/*! 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 9 In Exercises \(7-12,\) use the s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(7-12,\) use the surface integral in Stokes' Theorem to calculate the circulation of the field \(\mathbf{F}\) around the curve \(C\) in the indicated direction. \begin{equation} \begin{array}{l}{\mathbf{F}=y \mathbf{i}+x z \mathbf{j}+x^{2} \mathbf{k}} \\\ {\text { C: The boundary of the triangle cut from the plane } x+y+z=1} \\\ {\text { by the first octant, counterclockwise when viewed from above }}\end{array} \end{equation}

Short Answer

Expert verified
The circulation around the curve is \( \frac{1}{3} \).

Step by step solution

01

Identify Stokes' Theorem

According to Stokes' Theorem, the circulation of the vector field \( \mathbf{F} \) around the closed curve \( C \) is given by the surface integral of the curl of \( \mathbf{F} \), such that \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} \), where \( S \) is the surface bounded by the curve \( C \).
02

Calculate the Curl of F

First, compute the curl of \( \mathbf{F} = y \mathbf{i} + xz \mathbf{j} + x^2 \mathbf{k} \). The curl is calculated as:\[ abla \times \mathbf{F} = \left( \frac{\partial x^2}{\partial y} - \frac{\partial xz}{\partial z} \right) \mathbf{i} + \left( \frac{\partial y}{\partial z} - \frac{\partial x^2}{\partial x} \right) \mathbf{j} + \left( \frac{\partial xz}{\partial x} - \frac{\partial y}{\partial y} \right) \mathbf{k} \]Substituting the partial derivatives, we obtain:\[ abla \times \mathbf{F} = 0 \mathbf{i} - 2x \mathbf{j} + z \mathbf{k} \].
03

Describe the Surface S and Calculate dS

The surface \( S \) is the triangular region cut from the plane \( x + y + z = 1 \) by the first octant. We parametrize \( S \) using \( x \) and \( y \) such that the third variable \( z = 1 - x - y \).The differential surface area element is \( d\mathbf{S} = \mathbf{n} \, dS \) where \( \mathbf{n} \) is the unit normal vector. The area element is:\[ d\mathbf{S} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \, dx \, dy \].
04

Set Up the Surface Integral

Substitute the curl from Step 2 into the surface integral:\[ \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S (0 \mathbf{i} - 2x \mathbf{j} + z \mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) \, dx \, dy \].This simplifies to:\[ \iint_S (-2x + z) \, dx \, dy \].
05

Evaluate the Surface Integral

Substitute \( z = 1 - x - y \) into the integral:\[ \iint_S (-2x + 1 - x - y) \, dx \, dy = \iint_S (1 - 3x - y) \, dx \, dy \].The limits for integration are from 0 to 1 for \( x \), and from 0 to \( 1-x \) for \( y \). Compute the integral:\[ \int_0^1 \int_0^{1-x} (1 - 3x - y) \, dy \, dx \].First, evaluate with respect to \( y \):\[ \int_0^{1-x} (1 - 3x - y) \, dy = (1 - 3x)y - \frac{y^2}{2} \bigg|_0^{1-x} = (1 - 3x)(1-x) - \frac{(1-x)^2}{2} \].
06

Integrate with respect to x

Simplify and further integrate:\[(1-3x)(1-x) = 1 - 3x - x + 3x^2 = 1 - 4x + 3x^2, \]\[\text{and } \frac{(1-x)^2}{2} = \frac{1 - 2x + x^2}{2}. \]Combine and integrate:\[ \int_0^1 \left( 1 - 4x + 3x^2 - \frac{1}{2} + x - \frac{x^2}{2} \right) \, dx = \int_0^1 \left( \frac{1}{2} - 3x + \frac{5x^2}{2} \right) \, dx. \]Calculate the integral:\[ \left[ \frac{x}{2} - \frac{3x^2}{2} + \frac{5x^3}{6} \right]_0^1 = \frac{1}{2} - \frac{3}{2} + \frac{5}{6} = \frac{-3 + 5}{6} = \frac{2}{6} = \frac{1}{3}. \]
07

Conclusion: Calculate Circulation

According to Stokes' Theorem, the circulation of \( \mathbf{F} \) around \( C \) is \( \frac{1}{3} \). This is the value obtained from evaluating the surface integral of the curl of \( \mathbf{F} \).

