/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 In Exercises \(19-24,\) use the ... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 24

In Exercises \(19-24,\) use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(F\) across the surface \(S\) in the direction of the outward unit normal \(\mathbf{n}\) . \begin{equation} \begin{array}{l}{\mathbf{F}=y^{2} \mathbf{i}+z^{2} \mathbf{j}+x \mathbf{k}} \\\ {S : \quad \mathbf{r}(\phi, \theta)=(2 \sin \phi \cos \theta) \mathbf{i}+(2 \sin \phi \sin \theta) \mathbf{j}+(2 \cos \phi) \mathbf{k}} \\\ {0 \leq \phi \leq \pi / 2, \quad 0 \leq \theta \leq 2 \pi}\end{array} \end{equation}

Short Answer

Expert verified
The flux is zero.

Step by step solution

01

Understand Stokes' Theorem

Stokes' Theorem relates a surface integral over surface \(S\) to a line integral over the boundary curve \(C\) of \(S\). The theorem states that \(\iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}\). In this exercise, the objective is to find the surface integral of the curl of \(\mathbf{F}\) across the surface.
02

Compute the Curl of \(\mathbf{F}\)

The vector field is given by \(\mathbf{F} = y^2 \mathbf{i} + z^2 \mathbf{j} + x \mathbf{k}\). To compute the curl, use the formula for the curl, \(abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} - \left( \frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k}\). Calculate each component: \((2z - 0)\mathbf{i} + (1 - 0)\mathbf{j} + (0 - 2y)\mathbf{k} = 2z\mathbf{i} + \mathbf{j} - 2y\mathbf{k}\).
03

Parametrize the Surface \(S\)

The surface \(S\) is given parametrically by \(\mathbf{r}(\phi, \theta) = (2\sin\phi \cos\theta)\mathbf{i} + (2\sin\phi \sin\theta)\mathbf{j} + (2\cos\phi)\mathbf{k}\). This is the upper hemisphere of a sphere with radius 2.
04

Compute the Surface Element \(d\mathbf{S}\)

To find \(d\mathbf{S}\), compute the cross product \(\frac{\partial \mathbf{r}}{\partial \phi} \times \frac{\partial \mathbf{r}}{\partial \theta}\). Derive \(\mathbf{r}(\phi, \theta)\) with respect to \(\phi\) and \(\theta\), then take the cross product. The computation gives \(\frac{\partial \mathbf{r}}{\partial \phi} = 2\cos\phi\cos\theta\mathbf{i} + 2\cos\phi\sin\theta\mathbf{j} - 2\sin\phi\mathbf{k}\) and \(\frac{\partial \mathbf{r}}{\partial \theta} = -2\sin\phi\sin\theta\mathbf{i} + 2\sin\phi\cos\theta\mathbf{j}\). The cross product results in \(d\mathbf{S} = 4 \sin\phi \mathbf{r}(\phi,\theta)\,d\phi\,d\theta\).
05

Evaluate the Surface Integral

Substitute \(abla \times \mathbf{F} = 2z\mathbf{i} + \mathbf{j} - 2y\mathbf{k}\) and the surface element \(d\mathbf{S} = 4 \sin^2 \phi \mathbf{r}(\phi, \theta) \, d\phi \, d\theta\) into the integral \(\iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S}\). Use spherical symmetry and compute: \(\iint_S \left(2(2\cos\phi)\right) \cdot 4 \sin^2 \phi \,d\phi \,d\theta = 16 \int_0^{\pi/2} \cos \phi \sin^2 \phi \, d\phi \int_0^{2\pi} \,d\theta\). Evaluate the integrals, resulting in the value 0.

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Key Concepts

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

Surface Integral
Surface integrals are a fascinating concept in vector calculus. Imagine you have a vector field, like the wind blowing over a geographic region, and you want to measure some property of this field as it "flows" over a surface. The surface integral allows us to compute this interaction.

Typically denoted as \(\iint_S \mathbf{F} \cdot d\mathbf{S}\), a surface integral calculates the "flux" of the vector field \(\mathbf{F}\) through a surface \(S\).
  • The dot product \(\cdot\) looks at the component of the vector field that points perpendicularly through the surface.
  • The term \(d\mathbf{S}\) represents a tiny patch of the surface, much like how \(dx\) and \(dy\) represent tiny segments of the x and y axes in regular integrals.
Surface integrals are fundamental when applying Stokes' Theorem, as they equate the integral over a surface to a line integral around its boundary.
Curl of a Vector Field
Think of the curl of a vector field as a measure of how much and in what direction the field "rotates" at a given point. It's especially useful when analyzing fluid flow or rotating magnetic fields.

