Problem 51

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. \(\mathbf{F}=x^{-1} e^{y} \mathbf{i}+\left(e^{y} \ln x+2 x\right) \mathbf{j},\) \(C :\) The boundary of the region defined by \(y=1+x^{4}(\) below \()\) and \(y=2\) (above)

Step-by-Step Solution

Verified
Answer
The circulation of the field \( \mathbf{F} \) around \( C \) is \( \frac{16}{5} \).
1Step 1: Plot the Curve C
First, let's plot the curve \( C \) in the \(xy\)-plane as defined. The region is bounded by \( y = 1 + x^4 \) below and \( y = 2 \) above. This would mean plotting these two curves and the resulting region between these limits will be our area of interest. By using a CAS tool, such as Desmos, Python's matplotlib, or GeoGebra, plot the region in the xy-plane where these conditions hold.
2Step 2: Find Partial Derivatives of N and M
We are given the vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \) with \( M = x^{-1} e^y \) and \( N = e^y \ln x + 2x \). Use CAS to compute the partial derivatives needed for Green's Theorem:\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left( e^y \ln x + 2x \right) = \frac{e^y}{x} + 2 \]\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} \left( x^{-1} e^y \right) = x^{-1} e^y \]Now calculate the integrand for Green's Theorem, which is:\[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = \frac{e^y}{x} + 2 - x^{-1} e^y = 2 \]
3Step 3: Set Up and Evaluate Double Integral
According to Green's Theorem, the circulation is equal to the double integral over the region \( R \) of \( \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \). From Step 2, we know the integrand is a constant \(2\). The limits of integration are determined from the region bounded by \( y = 1 + x^4 \) and \( y = 2 \). Therefore, \( y \) ranges from \(1 + x^4\) to \(2\) and \(x\) should be bounded by the values where these functions intersect (solve \(1 + x^4 = 2\) to find these bounds):\[ \int_{-1}^{1} \int_{1 + x^4}^{2} 2 \, dy \, dx \]Calculate the integral:\[ = \int_{-1}^{1} \left[ 2y \right]_{1 + x^4}^{2} \, dx \]\[ = \int_{-1}^{1} \left(4 - 2(1 + x^4)\right) \, dx \]\[ = \int_{-1}^{1} (2 - 2x^4) \, dx \]\[ = \left[ 2x - 2 \frac{x^5}{5} \right]_{-1}^{1} \]\[ = \left( 2 \cdot 1 - \frac{2}{5}(1)^5 \right) - \left( 2 \cdot (-1) - \frac{2}{5}(-1)^5 \right) \]\[ = (2 - \frac{2}{5}) - (-2 + \frac{2}{5}) = \frac{16}{5} \]
4Step 4: Conclusion
The counterclockwise circulation of the field \( \mathbf{F} \) around the curve \( C \) using Green's Theorem is equal to the evaluated double integral, \( \frac{16}{5} \).

Key Concepts

Vector FieldCounterclockwise CirculationPartial DerivativesDouble Integral
Vector Field
A vector field is essentially a function that assigns a vector to each point in a space. Imagine a landscape where each point has a little arrow pointing in a certain direction and with a particular strength. In math and physics, vector fields are represented using unit vectors like \( \mathbf{i} \) and \( \mathbf{j} \) to indicate directions of the x and y axes, respectively.
For this problem, our vector field is \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \), where \( M = x^{-1} e^y \) and \( N = e^y \ln x + 2x \). This defines a field with components of flow along the x and y directions. The task is to analyze the behavior of these components around a given closed curve in the xy-plane. Understanding vector fields is crucial as it helps us calculate important properties like circulation and flux using the tools of vector calculus.
Counterclockwise Circulation
Circulation is a measure of how much a vector field "circulates" around a closed curve. In practical terms, it helps us understand the net 'twisting strength' of a field over a particular path.
This is akin to measuring how the water swirls around the edge of a drain. When we state 'counterclockwise circulation,' we are specifically looking at this movement going around in the opposite direction to a clock's hands. In the exercise, Green's Theorem helps us compute this by converting line integrals into double integrals over a region enclosed by the curve \( C \).
Considering counterclockwise circulation aligns with the standard positive orientation in vector calculus, making it a significant concept for evaluating properties like torque and vorticity in fields of thermodynamics and fluid dynamics.
Partial Derivatives
Partial derivatives are essential tools in multivariable calculus, allowing us to examine how a function changes as one of its variables changes, while keeping others constant.
In our vector field \( \mathbf{F} \), we find the partial derivatives of the components \( N \) and \( M \):
  • \( \frac{\partial N}{\partial x} = \frac{e^y}{x} + 2 \)
  • \( \frac{\partial M}{\partial y} = x^{-1} e^y \)
These derivatives tell us how the vector field's components vary in the X and Y directions.
The difference \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \) becomes the integrand in Green's Theorem to find circulation. Understanding partial derivatives is vital for comprehensively analyzing how complex functions behave in multiple dimensions.
Double Integral
A double integral extends the concept of an integral to functions of two variables, representing volume under a surface in three-dimensional space.
Green's Theorem uses the double integral to convert a line integral over a closed curve into an easier calculation over the region it encloses. For our problem, the evaluated region is bounded by \( y = 1 + x^4 \) below and \( y = 2 \) above:
  • The integral is \( \int_{-1}^{1} \int_{1 + x^4}^{2} 2 \, dy \, dx \)
  • Solving the integral provides the circulation value \( \frac{16}{5} \)
This approach simplifies understanding complex curves and regions, explaining phenomena like circulation in terms that are easier to visualize and compute.