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Chapter 17: Problem 31

Use Stokes' Theorem to find the circulation of the following vector fields around any simple closed smooth curve \(C\). $$\mathbf{F}=\nabla\left(x \sin y e^{z}\right)$$

Short Answer

Expert verified
Answer: The circulation of the given vector field \(\mathbf{F}\) around any simple closed smooth curve \(C\) is 0.

Step by step solution

01

Find the vector field \(\mathbf{F}\) from the given \(\nabla$$ \bullet \mathbf{F}\)

Given \(\mathbf{F}=\nabla\left(x \sin y e^{z}\right)\), so we need to find the gradient of the scalar function \(f(x, y, z) = x \sin y e^{z}\): $$\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle = \left\langle\sin y e^{z}, x e^{z} \cos y, x \sin y e^{z}\right\rangle$$ Therefore, the vector field \(\mathbf{F}\) is: $$\mathbf{F} = \langle\sin y e^{z}, x e^{z} \cos y, x \sin y e^{z}\rangle$$
02

Compute the curl of the vector field \(\mathbf{F}\)

Calculating the curl of the vector field \(\mathbf{F}\): $$\nabla \times \mathbf{F} = \text{det}\left|\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \sin y e^{z} & x e^{z} \cos y & x \sin y e^{z} \end{matrix}\right|$$ Expanding the determinant and differentiating, we get: $$\nabla \times \mathbf{F} = \left\langle\frac{\partial}{\partial y}(x \sin y e^{z})-\frac{\partial}{\partial z}(x e^{z} \cos y),\frac{\partial}{\partial z}(\sin y e^{z})-\frac{\partial}{\partial x}(x \sin y e^{z}), \frac{\partial}{\partial x}(x e^{z} \cos y)-\frac{\partial}{\partial y}(\sin y e^{z})\right\rangle$$ $$\nabla \times \mathbf{F} = \left\langle(x \cos y e^{z}+x \sin y e^{z})-(-x e^{z} \sin y), e^{z} \cos y-(\sin y e^{z}+x \sin y e^{z}), (-x \sin y e^{z} + e^{z} \cos y)-\cos y e^{z}\right\rangle$$ $$\nabla \times \mathbf{F} = \langle0, 0, 0\rangle$$
03

Compute the circulation using Stokes' Theorem

In this case, \(\nabla \times \mathbf{F} = \langle0,0,0\rangle\). According to Stokes' Theorem, the circulation of the vector field around a simple closed smooth curve \(C\) is given by: $$\text{Circulation}=\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$$ Since the curl is zero, the circulation is equal to zero as well: $$\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$$ Thus, the circulation of the given vector field \(\mathbf{F}\) around any simple closed smooth curve \(C\) is zero.

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Key Concepts

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

Circulation of a Vector Field
Imagine walking along a path and feeling the breeze swirling around you in a consistent direction. This swirl, or rotation, of a vector field along a closed curve is what mathematicians refer to as the circulation of a vector field. In essence, it measures how much the field wraps around or circulates the path you've taken.

To compute this, we use a line integral around the path, also known as a closed curve, which gives us a scalar value of circulation. If the vector field represents a fluid flow, the circulation tells us, in a sense, how much the fluid is spinning around the path. Stokes' Theorem is a powerful tool in this context, linking the circulation to another concept known as the curl of a vector field over a surface bounded by the curve. In our specific exercise, using Stokes' Theorem revealed that the circulation of the vector field was zero 鈥� this implies that the vector field has no 'twistiness' at all along the curve.
Gradient of a Scalar Field
When we talk about the gradient of a scalar field, we're looking at the multidimensional generalization of a derivative. It points in the direction of the steepest ascent of the scalar field and its magnitude tells us how steep the hill is at that point. In practical terms, if you were hiking and looking for the steepest path up a mountain, following the direction of the gradient at any point would lead you to the peak most directly.

In the given exercise, we found the gradient of a scalar function by taking partial derivatives with respect to each variable. For our function, the gradient was defined as a vector field that led us to explore the given vector field more deeply. Understanding the gradient is crucial for many fields such as physics, engineering, and even economics, as it provides critical insights into rates of change in various directions.
Curl of a Vector Field
Now, let's dive into the concept of the curl of a vector field. The curl is a measure of the rotation at a point in the vector field. Think of it like placing a tiny paddle wheel in a flowing river; the way the water makes the wheel turn is analogous to the vector field's curl at that point. Mathematically, we calculate this using a special operation involving partial derivatives, similar to taking the cross product of a gradient with the vector field.

In our exercise, the curl of the vector field was determined to be zero, indicating no rotation at any point, akin to a river flowing perfectly straight with no whirlpools. The zero curl helped us conclude via Stokes' Theorem that the circulation around a closed curve was also zero. The concept of curl is critically important in electromagnetism and fluid dynamics as it describes the rotation of fields and fluids.

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Most popular questions from this chapter

Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2},\) for \(-L \leq z \leq L\) a. Find the outward flux of the field \(\mathbf{F}=\langle x, y, 0\rangle\) across \(S\) b. Find the outward flux of the field \(\mathbf{F}=\frac{\langle x, y, 0\rangle}{\left(x^{2}+y^{2}\right)^{p / 2}}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}}\) across \(S\), where \(|\mathbf{r}|\) is the distance from the \(z\) -axis and \(p\) is a real number. c. In part (b), for what values of \(p\) is the outward flux finite as \(a \rightarrow \infty(\text { with } L\) fixed)? d. In part (b), for what values of \(p\) is the outward flux finite as \(L \rightarrow \infty\) (with \(a\) fixed)?

Area of a region in a plane Let \(R\) be a region in a plane that has a unit normal vector \(\mathbf{n}=\langle a, b, c\rangle\) and boundary \(C .\) Let \(\mathbf{F}=\langle b z, c x, a y\rangle\). a. Show that \(\nabla \times \mathbf{F}=\mathbf{n}\) b. Use Stokes' Theorem to show that $$\text { area of } R=\oint_{C} \mathbf{F} \cdot d \mathbf{r}$$ c. Consider the curve \(C\) given by \(\mathbf{r}=\langle 5 \sin t, 13 \cos t, 12 \sin t\rangle\) for \(0 \leq t \leq 2 \pi .\) Prove that \(C\) lies in a plane by showing that \(\mathbf{r} \times \mathbf{r}^{\prime}\) is constant for all \(t\) d. Use part (b) to find the area of the region enclosed by \(C\) in part (c). (Hint: Find the unit normal vector that is consistent with the orientation of \(C .\) )

The potential function for the force field due to a charge \(q\) at the origin is \(\varphi=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{|\mathbf{r}|},\) where \(\mathbf{r}=\langle x, y, z\rangle\) is the position vector of a point in the field, and \(\varepsilon_{0}\) is the permittivity of free space. a. Compute the force field \(\mathbf{F}=-\nabla \varphi\) b. Show that the field is irrotational; that is, show that \(\nabla \times \mathbf{F}=\mathbf{0}\)

Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. $$\begin{aligned} &\iint_{S} \frac{\langle x, 0, z\rangle}{\sqrt{x^{2}+z^{2}}} \cdot \mathbf{n} d S, \text { where } S \text { is the cylinder } x^{2}+z^{2}=a^{2}\\\ &|y| \leq 2 \end{aligned}$$

Surface integrals of vector fields Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an explicit or a parametric description of the surface. \(\mathbf{F}=\langle-y, x, 1\rangle\) across the cylinder \(y=x^{2},\) for \(0 \leq x \leq 1\) \(0 \leq z \leq 4 ;\) normal vectors point in the general direction of the positive y-axis.
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