Problem 50 - 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. \(\mathbf{F}=\left(2 x^{3}-y^{3}\right) \mathbf{i}+\left(x^{3}+y^{3}\right) \mathbf{j}, \quad C :\) The ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)

Step-by-Step Solution

Answer

The counterclockwise circulation is \(6\pi\).

Step 1: Plot the Ellipse

The given curve \( C \) is an ellipse described by the equation \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). This ellipse is centered at the origin with a semi-major axis of 3 along the \( y \)-axis and a semi-minor axis of 2 along the \( x \)-axis. Use a CAS tool to plot this ellipse in the \( xy \)-plane. This visualization will help identify the region \( R \) over which we'll apply Green's Theorem.

Step 2: Determine Partial Derivatives for Green's Theorem

Green's Theorem for circulation in tangential form states: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int \int_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \] where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \) and here \( M = 2x^3 - y^3 \) and \( N = x^3 + y^3 \). Compute the partial derivatives: \[ \frac{\partial N}{\partial x} = 3x^2, \quad \frac{\partial M}{\partial y} = -3y^2 \]. Thus, the integrand is \[ (3x^2 - (-3y^2)) = 3x^2 + 3y^2 \].

Step 3: Setup Double Integral and Determine Limits

The region \( R \) in the \( xy \)-plane is bounded by the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). For integration, set it up in either rectangular or polar form. Converting to polar: \( x = 2\cos\theta \) and \( y = 3\sin\theta \) for \( 0 \leq \theta < 2\pi \). Substitute these into \( 3x^2 + 3y^2 \), simplifying and multiplying by the Jacobian \( r \): \[ \int_0^{2\pi} \int_0^1 (12r^2(\cos^2\theta + \sin^2\theta)) \, r \, dr \, d\theta \].

Step 4: Evaluate the Double Integral

Simplify the integrand: \( 12r^2(\cos^2\theta + \sin^2\theta) = 12r^2 \). Evaluate: \[ \int_0^{2\pi} \int_0^1 12r^3 \, dr \, d\theta \]. Compute the inner integral: \( \int_0^1 12r^3 \, dr = [3r^4]_0^1 = 3 \). Compute the outer integral: \[ \int_0^{2\pi} 3 \, d\theta = 3\theta \bigg|_0^{2\pi} = 6\pi \].

Key Concepts

Circulation

The concept of circulation in vector calculus is essential when dealing with fields like physics or engineering. Circulation represents the total "flow" of a vector field around a closed curve. It's like measuring how much a river flows around the banks. This is typically considered in the context of Green's Theorem, which allows us to calculate circulation as a double integral over a region. The specific equation used is: \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), where \( C \) is a closed curve and \( \mathbf{F} \) is the vector field. In the exercise, the curve \( C \) is an ellipse. We aim to understand how the vector field circulates around this specific shape.

• Use Green's Theorem to transform line integrals into simpler area-based calculations.
• Circulation is crucial in fields like fluid dynamics and electromagnetism.

A thorough grasp of these interactions helps in analyzing real-life flow situations, providing clarity and precision in problem-solving.

Partial Derivatives

Partial derivatives are a cornerstone of calculus that help us understand how a function changes as its input variables change. In a function with multiple variables, a partial derivative is the derivative of the function with respect to one of these variables while holding the others constant. In the given exercise, we have functions of two variables, \( x \) and \( y \), and we calculate the partial derivatives \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \).
This leads to:

• \( \frac{\partial N}{\partial x} = 3x^2 \)
• \( \frac{\partial M}{\partial y} = -3y^2 \)

These derivatives form part of the integrand for the equation in Green's Theorem. Understanding partial derivatives is essential for evaluating how a vector field changes in different directions, setting the stage for further analysis of field behavior.

Double Integral

A double integral is an extension of an integral concept from calculus into two dimensions, often used to compute area or volume. It helps in evaluating the accumulation of quantities over a plane area. In the context of Green's Theorem, we use double integrals to compute the circulation of a vector field within a closed region, like the interior of the ellipse in our problem.
The double integral for the problem is set up using polar coordinates, which makes calculations over elliptical regions more manageable.
This is represented by:
\[ \int_0^{2\pi} \int_0^1 (12r^2(\cos^2\theta + \sin^2\theta)) \, r \, dr \, d\theta \]
Breaking it down, we integrate the function over radial and angular components, simplifying the calculations and helping define the behavior of vector fields over complex regions.

Ellipse

An ellipse is a geometric shape characterized by its oval shape, defined mathematically as the set of points where the sum of distances from two fixed points (foci) is constant. In Cartesian coordinates, it is expressed as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively. In this exercise, the ellipse is \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \).
This particular ellipse is horizontal, centered at the origin:

• The semi-major axis is 3 along the \( y \)-axis.
• The semi-minor axis is 2 along the \( x \)-axis.

Ellipses often feature in physics and engineering to describe orbits, antenna patterns, and other phenomena. Recognizing how they fit into problems is crucial for solving calculus-related challenges.

Vector Fields

Vector fields are a type of function where each point in space is associated with a vector. They're critical for visualizing phenomena where direction and magnitude are important, such as magnetic fields or fluid flow. In our exercise, the vector field \( \mathbf{F} = (2x^3 - y^3) \mathbf{i} + (x^3 + y^3) \mathbf{j} \) is defined in the plane.
This vector field shows:

• How vectors are distributed across a region
• How these vectors might "flow" across shapes like the given ellipse

Understanding vector fields helps predict how variables evolve over space. In practical applications, this can mean anything from optimizing airflow around aircraft wings to tuning electric circuits. By analyzing these fields, we gain insights into the continuous changes occurring within defined spaces.