Problem 43
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. $$\begin{array}{l}{\text { a. Plot } C \text { in the } x y \text { -plane. }} \\\ {\text { b. Determine the integrand }(\partial N / \partial x)-(\partial M / \partial y) \text { for the tangen- }} \\ {\text { tial form of Green's Theorem. }} \\ {\text { c. Determine the (double integral) limits of integration from }} \\ {\text { your plot in part (a) and evaluate the curl integral for the }} \\ {\text { circulation. }}\end{array}$$ $$\mathbf{F}=(2 x-y) \mathbf{i}+(x+3 y) \mathbf{j}, \quad C : \text { The ellipse } x^{2}+4 y^{2}=4$$
Step-by-Step Solution
Verified Answer
The counterclockwise circulation is \(4\pi\).
1Step 1: Plot the Curve C in the xy-plane
Consider the equation of the ellipse given by \(x^2 + 4y^2 = 4\). To plot this curve, rewrite it in the standard form by dividing each side by 4, which gives \(\frac{x^2}{4} + \frac{y^2}{1} = 1\). This represents an ellipse centered at the origin with semi-major axis 2 along the x-axis and semi-minor axis 1 along the y-axis.
2Step 2: Compute the Integrand Using Green's Theorem
Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. The curl of \( \mathbf{F} = (2x-y) \mathbf{i} + (x+3y) \mathbf{j} \) is computed as \(\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\). Here, \(M = 2x - y\) and \(N = x + 3y\). Therefore, the partial derivative \(\frac{\partial N}{\partial x} = 1\) and \(\frac{\partial M}{\partial y} = -1\). The integrand \(\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) = 1 - (-1) = 2\).
3Step 3: Determine Integral Limits and Evaluate the Double Integral
The curve C is the boundary of the ellipse \(x^2 + 4y^2 = 4\), which describes the region D as \(\frac{x^2}{4} + \frac{y^2}{1} \leq 1\). For integration purposes, consider the transformation: \(x = 2\cos\theta, \, y = \sin\theta\), with \(dA = dx \, dy = 2\cos\theta \, d\theta \, d\phi\), where \(\theta\) is from 0 to \(2\pi\). The integral becomes \( \int_{0}^{2\pi}\int_{0}^{1}(2) \, 2 \, d\phi \, d\theta = 4\pi\). This evaluates the circulation integral.
Key Concepts
Counterclockwise CirculationDouble IntegralEllipse in Calculus
Counterclockwise Circulation
Counterclockwise circulation in calculus is often evaluated using Green's Theorem, which provides a vital connection between the circulation along a closed curve and a double integral over the area it encloses. In simple terms, if you imagine a field of arrows (known as a vector field) across a plane, the counterclockwise circulation measures the twisting or rotating tendency of this field around a closed loop. By calculating this, you are essentially measuring how much and in what direction the fluid represented by the vector field rotates around the loop.
When using Green's Theorem, you compute an area integral instead of a line integral, which can often simplify calculations. This involves determining the integrand \( rac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\) and setting up the double integral over the region enclosed by the curve. By evaluating this double integral, you get the counterclockwise circulation. This technique is particularly useful because it beautifully simplifies complex calculations that would otherwise be required if approached directly through line integrals.
When using Green's Theorem, you compute an area integral instead of a line integral, which can often simplify calculations. This involves determining the integrand \( rac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\) and setting up the double integral over the region enclosed by the curve. By evaluating this double integral, you get the counterclockwise circulation. This technique is particularly useful because it beautifully simplifies complex calculations that would otherwise be required if approached directly through line integrals.
Double Integral
A double integral is a method of integrating over a two-dimensional area. It's an integral within an integral, allowing you to calculate a volume under a surface or more commonly, the accumulated quantities over a specific region. When applying Green's Theorem, the double integral is used to evaluate properties like circulation and flux across a particular region. This requires identifying the appropriate bounds of integration as derived from the curve or shape enclosing the area.
To solve a double integral, particularly for the circulation, you'll set up your integral with the determined integrand over the defined region. In the exercise we've looked at, the integrand was constant (2), making the integration straightforward over a defined elliptical area. The ellipse in question allows for a transformation using trigonometric identities, like converting coordinates to polar forms:
To solve a double integral, particularly for the circulation, you'll set up your integral with the determined integrand over the defined region. In the exercise we've looked at, the integrand was constant (2), making the integration straightforward over a defined elliptical area. The ellipse in question allows for a transformation using trigonometric identities, like converting coordinates to polar forms:
- Transform from rectangular \(x = 2\cos\theta\), \(y = \sin\theta\)
- Integrate with respect to \(\theta\) over \(0\) to \(2\pi\)
Ellipse in Calculus
The ellipse plays a crucial role in various calculus problems, especially when working with closed curves and regions. In the context of vector fields and Green's Theorem, an ellipse like
Understanding the geometry of the ellipse allows you to effectively set up the double integral limits required by Green's Theorem. This involves transforming the domain of integration into more manageable forms, often utilizing trigonometric substitutions as seen in the polar conversion. In our exercise, knowing the size and orientation of the ellipse permits efficiently bounding the region of interest. The transformation
- \(x^2 + 4y^2 = 4\)
Understanding the geometry of the ellipse allows you to effectively set up the double integral limits required by Green's Theorem. This involves transforming the domain of integration into more manageable forms, often utilizing trigonometric substitutions as seen in the polar conversion. In our exercise, knowing the size and orientation of the ellipse permits efficiently bounding the region of interest. The transformation
- \(x = 2\cos\theta\)
- \(y = \sin\theta\)
Other exercises in this chapter
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