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
Surface integrals allow us to extend the concept of integrating over curves to integrating over surfaces. Imagine draping a thin fabric over a surface. The surface integral helps us understand how a vector field interacts with this surface.
In Stokes' Theorem, we use surface integrals to connect the behavior of a vector field over a surface with its behavior around the boundary of that surface. Specifically, Stokes' Theorem states:
  • The circulation of a vector field \( \mathbf{F} \) over a closed curve \( C \) is equal to the surface integral of the curl of \( \mathbf{F} \) over a surface \( S \) that \( C \) bounds.
The surface integral component of Stokes' Theorem is expressed as: \[ \iint_S ( abla \times \mathbf{F} ) \cdot d\mathbf{S} \] Here, \( S \) is the surface bounded by \( C \), \( abla \times \mathbf{F} \) is the curl of the vector field, and \( d\mathbf{S} \) is the differential surface area element. Understanding surface integrals is crucial for visualizing and solving problems using Stokes' Theorem.
Curl of a Vector Field
The curl of a vector field is a measure of the field's rotation or "swirl" at each point. Think of how water might swirl around a drain; the curl represents that kind of rotational tendency.
The formula for the curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is:\[ abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \]The resulting vector gives the axis of rotation and the magnitude of the swirl.
When calculating the curl for the vector field \( \mathbf{F} = y \mathbf{i} + xz \mathbf{j} + x^2 \mathbf{k} \) as seen in the exercise, specific partial derivatives are calculated for \( P, Q, \) and \( R \). These calculations reveal where and how intensely the vector field rotates in space, an important step in integrating it over a surface using Stokes' Theorem.
Parametrization of Surfaces
Parametrization is the method by which we describe surfaces using variables or parameters. This is done to handle surfaces mathematically within integrals. In the given problem, the triangular surface created by the plane \( x + y + z = 1 \) in the first octant is considered.
This surface is parametrized using two variables, often \( x \) and \( y \), and expressing the third variable \( z\) as a function of the first two:
  • \( z = 1 - x - y \).
Parametrization simplifies complex surfaces into well-defined domains in the xy-plane, making the process of applying calculus more manageable.
It transforms the description of a surface, allowing us to integrate functions over it and to evaluate surface integrals, as is required by Stokes' Theorem.
Furthermore, by converting the surface into a two-dimensional region, it becomes easier to apply multivariable calculus techniques and compute necessary integrals.
Unit Normal Vector
A unit normal vector is a vector that is perpendicular (orthogonal) to a surface at a given point and has a magnitude of 1. Unit normal vectors are crucial in defining the surface's orientation, especially in the context of surface integrals within Stokes' Theorem.
For a surface given by the equation \( z = 1 - x - y \), the unit normal vector can be determined by cross product or can directly be influenced by surface orientation.
  • In this exercise, the normal vector is given as \( \mathbf{i} + \mathbf{j} + \mathbf{k} \), needing normalization.
The normalization process divides the normal vector by its magnitude, ensuring that it has unit length.
This vector plays a key role in calculating the differential area element \( d\mathbf{S} \) because it "weights" the surface integral by aligning it correctly along with the vector field.
In the context of the exercise, correctly determining the unit normal aids in accurately computing the surface integral, ensuring the accurate application of Stokes' Theorem which directly relates to finding the circulation of the field around the curve.

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

a. A torus of revolution (doughnut) is obtained by rotating a circle \(C\) in the \(x z\) -plane about the \(z\) -axis in space. (See the accompanying figure.) If \(C\) has radius \(r>0\) and center \((R, 0,0),\) show that a parametrization of the torus is $$ \begin{aligned} \mathbf{r}(u, \boldsymbol{v})=&((R+r \cos u) \cos v) \mathbf{i} \\ &+((R+r \cos u) \sin v) \mathbf{j}+(r \sin u) \mathbf{k} \end{aligned} $$ where \(0 \leq u \leq 2 \pi\) and \(0 \leq v \leq 2 \pi\) are the angles in the figure. b. Show that the surface area of the torus is \(A=4 \pi^{2} R r\)

In Exercises 9-20, use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) Cylinder and paraboloid \(\mathbf{F}=y \mathbf{i}+x y \mathbf{j}-z \mathbf{k}\) \(D :\) The region inside the solid cylinder \(x^{2}+y^{2} \leq 4\) between the plane \(z=0\) and the paraboloid \(z=x^{2}+y^{2}\)

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. Cone frustum \(\mathbf{F}=-x \mathbf{i}-y \mathbf{j}+z^{2} \mathbf{k}\) outward (normal away from the \(z\) -axis) through the portion of the cone \(z=\sqrt{x^{2}+y^{2}}\) between the planes \(z=1\) and \(z=2\)

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.)

Let \(S\) be the portion of the cylinder \(y=e^{x}\) in the first octant that projects parallel to the \(x\) -axis onto the rectangle \(R_{y z} : 1 \leq y \leq 2\) \(0 \leq z \leq 1\) in the \(y z\) -plane (see the accompanying figure). Let \(\mathbf{n}\) be the unit vector normal to \(S\) that points away from the \(y z\) -plane. Find the flux of the field \(\mathbf{F}(x, y, z)=-2 \mathbf{i}+2 y \mathbf{j}+z \mathbf{k}\) across \(S\) in the direction of \(\mathbf{n} .\)

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.