Problem 45 - Thomas Calculus

Question

In Exercises \(43-46,\) use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(F\) around the simple closed curve C. Perform the following CAS steps. $$\begin{array}{l}{\text { a. Plot } C \text { in the } x y \text { -plane. }} \\\ {\text { b. Determine the integrand }(\partial N / \partial x)-(\partial M / \partial y) \text { for the tangen- }} \\ {\text { tial form of Green's Theorem. }} \\ {\text { c. Determine the (double integral) limits of integration from }} \\ {\text { your plot in part (a) and evaluate the curl integral for the }} \\ {\text { circulation. }}\end{array}$$ $$\begin{array}{l}{\mathbf{F}=x^{-1} e^{y} \mathbf{i}+\left(e^{y} \ln x+2 x\right) \mathbf{j}} \\ {C : \text { The boundary of the region defined by } y=1+x^{4}(\text { below }) \text { and }} \\ {y=2 \text { (above) }}\end{array}$$

Step-by-Step Solution

Answer

The counterclockwise circulation is 3.2.

Step 1: Plot the Curve C

Using a graphing tool, plot the curve described by the boundary of the region: the curve given by \( y = 1 + x^4 \) as the lower boundary and the horizontal line \( y = 2 \) as the upper boundary. Observe the closed region bounded by these curves that will serve as our curve \( C \) in the xy-plane.

Step 2: Determine the Integrand for Green's Theorem

For Green's Theorem, we need to find \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \) for the vector field \( \mathbf{F} = x^{-1} e^y \mathbf{i} + (e^y \ln x + 2x) \mathbf{j} \). Here, \( M = x^{-1} e^y \) and \( N = e^y \ln x + 2x \). Calculate the partial derivatives: \( \frac{\partial N}{\partial x} = \frac{1}{x} e^y + 2 \) and \( \frac{\partial M}{\partial y} = x^{-1} e^y \). The integrand becomes \( \frac{1}{x} e^y + 2 - x^{-1} e^y = 2 \).

Step 3: Determine Limits of Integration and Evaluate the Integral

The region is bounded by \( y = 1 + x^4 \) and \( y = 2 \). The integral will be a double integral over this region. For each fixed \( x \), \( y \) ranges from \( 1 + x^4 \) to \( 2 \). Thus, our limits of integration are from \( y = 1 + x^4 \) to \( y = 2 \) and for \( x \), we determine by setting \( 1 + x^4 = 2 \), giving \( x = 0 \), so \( -1 \leq x \leq 1 \). Evaluate the double integral: \[ \iint_R 2 \, dy \, dx = \int_{-1}^{1} \int_{1+x^4}^{2} 2 \, dy \, dx. \] Evaluate the inner integral: \[ \int_{1+x^4}^{2} 2 \, dy = [2y]_{1+x^4}^{2} = 2(2 - (1+x^4)) = 2(1-x^4) = 2 - 2x^4. \] The integral becomes: \[ \int_{-1}^{1} (2 - 2x^4) \, dx = [2x - 2\frac{x^5}{5}]_{-1}^{1}. \] Calculate: \[ = (2(1) - 2\frac{1^5}{5}) - (2(-1) - 2\frac{(-1)^5}{5}) = (2 - \frac{2}{5}) - (-2 + \frac{2}{5}) = 3.2. \]

Key Concepts

Vector Field

A vector field is a way of assigning a vector to every point in a region of space. In simpler terms, imagine a field of arrows, where each arrow has a direction and a magnitude, pointing somewhere in space. For the vector field given in the problem, \( \mathbf{F} = x^{-1} e^y \mathbf{i} + (e^y \ln x + 2x) \mathbf{j} \), each point \((x, y)\) in the plane has a corresponding vector.
This vector field is composed of two components \(M\) and \(N\), corresponding to the \(i\) and \(j\) components, respectively. Here, \(M = x^{-1} e^y\) and \(N = e^y \ln x + 2x\).
Understanding the vector's behavior across the plane is crucial when applying Green's Theorem, which evaluates the circulation or flow around a closed curve. In this exercise, you're examining how the vectors in \( \mathbf{F} \) circulate within a specific region defined by the curve \( C \).

Double Integral

A double integral extends the concept of a single integral to functions of two variables. It is used to calculate the volume under a surface defined by a function over a region in the plane. In the context of this problem, we use a double integral to compute the integral of the function formed by applying Green's Theorem.
The formula is expressed as \( \iint_R f(x, y) \, dy \, dx \), where \(R\) represents the region over which the integration is performed. In our case, \( f(x, y) = 2 \), which was derived from the integrand.
The limits of the double integral are determined by the boundaries of the closed region defined by the curves \(y = 1 + x^4\) and \(y = 2\), together with the \(x\)-limit \(-1 \le x \le 1\). This whole integral helps evaluate the total circulation within the area.

Partial Derivatives

Partial derivatives are used to find the rate of change of a function with respect to one variable while keeping the others constant. They are fundamental in multivariable calculus. In this exercise, calculating partial derivatives is crucial for applying Green's Theorem.
Specifically, you need to determine the partial derivatives \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \), where \(N = e^y \ln x + 2x\) and \(M = x^{-1} e^y\).
- For \( \frac{\partial N}{\partial x} \), treat \(y\) as constant, giving \( \frac{1}{x} e^y + 2 \).- For \( \frac{\partial M}{\partial y} \), treat \(x\) as constant, resulting in \( x^{-1} e^y \).
These derivatives help form the integrand \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 2 \), used in the double integral to evaluate the circulation.

Curve Plotting

Curve plotting involves drawing the visual representation of functions or equations on a graph. This exercise begins by plotting the simple closed curve \(C\) in the xy-plane. Knowing the shape and area of this curve is essential because it defines the limits of integration for the problem.
For the given problem, \(C\) is delineated by the curve \(y = 1 + x^4\) (the lower boundary) and the line \(y = 2\) (the upper boundary). These two curves enclose a region which you will analyze using Green's Theorem.
By plotting these curves:

• You'll observe that \(y = 1 + x^4\) forms a wave-like curve from left to right.
• The boundary \(y = 2\) is a straightforward horizontal line.

This region lies symmetrically across the y-axis, with boundaries going from \(-1\) to \(1\) in the x-direction. An accurate plot is vital, as it assists in visualizing and setting up the correct limits for the integral calculations.