavior over a surface to its behavior along the boundary of that surface. For instance:
In connection with Green's Theorem, double integrals are used to relate a function's behavior over a surface to its behavior along the boundary of that surface. For instance:
- \( \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \) illustrates the relationship between the vector field inside the region \(D\) and its boundary \(C\).
- Through this, Green's Theorem equates a line integral around the boundary to a double integral over the region it encloses.
Circle Area Calculation
Calculating the area of a circle is a classical mathematical problem. The formula \( \pi a^2 \) provides the area, where \(a\) is the radius of the circle. Green's Theorem offers a creative method to reconfirm this.
In this exercise, Green's Theorem is utilized by selecting suitable functions \(M\) and \(N\) to simplify calculations:
In this exercise, Green's Theorem is utilized by selecting suitable functions \(M\) and \(N\) to simplify calculations:
- Assign \(M = -\frac{y}{2}\) and \(N = \frac{x}{2}\) for easing integration.
- The theorem rewrites the circle area as a line integral, resulting in \( A = \frac{1}{2} \oint_{C} (x \, dy - y \, dx) \).
Parameterization of Curves
Parameterization involves expressing a curve in terms of a parameter, often \(t\). For circles, a common parameterization is:
In this specific case, parameterization helps break down the circle into understandable components:
- \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \).
- Here \(0 \leq t \leq 2\pi\), which covers the entire circle.
In this specific case, parameterization helps break down the circle into understandable components:
- The \(x\) and \(y\) components of the graph, \(x(t) = a \cos t\) and \(y(t) = a \sin t\), are easily manipulated.
- By computing derivatives, \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = a \cos t \), we advance the integration process.
