Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In Exercises \(17-20\), a closed curve \(C\) that is the boundary of a surface \(S\) is given along with a vector field \(\vec{F}\). Find the circulation of \(\vec{F}\) around \(C\) either through direct computation or through Stokes' Theorem. \(C\) is the curve whose \(x\) - and \(y\) -values are determined by the three sides of a triangle with vertices at (-1,0),(1,0) and \((0,1),\) traversed in that order, and the \(z\) -values are determined by the function \(z=x y ; \vec{F}=\left\langle z-y^{2}, x, z\right\rangle .\)

Short Answer

Expert verified
The circulation of \( \vec{F} \) around \( C \) can be found via direct computation or Stokes' Theorem, yielding the same result.

Step by step solution

01

Parameterize the Curve C

To find the circulation of \( \vec{F} \) around the curve \( C \), we first parameterize the sides of the triangle. For each segment, we determine the parameterization that respects the direction of traversal.1. Segment 1: From \( (-1, 0) \) to \( (1, 0) \): \[ \vec{r}_1(t) = (t, 0, 0), -1 \leq t \leq 1 \]2. Segment 2: From \( (1, 0) \) to \( (0, 1) \): \[ \vec{r}_2(t) = (1-t, t, (1-t)t), 0 \leq t \leq 1 \]3. Segment 3: From \( (0, 1) \) back to \( (-1, 0) \): \[ \vec{r}_3(t) = (-t, 1-t, -t(1-t)), 0 \leq t \leq 1 \]
02

Compute \( \vec{F} \cdot d\vec{r} \) for Each Segment

For each segment, we compute \( \vec{F}(\vec{r}(t)) \cdot d\vec{r} \).1. Segment 1: \[ \vec{F}_1 = \langle 0, t, 0 \rangle \] \[ d\vec{r}_1 = \langle 1, 0, 0 \rangle \, dt \] \[ \vec{F}_1 \cdot d\vec{r}_1 = 0 \cdot 1 + t \cdot 0 + 0 \cdot 0 = 0 \, dt \]2. Segment 2: \[ \vec{F}_2 = \langle (1-t)t - t^2, 1-t, (1-t)t \rangle \] \[ d\vec{r}_2 = \langle -1, 1, 1-t \rangle \, dt \] \[ \vec{F}_2 \cdot d\vec{r}_2 = -((1-t)t - t^2) + (1-t) + (1-t)t(1-t) \, dt \] Simplify to obtain its expression.3. Segment 3: \[ \vec{F}_3 = \langle -t(1-t) - (1-t)^2, -t, -t(1-t) \rangle \] \[ d\vec{r}_3 = \langle -1, -1, 1-t \rangle \, dt \] \[ \vec{F}_3 \cdot d\vec{r}_3 = ((1-t)^2 - t(1-t)) + t + t(1-t)(1-t) \, dt \] Simplify to obtain the expression.
03

Integrate Each Segment

Integrate \( \vec{F} \cdot d\vec{r} \) over each segment.1. Segment 1: \[ \int_{-1}^{1} 0 \, dt = 0 \]2. Segment 2: Perform the integral from 0 to 1 using the simplified result from Step 2.3. Segment 3: Perform the integral from 0 to 1 using the simplified result from Step 2.
04

Apply Stokes' Theorem (Optional Method)

If direct computation seems tedious, apply Stokes' Theorem:\[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S (abla \times \vec{F}) \cdot \vec{n} \, dS \]1. Compute the curl of \( \vec{F} \): \[ abla \times \vec{F} = \left( \frac{\partial z}{\partial y} - \frac{\partial (z-y^2)}{\partial z}, \frac{\partial (z-y^2)}{\partial x} - \frac{\partial x}{\partial z}, \frac{\partial x}{\partial y} - \frac{\partial (z-y^2)}{\partial x} \right) \]2. Since \( \vec{n} = \langle 0, 0, 1 \rangle \), only the \( z \)-component of the curl contributes: Simplify to find the integral over the region.
05

Combine Results for Total Circulation

Add the results from each integral to find the total circulation around \( C \). Direct computation gives: Combine the results from Segment 1, Segment 2, and Segment 3 to get the total circulation: \( \oint_C \vec{F} \cdot d\vec{r} = \) total sum. If using Stokes' Theorem, evaluate the surface integral instead.

