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

Let \(C\) and \(S\) be as in Stokes's theorem. Prove, under appropriate assumptions: a) \(\int_{C} f \mathbf{T} \cdot \mathbf{i} d s=\iint_{S} \mathbf{n} \times \nabla f \cdot \mathbf{i} d \sigma\); b) \(\int_{C} f \mathbf{T} d s=\iint_{S} \mathbf{n} \times \nabla f d \sigma\).

Short Answer

Expert verified
Question: Prove the following statements based on Stokes's theorem: a) \(\int_{C} f \mathbf{T} \cdot \mathbf{i} d s=\iint_{S} \mathbf{n} \times \nabla f \cdot \mathbf{i} d \sigma\) b) \(\int_{C} f \mathbf{T} d s = \iint_{S} \mathbf{n} \times \nabla f d \sigma\) Answer: We have proven that: a) \(\int_{C} f \mathbf{T} \cdot \mathbf{i} d s = \iint_{S} \mathbf{n} \times \nabla f \cdot \mathbf{i} d \sigma\) b) \(\int_{C} f \mathbf{T} d s = \iint_{S} \mathbf{n} \times \nabla f d \sigma\)

Step by step solution

01

Apply Stokes's theorem

According to Stokes's theorem, we have: \(\int_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (\nabla \times \mathbf{F}) \cdot \mathbf{n} d\sigma\) To make use of this theorem, we need to write \(\mathbf{F}\) in terms of the given functions and vectors.
02

Rewrite the given vectors as a force field

We can rewrite the vector function \(f\mathbf{T}\) as: \(\mathbf{F} = f \mathbf{T}\) Now, we can apply Stokes's theorem to this new force field \(\mathbf{F}\).
03

Apply Stokes's theorem to the new force field

Applying Stokes's theorem to \(\mathbf{F} = f \mathbf{T}\), we get: \(\int_{C} f \mathbf{T} \cdot d \mathbf{r} = \iint_{S} (\nabla \times (f \mathbf{T})) \cdot \mathbf{n} d\sigma\)
04

Calculate the curl of the force field

Now, we need to calculate \(\nabla \times (f \mathbf{T})\) using the properties of the gradient and cross product. We have: \(\nabla \times (f \mathbf{T}) = \nabla f \times \mathbf{T} + f(\nabla \times \mathbf{T})\)
05

Simplify the integral using the curl calculated in step 4

Using the curl we found in step 4, we can rewrite the integral as: \(\iint_{S} (\nabla f \times \mathbf{T} + f(\nabla \times \mathbf{T})) \cdot \mathbf{n} d\sigma\) Since \(f(\nabla \times \mathbf{T})\) is not relevant to part a), we can ignore it for now.
06

Rewrite the integral in terms of \(\mathbf{i}\)

Now we rewrite the remaining term of the integral in terms of \(\mathbf{i}\): \(\iint_{S} (\mathbf{n} \times \nabla f) \cdot \mathbf{i} d\sigma\) This completes the proof of part a), as we have shown that: \(\int_{C} f \mathbf{T} \cdot \mathbf{i} d s = \iint_{S} \mathbf{n} \times \nabla f \cdot \mathbf{i} d \sigma\) For b), we want to prove that: \(\int_{C} f \mathbf{T} d s = \iint_{S} \mathbf{n} \times \nabla f d \sigma\)
07

Use the result from part a)

We already have the result: \(\int_{C} f \mathbf{T} \cdot \mathbf{i} d s = \iint_{S} \mathbf{n} \times \nabla f \cdot \mathbf{i} d \sigma\)
08

Take the integral with respect to \(s\)

Now, we can take the integral with respect to \(s\) on both sides: \(\int_{C} f \mathbf{T} d s = \iint_{S} \mathbf{n} \times \nabla f d \sigma\) This completes the proof of part b), as we have shown that: \(\int_{C} f \mathbf{T} d s = \iint_{S} \mathbf{n} \times \nabla f d \sigma\)

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.

