/*! 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 20 In Exercises \(19-24,\) use the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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-z) \mathbf{i}+(z-x) \mathbf{j}+(x+z) \mathbf{k}} \\ {S : \quad \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+\left(9-r^{2}\right) \mathbf{k},} \\ {0 \leq r \leq 3, \quad 0 \leq \theta \leq 2 \pi}\end{array} \end{equation}

Short Answer

Expert verified
The flux of the curl of \(\mathbf{F}\) across \(S\) is \(18\pi\).

Step by step solution

01

State Stoke's Theorem

Stokes' Theorem relates a surface integral over a surface \(S\) to a line integral over its boundary \(C\). Mathematically, it is expressed as \( \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r} \). Here, \(abla \times \mathbf{F}\) is the curl of the field \(\mathbf{F}\) and \(d\mathbf{S}\) is the oriented surface element.
02

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

The vector field is \(\mathbf{F} = (y-z) \mathbf{i} + (z-x) \mathbf{j} + (x+z) \mathbf{k}\). To find the curl, compute \(abla \times \mathbf{F}\) using the determinant form:\[abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ y-z & z-x & x+z \end{vmatrix}\]Calculating the determinant gives:\[ abla \times \mathbf{F} = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \]
03

Parameterize the Surface \(S\)

The surface \(S\) is parametrized by \(\mathbf{r}(r, \theta) = (r\cos\theta)\mathbf{i} + (r\sin\theta)\mathbf{j} + (9-r^2)\mathbf{k}\). This represents a portion of a paraboloid, for \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2\pi\).
04

Find \(d\mathbf{S}\) for the Surface Integral

To find \(d\mathbf{S} = \mathbf{n}\, dS\), compute the partial derivatives \(\frac{\partial \mathbf{r}}{\partial r}\) and \( \frac{\partial \mathbf{r}}{\partial \theta} \), and then their cross product:\[ \frac{\partial \mathbf{r}}{\partial r} = (\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} - 2r\mathbf{k} \]\[ \frac{\partial \mathbf{r}}{\partial \theta} = (-r\sin\theta)\mathbf{i} + (r\cos\theta)\mathbf{j} \]The cross product is:\[ \frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta} = (2r^2\cos\theta)\mathbf{i} + (2r^2\sin\theta)\mathbf{j} + r\,\mathbf{k} \]Normalize this vector to get the unit normal \(\mathbf{n}\).
05

Compute the Surface Integral

The surface integral is given by:\[ \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S (2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \cdot ((2r^2\cos\theta)\mathbf{i} + (2r^2\sin\theta)\mathbf{j} + r\,\mathbf{k}) \, dr\, d\theta \]Evaluating the dot product:\[ = \iint (4r^2 \cos\theta + 4r^2 \sin\theta + 2r) \, dr \, d\theta \]Now set up the double integral with the given bounds:\[ \int_0^{2\pi} \int_0^3 (4r^2 \cos\theta + 4r^2 \sin\theta + 2r) \, dr \, d\theta \]
06

Evaluate the Integral

Separate the integral into manageable parts. Notice the \(\cos\theta\) and \(\sin\theta\) terms integrate to zero over the full \(0 \leq \theta \leq 2\pi\) range.So we just focus on the \(2r\) term:\[ \int_0^{2\pi} \int_0^3 (2r) \, dr \, d\theta = \int_0^{2\pi} \left[ r^2 \right]_0^3 \, d\theta = \int_0^{2\pi} 9 \, d\theta = 18\pi \]
07

