Problem 44
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. $$\begin{array}{l}{\text { a. Plot } C \text { in the } x y \text { -plane. }} \\\ {\text { b. Determine the integrand }(\partial N / \partial x)-(\partial M / \partial y) \text { for the tangen- }} \\ {\text { tial form of Green's Theorem. }} \\ {\text { c. Determine the (double integral) limits of integration from }} \\ {\text { your plot in part (a) and evaluate the curl integral for the }} \\ {\text { circulation. }}\end{array}$$ $$\mathbf{F}=\left(2 x^{3}-y^{3}\right) \mathbf{i}+\left(x^{3}+y^{3}\right) \mathbf{j}, \quad C : \text { The ellipse } \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$$
Step-by-Step Solution
Verified Answer
The counterclockwise circulation of the field \( F \) is \( 84\pi \).
1Step 1: Plot the Curve C
The curve \( C \) is given as an ellipse described by the equation \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). To plot this in the \( xy \)-plane, recognize that this represents an ellipse centered at the origin with semi-major axis 3 on the \( y \)-axis and semi-minor axis 2 on the \( x \)-axis.
2Step 2: Determine the Integrand for Green's Theorem
Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the region \( R \) enclosed by \( C \). The integrand for the circulation form of Green's Theorem is \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \) where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \). From \( \mathbf{F} \), identify \( M = 2x^3 - y^3 \) and \( N = x^3 + y^3 \). Thus, \( \frac{\partial N}{\partial x} = 3x^2 \) and \( \frac{\partial M}{\partial y} = -3y^2 \). Therefore, the integrand becomes \( 3x^2 + 3y^2 \).
3Step 3: Determine the Integration Limits and Evaluate the Double Integral
The region \( R \) is the interior of the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). Change to elliptical coordinates: \( x = 2u \) and \( y = 3v \), transforming the ellipse equation to \( u^2 + v^2 = 1 \) with limits \( u^2 + v^2 \leq 1 \). The Jacobian of the transformation is \( 6 \). Thus, we evaluate the integral \[ \int_{0}^{2\pi}\int_{0}^{1} 3(4u^2 + 9v^2) \cdot 6 \cdot r \, dr \, d\theta \] using polar coordinates, to obtain the circulation. Solving this integral provides the result \( 84\pi \).
Key Concepts
Line IntegralsDouble IntegralsElliptical CoordinatesVector Fields
Line Integrals
Line integrals are a type of integral where the function to be integrated is evaluated along a curve, or path, in a coordinate space. This concept is especially useful in the context of vector fields where you want to measure quantities like work done by a force along a path.
In our exercise, we use a line integral to find the circulation of the vector field \( \mathbf{F} \) around the curve \( C \). Circulation measures the tendency for the vector field to "rotate" around a path. Green's Theorem allows us to transform this line integral around the closed curve \( C \) into a double integral over the region enclosed by \( C \).
Understanding line integrals is crucial in physics and engineering, particularly when dealing with problems involving conservative forces or fluid flow.
In our exercise, we use a line integral to find the circulation of the vector field \( \mathbf{F} \) around the curve \( C \). Circulation measures the tendency for the vector field to "rotate" around a path. Green's Theorem allows us to transform this line integral around the closed curve \( C \) into a double integral over the region enclosed by \( C \).
Understanding line integrals is crucial in physics and engineering, particularly when dealing with problems involving conservative forces or fluid flow.
Double Integrals
Double integrals extend the concept of integrating a single function over an interval to integrating over a two-dimensional region. This is like summing up the value of a function over an entire area, rather than just along a line.
With Green's Theorem, a line integral over a closed curve expressing circulation can be converted into a double integral over the plane region that the curve encloses. In this problem, the conversion from line to double integral lets us simplify calculations and provides insights into the vector field's behavior over a larger area.
With Green's Theorem, a line integral over a closed curve expressing circulation can be converted into a double integral over the plane region that the curve encloses. In this problem, the conversion from line to double integral lets us simplify calculations and provides insights into the vector field's behavior over a larger area.
- We compute the double integral for the region inside the ellipse transformed to a circle using elliptical coordinates to handle symmetry and calculation simplicity.
Elliptical Coordinates
Elliptical coordinates can be used to make calculations easier when dealing with ellipses, just like how polar coordinates simplify circles. In the given exercise, the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) is manipulated into a circular form.
By setting \( x = 2u \) and \( y = 3v \), you transform the ellipse into \( u^2 + v^2 = 1 \), which is much simpler to handle mathematically. This transformation allows us to easily set up the limits of integration and simplifies the calculation of the double integral.
By setting \( x = 2u \) and \( y = 3v \), you transform the ellipse into \( u^2 + v^2 = 1 \), which is much simpler to handle mathematically. This transformation allows us to easily set up the limits of integration and simplifies the calculation of the double integral.
- This transformation reduces the complexity of evaluating the double integral by leveraging the symmetry of the unit circle.
- The Jacobian of this coordinate transformation is used to adjust the scale of integration properly.
Vector Fields
A vector field assigns a vector to every point in a subset of space. Vector fields are used to model a variety of physical phenomena, such as fluid flow, electrostatic fields, or the magnetic fields around a magnet.
In this exercise, the vector field \( \mathbf{F} = (2x^3 - y^3) \mathbf{i} + (x^3 + y^3) \mathbf{j} \) is given, and we analyze it to compute its circulation along an ellipse. By breaking it down, we identify the components \( M = 2x^3 - y^3 \) and \( N = x^3 + y^3 \) to apply Green's Theorem.
In this exercise, the vector field \( \mathbf{F} = (2x^3 - y^3) \mathbf{i} + (x^3 + y^3) \mathbf{j} \) is given, and we analyze it to compute its circulation along an ellipse. By breaking it down, we identify the components \( M = 2x^3 - y^3 \) and \( N = x^3 + y^3 \) to apply Green's Theorem.
- Vector fields are useful for understanding many natural and engineered systems where direction and magnitude vary over a space.
- Analyzing the behavior of these fields can reveal insights about the system's dynamics, such as rotation and divergence.
Other exercises in this chapter
Problem 43
In Exercises \(43-46,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array}{l}{\text { a. Find } d s=|\mathbf{v}(t)| d t \
View solution Problem 44
Find the area of the upper portion of the cylinder \(x^{2}+z^{2}=1\) that lies between the planes \(x=\pm 1 / 2\) and \(y=\pm 1 / 2.\)
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Two "central" fields Find a field \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) in the \(x y\) -plane with the property that at each point \((x, y) \neq(
View solution Problem 44
In Exercises \(43-46,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array}{l}{\text { a. Find } d s=|\mathbf{v}(t)| d t \
View solution