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 Vaia!

Key Concepts

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

Understanding Vector Fields
When studying calculus and vector fields, it's essential to grasp what a vector field is. Picture a vector field as being like a field of arrows. In mathematical terms, a vector field assigns a vector to every point in space. These vectors can represent anything, from wind speeds in a weather model to magnetic forces around a magnet.

The vector field given in the exercise, \(\vec{F} = \langle z - y^2, x, z \rangle\), associates each point in space with a vector based on its coordinates \(x, y, z\). Here, \(z\) is dependent on \(x\) and \(y\), given by the formula \(z = xy\). This means that as you move through the three-dimensional space, the vector changes its direction and magnitude depending on the location.
  • Vector fields are commonly written as \(\vec{F} = \langle P(x, y, z), Q(x, y, z), R(x, y, z) \rangle\).
  • They are used to describe physical quantities, such as force and velocity, that have both magnitude and direction.

Understanding this representation helps in visualizing the behavior of physical systems in fields like physics and engineering.
Parameterization of a Curve
Parameterizing a curve helps in describing its movement through space with respect to a chosen parameter, typically denoted as \(t\). This process involves expressing the coordinates of points on the curve as functions of \(t\). In the exercise, the curve \(C\) is parameterized in three segments corresponding to the sides of a triangle.

Each segment of the curve \(C\) has its own parameterization:
  • Segment 1, moving from \((-1, 0)\) to \((1, 0)\), is parameterized by \( \vec{r}_1(t) = (t, 0, 0) \), where \(-1 \leq t \leq 1\).
  • Segment 2, covering the path from \((1, 0)\) to \((0, 1)\), is described as \( \vec{r}_2(t) = (1-t, t, (1-t)t) \) with \(0 \leq t \leq 1\).
  • Segment 3, returning to the start at \((-1, 0)\) from \((0, 1)\), follows \( \vec{r}_3(t) = (-t, 1-t, -t(1-t)) \) for \(0 \leq t \leq 1\).


The parameterization respects the direction of traversal and simplifies the process of calculating integrals along the curve. This technique turns the geometric problem into an algebraic one, making it easier to manipulate and solve.
Exploring Curl of a Vector Field
The curl of a vector field provides insight into the field's rotation. In simple terms, it measures how much a vector field is 'twisting' around a point. Mathematically, the curl of a vector field \(\vec{F} = \langle P, Q, R \rangle\) is described by the formula:
\[abla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\]
This operation results in another vector field. The direction of this new field represents the axis of rotation, and its magnitude indicates the speed of rotation.
  • The curl is particularly useful in fluid dynamics and electromagnetism to describe rotational motion.
  • A non-zero curl implies that the field has some form of swirling or rotational behavior.

In the exercise, calculating the curl of \( \vec{F} \) is a step towards applying Stokes' Theorem, which relates the circulation around a closed loop to the rotation within the surface enclosed by that loop.
Understanding Surface Integrals
A surface integral extends the concept of a single-variable integral to the analysis of a function over a surface. If you imagine a sheet in three-dimensional space, a surface integral evaluates how a given function interacts across that sheet.

In the context of Stokes' Theorem, a surface integral of the curl of a vector field over a surface relates to the circulation of the field along the surface boundary. Mathematically expressed as:
\[\iint_S (abla \times \vec{F}) \cdot \vec{n} \, dS\]- \(abla \times \vec{F}\) is the curl of the vector field.- \(\vec{n}\) is the unit normal vector pointing outward from the surface.- \(dS\) represents a tiny piece of the surface.

Why Use a Surface Integral?

Stokes' Theorem provides a powerful tool by simplifying complex problems:
  • Calculates circulation without breaking down into line segments.
  • Handles vector fields that can swirl around a point within the surface.
In practical applications, surface integrals help solve real-world problems in areas such as fluid flow, electromagnetism, and even some areas of finance.

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

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.

Sign-up for free