Conclusion

Finally, according to Stoke's Theorem, the flux of the curl of \(\mathbf{F}\) across the surface \(S\) is the value of the surface integral we calculated, which is \(18\pi\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Surface Integral
The surface integral is a crucial concept in vector calculus. It involves integrating a vector field across a surface. This means we are essentially summing up vector values over a given surface area.
In the context of Stokes' Theorem, the surface integral is used to equate a line integral around a closed curve with an integral over the surface bounded by that curve. The expression \( \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} \) calculates such an integral, measuring the total effect of the vector field's curl across the surface.
The result can be interpreted as the total 'twisting' effect of the field as it passes through the surface. This provides a powerful method to convert complex three-dimensional surface calculations into simpler path calculations.
Curl of a Vector Field
The curl of a vector field measures how much and in what way the field 'twists' or 'rotates'. To find the curl, you use the operator \( abla \times \mathbf{F} \), which is a form of a vector product (cross product).
In our exercise, the vector field \( \mathbf{F} = (y-z) \mathbf{i} + (z-x) \mathbf{j} + (x+z) \mathbf{k} \) had the curl calculated using a determinant involving unit vectors and partial derivatives. The computation yields \( abla \times \mathbf{F} = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \).
This result tells us there is a uniform twisting effect in all three coordinate directions, which simplifies the later surface integration process.
Parametrized Surface
Parametrizing a surface involves describing it using a set of equations that depend on two parameters, often labeled \(r\) and \(\theta\). In our problem, the surface was given by \( \mathbf{r}(r, \theta) = (r \cos \theta)\mathbf{i} + (r \sin \theta)\mathbf{j} + (9-r^2)\mathbf{k} \).
Parametrized surfaces are beneficial as they transform complex geometric objects into more manageable mathematical descriptions.
  • The parameters \(r\) and \(\theta\) vary over chosen ranges to capture the full geometry of the surface.
  • In this case, it's a paraboloid surface with circular symmetry around the z-axis.

Proper parametrization is vital for performing surface integrals, as it defines how the surface spans the space and how calculations can be mapped onto it.
Flux Calculation
Flux calculation through a surface involves determining how much of a field passes through that surface. In context, it's part of computing the surface integral for Stokes' Theorem.
We start by finding the differential surface area vector \(dS\), which we then dot with the field we want to integrate, such as the curl of a vector field.
The given integral is \( \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} \), which becomes \( \int_0^{2\pi} \int_0^3 (4r^2 \cos\theta + 4r^2 \sin\theta + 2r) \, dr \, d\theta \). The symmetry of \(\cos\theta\) and \(\sin\theta\) makes their contributions over \( \theta \) negligible, allowing simplification to \( \int_0^{2\pi} 9 \, d\theta = 18\pi \).
Understanding the concept of flux is essential for analyzing how the result of the calculation reflects the physical interpretation of field interaction through the surface.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Conservation of mass Let \(\mathbf{v}(t, x, y, z)\) be a continuously differentiable vector field over the region \(D\) in space and let \(p(t, x)\) \(y, z )\) be a continuously differentiable scalar function. The variable \(t\) represents the time domain. The Law of Conservation of Mass asserts that $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=-\iint_{S} p \mathbf{v} \cdot \mathbf{n} d \sigma$$ where \(S\) is the surface enclosing \(D\) a. Give a physical interpretation of the conservation of mass law if \(\mathbf{v}\) is a velocity flow field and \(p\) represents the density of the fluid at point \((x, y, z)\) at time \(t\) b. Use the Divergence Theorem and Leibniz's Rule, $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=\iiint_{D} \frac{\partial p}{\partial t} d V$$ to show that the Law of Conservation of Mass is equivalent to the continuity equation, $$\nabla \cdot p \mathbf{v}+\frac{\partial p}{\partial t}=0$$ (In the first term \(\nabla \cdot p \mathbf{v},\) the variable \(t\) is held fixed, and in the second term \(\partial p / \partial t,\) it is assumed that the point \((x, y, z)\) in \(D\) is held fixed.)

A revealing experiment By experiment, you find that a force field \(\mathbf{F}\) performs only half as much work in moving an object along path \(C_{1}\) from \(A\) to \(B\) as it does in moving the object along path \(C_{2}\) from \(A\) to \(B\) . What can you conclude about \(\mathbf{F}\) ? Give reasons for your answer.

Find the area of the surface \(2 x^{3 / 2}+2 y^{3 / 2}-3 z=0\) above the square \(R : 0 \leq x \leq 1,0 \leq y \leq 1,\) in the \(x y\) -plane.

Does Stokes' Theorem say anything special about circulation in a field whose curl is zero? Give reasons for your answer.

Area as a line integral Show that if \(R\) is a region in the plane bounded by a piecewise smooth, simple closed curve \(C,\) then $$\begin{array}{l}{\text {Area of}}\end{array}R=\oint_{C} x d y=-\oint_{C} y d x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.