Problem 46 - 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. $$\begin{array}{l}{\text { a. Plot } C \text { in the } x y \text { -plane. }} \\\ {\text { b. Determine the integrand }(\partial N / \partial x)-(\partial M / \partial y) \text { for the tangen- }} \\ {\text { tial form of Green's Theorem. }} \\ {\text { c. Determine the (double integral) limits of integration from }} \\ {\text { your plot in part (a) and evaluate the curl integral for the }} \\ {\text { circulation. }}\end{array}$$ $$\begin{array}{l}{\mathbf{F}=x e^{y} \mathbf{i}+\left(4 x^{2} \ln y\right) \mathbf{j}} \\ {C : \text { The triangle with vertices }(0,0),(2,0), \text { and }(0,4)}\end{array}$$

Step-by-Step Solution

Answer

The circulation is determined by the double integral of \( 8x \ln y - x e^y \) over the triangle.

Step 1: Plot the Curve C in the xy-plane

The curve \( C \) is a triangle with vertices \((0, 0)\), \((2, 0)\), and \((0, 4)\). Plot these points on the xy-plane and draw lines to connect these vertices, forming a triangle. Make sure the triangle is oriented counterclockwise from \((0, 0)\) to \((2, 0)\), to \((0, 4)\), and back to \((0, 0)\).

Step 2: Compute the Integrand for Green's Theorem

For the vector field \( \mathbf{F} = x e^y \mathbf{i} + 4x^2 \ln y \mathbf{j} \), we identify \( M = x e^y \) and \( N = 4x^2 \ln y \). Using Green's Theorem, the integrand is given by \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \). Solve:- \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(4x^2 \ln y) = 8x \ln y \)- \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x e^y) = x e^y \)Thus, the integrand is \( 8x \ln y - x e^y \).

Step 3: Determine the Limits of Integration and Evaluate the Double Integral

From Step 1, the triangle \( C \) is bounded by the lines \( y = 0 \) to \( y = 4 - 2x \). Thus, for any horizontal slice, \( x \) ranges from \( 0 \) to \( 2 - y/2 \) when \( y \) varies from 0 to 4. The double integral is:\[ \int_{0}^{4} \int_{0}^{2 - y/2} (8x \ln y - x e^y) \, dx \, dy \]First, compute \( \int_{0}^{2 - y/2} (8x \ln y - x e^y) \, dx \) for each \( y \). This requires integrating with respect to \( x \) while treating \( y \) as a constant:- \( \int 8x \ln y \, dx = 4x^2 \ln y |_{0}^{2 - y/2} \)- \( \int x e^y \, dx = \frac{x^2}{2} e^y |_{0}^{2 - y/2} \)Substitute and simplify the expression, then integrate with respect to \( y \). Substitute back the limits to evaluate the entire double integral.

Key Concepts

Counterclockwise Circulation

The concept of counterclockwise circulation in the context of Green's Theorem is important to understand because it determines how we evaluate the path integral of a vector field around a closed curve. Circulation measures how much of the vector field's flow wants to "rotate" around the curve. For Green's Theorem, we specifically consider curves that are oriented in a counterclockwise direction. This is because the theorem relates the line integral around a closed curve to a double integral over the region it encloses, and the standard is to consider the curve in a counterclockwise manner.

In our exercise, the curve \( C \) is a triangle defined by the vertices \((0, 0)\), \((2, 0)\), and \((0, 4)\). When plotting these points and connecting them, ensure that you trace the triangle counterclockwise: start from \((0, 0)\) to \((2, 0)\), to \((0, 4)\), and finally back to \((0, 0)\). This direction maintains the proper orientation needed for applying Green's Theorem.

Vector Field

A vector field assigns a vector to every point in a region of space. In our exercise, the vector field \( \mathbf{F} \) is given by \( \mathbf{F} = x e^y \mathbf{i} + 4x^2 \ln y \mathbf{j} \). Here, \( \mathbf{i} \) and \( \mathbf{j} \) denote the unit vectors in the horizontal and vertical directions, respectively.

A crucial aspect of this vector field is identifying the components \( M \) and \( N \) for the application of Green's Theorem. In this vector field:

• \( M = x e^y \) corresponds to the \( \mathbf{i} \) component
• \( N = 4x^2 \ln y \) corresponds to the \( \mathbf{j} \) component

Using these components, we can find the partial derivatives necessary for the integrand in Green's Theorem. Calculate \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \) first, to form the expression \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \), which simplifies the integrand used within the double integral.

Double Integral

Double integrals are used to calculate the volume under a surface over a certain region and are applied here to find the counterclockwise circulation via Green's Theorem. Essentially, you are taking an area integral of the curl of the vector field over the region enclosed by the curve \( C \).

For our exercise, we need to compute the double integral of the integrand \( 8x \ln y - x e^y \) over the triangular region enclosed by \( C \). This is mathematically expressed as:

\[\int_{0}^{4} \int_{0}^{2 - y/2} (8x \ln y - x e^y) \, dx \, dy\]
First, integrate with respect to \( x \), considering that \( y \) acts as a constant. This requires calculating \(\int 8x \ln y \, dx\) and \(\int x e^y \, dx\). Simplify these integrals and then substitute the limits of integration for \( x \) from \( 0 \) to \( 2 - y/2 \). This intermediate result is then integrated with respect to \( y \) over the range specified by \( 0 \) to \( 4 \) to find the result of the circulation.

Limits of Integration

Understanding the limits of integration is crucial when solving double integrals, particularly in the context of Green's Theorem. The limits define the region over which the integration is performed, ensuring we capture the entire area enclosed by the curve \( C \).

In this problem, the triangular region \( C \) has boundaries that affect both the \( x \) and \( y \) limits. The triangle extends vertically from \( y = 0 \) to \( y = 4 \). For a given \( y \), the \( x \) values range from 0 up to \( 2 - y/2 \), based on the equations of the triangle's sides.
These relationships translate into the integral's limits:

• The outer integral for \( y \) spans from \( 0 \) to \( 4 \).
• The inner integral for \( x \) varies from \( 0 \) to \( 2 - y/2 \), reflecting the triangle's width across horizontal slices.

This setup ensures the accurate numerical evaluation of the double integral, capturing the entire region required for calculating the circulation of the vector field around \( C \).