For a given vector field \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\), the curl is represented as \(abla \times \mathbf{F}\) and is computed using the derivatives of \(P\), \(Q\), and \(R\):
  • \(\left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i}\)
  • \(\left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\mathbf{j}\)
  • \(\left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k}\)
In the exercise, for the vector field \(\mathbf{F} = y^2 \mathbf{i} + z^2 \mathbf{j} + x \mathbf{k}\), the curl works out to \(2z\mathbf{i} + \mathbf{j} - 2y\mathbf{k}\), revealing how the field's rotational aspects can be calculated.
Parametrized Surface
Parametrizing a surface involves expressing it in terms of variables, usually \(\phi\) and \(\theta\), much like coordinates on a map. This process is crucial because it allows us to work with complex surfaces using more manageable mathematical expressions.

Take for example the surface described by \(\mathbf{r}(\phi, \theta) = (2\sin\phi\cos\theta)\mathbf{i} + (2\sin\phi\sin\theta)\mathbf{j} + (2\cos\phi)\mathbf{k}\).
  • This surface represents the upper hemisphere of a sphere with a radius of 2.
  • \(\phi\) and \(\theta\) become the new variables that govern positions on the surface, much like latitude and longitude do on Earth.
  • When \(\phi\) ranges from 0 to \(\pi/2\) and \(\theta\) from 0 to \(2\pi\), we cover the entire surface of the upper hemisphere.
Flux Calculation
Flux calculation helps us understand how a vector field passes through a surface and is a central concept in Stokes' Theorem. It relates to the flow of a fluid, heat, or electricity through a particular boundary.

When calculating flux using a surface integral, it becomes necessary to use the parametrization of the surface to find \(d\mathbf{S}\), the surface element.
  • From the exercise, we had \(d\mathbf{S} = 4 \sin^2 \phi \mathbf{r}(\phi, \theta) \, d\phi \, d\theta\).
  • Thus, the flux is \(\iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S}\).
  • Applying spherical symmetry and solving the double integral involves evaluating the fields' components and then integrating over the specified bounds.
In this example, the integral simplifies and evaluates to zero, showing there is no net flux of \(abla \times \mathbf{F}\) across the surface, indicating a rotational symmetry with no net outward flow.

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Most popular questions from this chapter

Parametrization of a surface of revolution Suppose that the parametrized curve \(C :(f(u), g(u))\) is revolved about the \(x\) -axis, where \(g(u)>0\) for \(a \leq u \leq b .\) a. Show that $$ \mathbf{r}(u, v)=f(u) \mathbf{i}+(g(u) \cos v) \mathbf{j}+(g(u) \sin v) \mathbf{k} $$ is a parametrization of the resulting surface of revolution, where \(0 \leq v \leq 2 \pi\) is the angle from the \(x y\) -plane to the point \(\mathbf{r}(u, v)\) on the surface. (See the accompanying figure.) Notice that \(f(u)\) measures distance along the axis of revolution and \(g(u)\) measures distance from the axis of revolution. b. Find a parametrization for the surface obtained by revolving the curve \(x=y^{2}, y \geq 0,\) about the \(x\) -axis.

In Exercises \(9-20\) , use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) Thick sphere \(\quad \mathbf{F}=\left(5 x^{3}+12 x y^{2}\right) \mathbf{i}+\left(y^{3}+e^{y} \sin z\right) \mathbf{j}+\) \(\left(5 z^{3}+e^{y} \cos z\right) \mathbf{k}\) \(D :\) The solid region between the spheres \(x^{2}+y^{2}+z^{2}=1\) and \(x^{2}+y^{2}+z^{2}=2\)

The base of the closed cubelike surface shown here is the unit square in the \(x y\) -plane. The four sides lie in the planes \(x=0\) , \(x=1, y=0,\) and \(y=1 .\) The top is an arbitrary smooth surface whose identity is unknown. Let \(\mathbf{F}=x \mathbf{i}-2 y \mathbf{j}+(z+3) \mathbf{k}\) and suppose the outward flux of \(\mathbf{F}\) through Side \(A\) is 1 and through Side \(B\) is \(-3 .\) Can you conclude anything about the outward flux through the top? Give reasons for your answer.

Gradient of a line integral Suppose that \(\mathbf{F}=\nabla f\) is a conservative vector field and $$g(x, y, z)=\int_{(0,0,0)}^{(x, y z)} \mathbf{F} \cdot d \mathbf{r}$$ Show that \(\nabla g=\mathbf{F}\).

In Exercises 9-20, use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) Sphere \(\quad \mathbf{F}=x^{3} \mathbf{i}+y^{3} \mathbf{j}+z^{3} \mathbf{k}\) \(D :\) The solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\)
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