Problem 49 - Thomas Calculus

Question

Use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(\mathbf{F}\) around the simple closed curve \(C\). Perform the following CAS steps. a. Plot \(C\) in the \(x y\)-plane. b. Determine the integrand \((\partial N / \partial x)-(\partial M / \partial y)\) for the tangential form of Green's Theorem. c. Determine the (double integral) limits of integration from your plot in part (a) and evaluate the curl integral for the circulation. \(\mathbf{F}=(2 x-y) \mathbf{i}+(x+3 y) \mathbf{j}, \quad C :\) The ellipse \(x^{2}+4 y^{2}=4\)

Step-by-Step Solution

Answer

The counterclockwise circulation of \(\mathbf{F}\) around \(C\) is \(8\pi\).

Step 1: Plot the Curve C

Plot the ellipse given by the equation \(x^{2} + 4y^{2} = 4\) in the \(xy\)-plane. This ellipse is centered at the origin with semi-major axis 2 along the \(x\)-axis and semi-minor axis 1 along the \(y\)-axis. This is because \(x/2\) and \(y/1\) are the terms in the standard form of an ellipse.

Step 2: Calculate Partial Derivatives for Green's Theorem

The field \(\mathbf{F} = (2x - y)\mathbf{i} + (x + 3y)\mathbf{j}\) can be expressed as \(M = 2x - y\) and \(N = x + 3y\). Compute \(\partial N/\partial x = 1\) and \(\partial M/\partial y = -1\). The integrand for Green's Theorem is \((\partial N / \partial x) - (\partial M / \partial y) = 1 - (-1) = 2\).

Step 3: Set Integral Limits and Evaluate Double Integral

The region inside the ellipse \(x^{2} + 4y^{2} = 4\) is described in terms of \(x\) and \(y\) limits. In polar coordinates with substitution \(x = 2\cos(\theta)\) and \(y = \sin(\theta)\), \(dA = rdrd\theta\) and \(r\) is between 0 to 2. \(\theta\) ranges from 0 to \(2\pi\). Thus, the double integral \(\iint_{R} 2 \, dA\) becomes \(\iint_{R} 2 \, rdrd\theta\). Evaluate this as \(\int_{0}^{2\pi} \int_{0}^{2} 2r \, dr \, d\theta = 2 \cdot (4\pi) = 8\pi\).

Key Concepts

Vector Field

A vector field is like a collection of arrows in the space, where each arrow represents both a direction and a magnitude. For example, in this exercise, we are given the vector field \( \mathbf{F} = (2x - y)\mathbf{i} + (x + 3y)\mathbf{j} \). This means that at every point \((x, y)\) in the plane, there is an associated vector \((2x-y, x+3y)\).

Understanding vector fields is important because they are used to model various physical situations like wind flows, magnetic fields, and fluid dynamics.

With vector fields, we can use calculus to explore these multidimensional functions that give us more than just a single output; they give us directionality and strength at every point.

Ellipse

An ellipse is a closed, symmetric curve that appears like a stretched-out circle. In the exercise, we consider the ellipse given by the equation \( x^{2} + 4y^{2} = 4 \).

This form tells us that the maximum extent along the \( x \)-axis is 2 (semi-major axis), and along the \( y \)-axis is 1 (semi-minor axis).

The center of this ellipse is at the origin \((0,0)\). It's crucial to recognize and plot ellipses accurately when solving problems involving areas, such as those addressed by Green's Theorem.

In practical applications, ellipses are significant in fields ranging from astronomy, where planetary orbits are often elliptical, to engineering, where they describe cross-sections of machined parts.

Partial Derivatives

Partial derivatives are essential in understanding how a multivariable function changes as we vary one of its variables while keeping the others constant. In our problem, we work with the vector field \( \mathbf{F} \) split into two functions: \( M(x, y) = 2x - y \) and \( N(x, y) = x + 3y \).

Calculating partial derivatives involves the following:

• \( \frac{\partial N}{\partial x} = 1 \)
• \( \frac{\partial M}{\partial y} = -1 \)

These computations are critical in forming the integrand \( (\partial N/\partial x) - (\partial M/\partial y) = 2 \) for Green's Theorem.

Such operations allow us to simplify otherwise complex multivariable problems and are widely used in various fields such as economics, where they model changes in output or cost with respect to one factor.

Double Integral

A double integral extends the concept of single-variable integration to functions of two variables. It allows us to compute the area, volume, or other properties over a two-dimensional region.

In this exercise, the region of interest is the interior of an ellipse \( x^{2} + 4y^{2} = 4 \). By transforming the traditional \((x, y)\) coordinates into polar coordinates, the double integral becomes easier to manage, especially when symmetry is present.

The double integral \( \iint_{R} 2 \, dA \) turns into \( \iint_{R} 2r \, dr \, d\theta \) in polar form, where:

• \( r \) ranges from 0 to 2
• \( \theta \) from 0 to \( 2\pi \)

This integration yields \( 8\pi \), showing the circulation using Green's Theorem. Double integrals are powerful tools for solving more complex problems in physics and engineering where three-dimensional aspects are involved.

Polar Coordinates

Polar coordinates offer a way to describe a point in the plane using a distance and an angle relative to a reference direction, unlike Cartesian coordinates which use \(x\) and \(y\) coordinates.

In this exercise, the use of polar coordinates is valuable in simplifying the integration within an ellipse. We convert \((x, y)\) to \((r, \theta)\) using:

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

This transformation allows the boundaries of the ellipse to translate into simple radial and angular limits. The integration, expressed as \( rdrd\theta \), leverages this simplicity to evaluate the integral efficiently.

Polar coordinates are particularly useful in scenarios with rotational symmetry, such as in physics when dealing with problems involving circular motions or waves.