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Chapter 16: Problem 14

Use Green's Theorem to find the counterclockwise circulation and outward flux for the field \(\mathbf{F}\) and curve \(C.\) $$\mathbf{F}=\left(\tan ^{-1} \frac{y}{x}\right) \mathbf{i}+\ln \left(x^{2}+y^{2}\right) \mathbf{j}$$ C: The boundary of the region defined by the polar coordinate inequalities \(1 \leq r \leq 2,0 \leq \theta \leq \pi\)

Short Answer

Expert verified
The counterclockwise circulation is 0 and the outward flux is 2.

Step by step solution

01

Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) and a double integral over the plane region \(R\) bounded by \(C\). For a vector field \(\mathbf{F} = M\mathbf{i} + N\mathbf{j}\), the circulation is given by \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \ dA\). The outward flux is \(\oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left(\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y}\right) \ dA\).
02

Identify M and N

The given vector field is \(\mathbf{F} = \left(\tan^{-1} \frac{y}{x}\right) \mathbf{i} + \ln(x^2 + y^2) \mathbf{j}\). Thus, \(M = \tan^{-1} \frac{y}{x}\) and \(N = \ln(x^2 + y^2)\).
03

Compute Partial Derivatives for Circulation

Find \(\frac{\partial N}{\partial x}\) and \(\frac{\partial M}{\partial y}\). The partial derivative \(\frac{\partial N}{\partial x}\) is \(\frac{2x}{x^2 + y^2}\) and \(\frac{\partial M}{\partial y}\) is \(\frac{x}{x^2 + y^2}\). The expression for circulation is thus \(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = \frac{2x}{x^2 + y^2} - \frac{x}{x^2 + y^2} = \frac{x}{x^2 + y^2}\).
04

Compute the Double Integral for Circulation

Convert the problem to polar coordinates, where \(x = r\cos\theta\) and \(y = r\sin\theta\). The Jacobian determinant for polar coordinates is \(r\). The circulation becomes \(\iint_R \frac{r\cos\theta}{r^2} r \, dr \, d\theta = \int_0^{\pi} \int_1^2 \frac{r\cos\theta}{r} \, dr \, d\theta = \int_0^{\pi} \cos\theta \int_1^2 \, dr \, d\theta\). The inner integral \(\int_1^2 \, dr\) evaluates to \(1\). Hence, the outer integral becomes \(\int_0^{\pi} \cos\theta \, d\theta = 0\).
05

Compute Partial Derivatives for Flux

Find \(\frac{\partial M}{\partial x}\) and \(\frac{\partial N}{\partial y}\). The partial derivative \(\frac{\partial M}{\partial x}\) is \(-\frac{y}{x^2 + y^2}\) and \(\frac{\partial N}{\partial y}\) is \(\frac{2y}{x^2 + y^2}\). The expression for flux is thus \(\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} = -\frac{y}{x^2 + y^2} + \frac{2y}{x^2 + y^2} = \frac{y}{x^2 + y^2}\).
06

