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Chapter 17: Problem 46

Circulation and flux For the following vector fields, compute (a) the circulation on, and (b) the outward flux across, the boundary of the given region. Assume boundary curves are oriented counterclockwise. $$\begin{array}{l} \mathbf{F}=\left\langle\ln \left(x^{2}+y^{2}\right), \tan ^{-1} \frac{y}{x}\right\rangle ; R \text { is the eighth-annulus } \\ \\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi / 4\\} \end{array}$$

Short Answer

Expert verified
In summary, for the given vector field $\mathbf{F}=\langle\ln \left(x^{2}+y^{2}\right), \tan ^{-1} \frac{y}{x}\rangle$ and the region $R$ defined by ${(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi / 4}$, (a) the circulation on the boundary of the region is 0. (b) the outward flux across the boundary of the region is $\frac{\pi}{2}$.

Step by step solution

01

Compute the divergence of the vector field

We will compute the divergence of the vector field \(\mathbf{F}=\langle \ln(x^2+y^2), \tan^{-1}\frac{y}{x}\rangle\). In polar coordinates, the vector field can be written as \(\mathbf{F} = \langle \ln(r^2), \theta \rangle\). So, the divergence of the vector field in polar coordinates can be given as: $$\nabla \cdot \mathbf{F} = \frac{1}{r}\frac{\partial}{\partial r}(rF_r) + \frac{1}{r}\frac{\partial}{\partial \theta}(F_{\theta}).$$ Now, find the divergence: $$\nabla \cdot \mathbf{F} = \frac{1}{r}\frac{\partial}{\partial r}(r\ln(r^2)) + \frac{1}{r}\frac{\partial}{\partial \theta}(\theta) = \frac{1}{r}(2) = \frac{2}{r}.$$
02

Finding the circulation

To compute the circulation on the boundary of the region \(R\), we need to integrate the vector field \(\mathbf{F}\) along the boundary curves in the counterclockwise direction. The boundary of \(R\) can be parametrized as four arcs as follows: 1. \(C_1: r = 1, 0\leq\theta\leq\frac{\pi}{4}\) 2. \(C_2: \theta = \frac{\pi}{4}, 1\leq r \leq 2\) 3. \(C_3: r = 2, 0 \leq \theta \leq \frac{\pi}{4}\) 4. \(C_4: \theta = 0, 1\leq r \leq 2\) Now, calculate the circulation integrals for each arc: 1. For \(C_1\): \( \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_0^{\frac{\pi}{4}} \langle \ln(1),\theta\rangle \cdot \langle -1,0\rangle d\theta = 0\) 2. For \(C_2\): \( \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int_1^{2} \langle \ln(r^2),\frac{\pi}{4}\rangle \cdot \langle 0,r\rangle dr = 0\) 3. For \(C_3\): \( \int_{C_3} \mathbf{F} \cdot d\mathbf{r} = \int_0^{\frac{\pi}{4}} \langle \ln(4),\theta\rangle \cdot \langle 2,0\rangle d\theta = 0\) 4. For \(C_4\): \( \int_{C_4} \mathbf{F} \cdot d\mathbf{r} = \int_1^{2} \langle \ln(r^2),0\rangle \cdot \langle 0,-r\rangle dr = 0\) So, the circulation is 0.
03

Finding the outward flux

Using the divergence theorem, we can compute the outward flux across the boundary as the double integral of the divergence of \(\mathbf{F}\) over the region \(R\). From Step 1, we found that the divergence of the vector field is \(\frac{2}{r}\). Now we can compute the double integral over the region \(R\) as follows: $$\int\int_R \nabla\cdot\mathbf{F}\,dA = \int_0^{\pi/4} \int_1^2 \frac{2}{r}\,r\, dr\,d\theta = \int_0^{\pi/4} \int_1^2 2\, dr\,d\theta$$ Now evaluate this integral: $$\int_0^{\pi/4} \int_1^2 2\, dr\,d\theta = 2\int_0^{\pi/4}(2-1)\,d\theta = \left[2\theta\right]_0^{\pi_{4}} = \frac{\pi}{2}$$ Hence, the outward flux across the boundary is \(\frac{\pi}{2}\).

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Key Concepts

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

Divergence Theorem
The Divergence Theorem is a fundamental tool in vector calculus that connects the flow (flux) of a vector field through a surface to the behavior of the field inside the surface.
It states that the outward flux of a vector field across a closed surface is equal to the volume integral of the divergence within the surface. In mathematical terms, it can be expressed as: \[ \int_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \int_{V} (abla \cdot \mathbf{F}) \, dV\]where:
  • \( S \) represents the closed surface,
  • \( \mathbf{F} \) is the vector field,
  • \( \mathbf{n} \) is the unit outward normal to the surface,
  • and \( V \) is the volume enclosed by the surface \( S \).
This theorem essentially transforms a difficult surface integral into a potentially easier volume integral by leveraging the internal properties of the vector field. In our exercise, the divergence of the given vector field \( \mathbf{F} \) is calculated in polar coordinates, where it simplifies the process of determining the flux over the eighth-annulus region. The divergence calculated is \( \frac{2}{r} \) and subsequently used in integrated form to find the outward flux, which results in \( \frac{\pi}{2} \).
Polar Coordinates
Polar coordinates offer an alternative way to describe points in the plane using a distance from a reference point and an angle from a reference direction.
This system is particularly useful in problems involving circular or rotational symmetry. In polar coordinates, any point is expressed as \( (r, \theta) \), where:
  • \( r \) is the radial distance from the origin, and
  • \( \theta \) is the angle measured from the positive x-axis.
For vector calculus, converting rectangular coordinates to polar can often simplify calculations for problems that have inherent circular symmetry.
This was evident in the original problem where the region provided was an eighth-annulus described in terms of \( r \) and \( \theta \). The vector field \( \mathbf{F} \), originally given in rectangular form, is rewritten in polar coordinates as \( \langle \ln(r^2), \theta \rangle \).
This transformation is crucial because it makes computations like integration over regarded circular paths much more straightforward. Understanding polar coordinates helps in visualizing and effectively solving problems related to circular boundaries and radial distances.
Vector Fields
A vector field assigns a vector to every point in a subset of space. It is a way to represent spatial variation of a quantity that has both magnitude and direction, like velocity fields or force fields.
A vector field \( \mathbf{F} \) is commonly represented as \( \langle F_1(x, y), F_2(x, y) \rangle \) in two dimensions, where:
  • \( F_1 \) and \( F_2 \) are the component functions of \( \mathbf{F} \).
The given vector field \( \mathbf{F} = \langle \ln(x^2 + y^2), \tan^{-1}(\frac{y}{x}) \rangle \), consists of such components,
and defines a unique vector at every point in its domain. To analyze a vector field, one often looks at derivatives, such as divergence and curl, to study how the field behaves.
In this exercise, exploring the vector field \( \langle \ln(r^2), \theta \rangle \) in polar form simplifies understanding its divergence, making it manageable to compute properties like circulation and flux across a specified boundary.
Understanding vector fields is essential in vector calculus as they are applied in various physics and engineering contexts to model natural phenomena such as fluid flows, electromagnetic fields, and gravitational forces.

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Most popular questions from this chapter

Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. \(\iint_{S}|\mathbf{r}| d S,\) where \(S\) is the cylinder \(x^{2}+y^{2}=4,\) for \(0 \leq z \leq 8\) where \(\mathbf{r}=\langle x, y, z\rangle\)

Within the cube \(\\{(x, y, z):|x| \leq 1\) \(|y| \leq 1,|z| \leq 1\\},\) where does div \(\mathbf{F}\) have the greatest magnitude when \(\mathbf{F}=\left\langle x^{2}-y^{2}, x y^{2} z, 2 x z\right\rangle ?\)

Green's Second Identity Prove Green's Second Identity for scalar-valued functions \(u\) and \(v\) defined on a region \(D:\) $$\iiint_{D}\left(u \nabla^{2} v-v \nabla^{2} u\right) d V=\iint_{S}(u \nabla v-v \nabla u) \cdot \mathbf{n} d S$$ (Hint: Reverse the roles of \(u\) and \(v\) in Green's First Identity.)

A beautiful flux integral Consider the potential function \(\varphi(x, y, z)=G(\rho),\) where \(G\) is any twice differentiable function and \(\rho=\sqrt{x^{2}+y^{2}+z^{2}} ;\) therefore, \(G\) depends only on the distance from the origin. a. Show that the gradient vector field associated with \(\varphi\) is \(\mathbf{F}=\nabla \varphi=G^{\prime}(\rho) \frac{\mathbf{r}}{\rho},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(\rho=|\mathbf{r}|\)b. Let \(S\) be the sphere of radius \(a\) centered at the origin and let \(D\) be the region enclosed by \(S\). Show that the flux of \(\mathbf{F}\) across \(S\) is \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=4 \pi a^{2} G^{\prime}(a)\) c. Show that \(\nabla \cdot \mathbf{F}=\nabla \cdot \nabla \varphi=\frac{2 G^{\prime}(\rho)}{\rho}+G^{\prime \prime}(\rho)\) d. Use part (c) to show that the flux across \(S\) (as given in part (b)) is also obtained by the volume integral \(\iiint_{D} \nabla \cdot \mathbf{F} d V\) (Hint: Use spherical coordinates and integrate by parts.)

Maximum surface integral Let \(S\) be the paraboloid \(z=a\left(1-x^{2}-y^{2}\right),\) for \(z \geq 0,\) where \(a>0\) is a real number. Let \(\mathbf{F}=\langle x-y, y+z, z-x\rangle .\) For what value(s) of \(a\) (if any) does \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) have its maximum value?
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