Compute the Double Integral for Flux

Convert to polar coordinates: the flux becomes \(\iint_R \frac{y}{x^2 + y^2} r \, dr \, d\theta = \int_0^{\pi} \int_1^2 \frac{r\sin\theta}{r^2} r \, dr \, d\theta = \int_0^{\pi} \sin\theta \int_1^2 \, dr \, d\theta\). Evaluating the integral with respect to \(r\) gives \(1\). Thus, evaluate \(\int_0^{\pi} \sin\theta \, d\theta = 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Field
In mathematics, a vector field is a function that assigns a vector to each point in space. Think of it like a breeze flowing over a surface that radiates out different speeds and directions at every point. For our exercise, the vector field \[ \mathbf{F} = \left( \tan^{-1} \frac{y}{x} \right) \mathbf{i} + \ln \left(x^2 + y^2 \right) \mathbf{j} \]assigns vectors throughout a region in the plane. Each vector in this field has two components:
  • \(M = \tan^{-1} \frac{y}{x}\) - Acts as the \(x\)-component.
  • \(N = \ln(x^2 + y^2)\) - Acts as the \(y\)-component.
Vector fields are central to understanding a variety of physical phenomena, from gravity to electromagnetics, making them essential in calculus and physics courses.
Partial Derivatives
Partial derivatives help us understand how a function changes as we tweak one variable, holding others constant. Imagine looking at how the slope changes as you walk along the side of a hill.
For our function, we need partial derivatives to apply Green's Theorem efficiently. These are the derivatives involved:
  • \(\frac{\partial N}{\partial x} = \frac{2x}{x^2 + y^2}\)
  • \(\frac{\partial M}{\partial y} = \frac{x}{x^2 + y^2}\)
  • \(\frac{\partial M}{\partial x} = -\frac{y}{x^2 + y^2}\)
  • \(\frac{\partial N}{\partial y} = \frac{2y}{x^2 + y^2}\)
Using these derivatives, we can simplify and compute circulation and flux through the region more effectively.
Polar Coordinates
Polar coordinates provide a handy alternative to Cartesian (\(x, y\)) coordinates, ideal for circular or radial problems. They use the distance from the origin (\(r\)) and the angle (\(\theta\)) from the positive x-axis. Here's the conversion:
  • \(x = r\cos\theta\)
  • \(y = r\sin\theta\)
In this problem, the boundary is defined by \(1 \leq r \leq 2\) and \(0 \leq \theta \leq \pi\). This means we have a semi-annular region. By switching to polar coordinates, we simplify the integration process, especially with circular regions.
Line Integral
Line integrals calculate the total of a field along a specific path or curve. In our task, the circulation or line integral is derived using Green鈥檚 Theorem:\[\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA\]This integral tells us about the field's rotation or "twirl" around the loop \(C\). For \(C\), the path encircles the area determined by \(r = 1\) to \(r = 2\) and \(0 \leq \theta \leq \pi\), resulting in the integral's final outcome.
Double Integral
A double integral computes the accumulation of a function across a two-dimensional region. It is like piling up small volumes into larger mounds. Green鈥檚 Theorem uses:
  1. \[ \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \] for circulation, and
  2. \[ \iint_R \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) dA \] for flux.
In our case, switching to polar coordinates eases calculation due to symmetry, simplifying the region \(R\) as a radial slice. Here, the Jacobian determinant \(r\) aids in transforming the area element from rectangular to circular space, avoiding complications in traditional Cartesian approach.

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Most popular questions from this chapter

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) The lower portion cut from the sphere \(x^{2}+y^{2}+z^{2}=2\) by the cone \(z=\sqrt{x^{2}+y^{2}}\)

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{n}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0} .\) In Exercises \(27-30,\) find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. The parabolic cylinder surface \(\mathbf{r}(x, y)=\) \(x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty

Let \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} .\) Show that the clockwise circulation of the field \(\mathbf{F}=\nabla f\) around the circle \(x^{2}+y^{2}=a^{2}\) in the \(x y\) -plane is zero a. by taking \(\mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi,\) and integrating \(\mathbf{F} \cdot d \mathbf{r}\) over the circle. b. by applying Stokes' Theorem.

Apply Green's Theorem to evaluate the integrals. \(\oint_{C}\left(y^{2} d x+x^{2} d y\right)\) \(C:\) The triangle bounded by \(x=0, x+y=1, y=0\)

Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2}, 0 \leq z \leq h,\) together with its top, \(x^{2}+y^{2} \leq a^{2}, z=h .\) Let \(\mathbf{F}=-y \mathbf{i}+x \mathbf{j}+x^{2} \mathbf{k} .\) Use Stokes' Theorem to find the flux of \(\nabla \times \mathbf{F}\) outward through \(S.\)
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