Vector Calculus
Vector calculus is a branch of mathematics that extends calculus to vector fields. It is fundamentally used to study vector quantities that vary in space. Central to vector calculus are concepts like gradient, divergence, and curl, which help in understanding changes in multivariable and vector fields.
  • Gradient: Describes how a scalar field changes at a point, indicating the steepest increase in the field.
  • Divergence: Measures a vector field's tendency to converge or diverge at a point, acting as a source or sink.
  • Curl: Reveals the rotation or swirling strength of a vector field around a point.
Vector calculus forms the foundation for advanced physics and engineering domains, such as electromagnetism and fluid dynamics. This branch ensures that vector fields are analyzed through integrals, derivatives, and coordinates, offering deep insights into spatial behaviors.
Line Integrals
Line integrals are fundamental in vector calculus, involving the integration of a function along a curve. They are used to calculate physical quantities such as work done by a force field along a path.
In a line integral, we evaluate the integral of a scalar or vector field along a designated curve, denoted as \(C\). This means summing up the field values along the path to capture the accumulated effect. For vector fields, it can be represented as:\[\int_{C} \mathbf{F} \cdot d\mathbf{r}\]where \(\mathbf{F}\) is the vector field and \(d\mathbf{r}\) is the differential path element.
  • Used in physics to compute work, flux, and circulation.
  • Interpretation depends on the vector field direction in relation to the curve.
  • Incorporates both the magnitude and direction of the field along \(C\).
By bridging scalar and vector fields along curves, line integrals allow for the analysis of dynamic and static fields in various scientific applications.
Surface Integrals
Surface integrals extend the concepts of line integrals into two dimensions. They involve integrating a function over a surface \(S\) in three-dimensional space and are crucial for understanding flux in vector fields.
The surface integral of a vector field is expressed as:\[\iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma\]where \(\mathbf{F}\) is the vector field, \(\mathbf{n}\) is the unit normal vector to the surface, and \(d\sigma\) is an infinitesimal surface area element.
  • Quantifies the total field passing through a surface.
  • Vital in contexts like fluid flow and Maxwell's equations in electromagnetism.
  • Requires the surface to be parametrized for evaluation.
Surface integrals provide an aggregate measure of a vector field across a surface, extending our ability to study complex systems and their interactions.
Curl
Curl is a vector operator used in vector calculus to describe the rotation of a three-dimensional vector field. It is an essential concept for understanding the "twist" or rotational motion within fields, often used in conjunction with Stokes's theorem.
Mathematically, the curl of a vector field \(\mathbf{F}\) is denoted as:\[abla \times \mathbf{F}\]
  • The result is a vector that points in the direction of the field's expected axis of rotation.
  • Magnitude reveals the strength of the rotation.
  • In physics, it plays a significant role in fields like fluid mechanics and electromagnetism.
The curl is instrumental in identifying rolling or rotating movement within a field, thus providing insights into flow patterns and making it indispensable in analyzing dynamic systems.

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) Prove (5.174) with the conditions stated. [Hint: Use (5.164) to obtain \(F_{\text {rad }}=\mathbf{F} \cdot \mathbf{u}\), where \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}=(x \mathbf{i}+y \mathbf{j}) / R\). Show by \((5.174)\) that \(q\) can be written as $$ \iint_{E} \mu\left[\frac{R^{2}}{r^{2}}(\xi \mathbf{i}+\eta \mathbf{j}) \cdot \mathbf{u}+R\left(1-\frac{R}{r}\right)\left(1+\frac{R}{r}\right)\right] d \xi d \eta . $$ Show by (5.173) that each of the terms inside the brackets is bounded by a constant for \(\left.R \geq r_{0}=2 d .\right]\) b) Prove (5.175) with the conditions stated. [Hint: Show by (5.175) that $$ s(x, y)=\iint_{E} \mu \frac{R^{2}}{r^{2}}(\xi \mathbf{i}+\eta \mathbf{j}) \cdot \mathbf{v} d \xi d \eta $$ and use (5.173) to show that the integrand is bounded for \(R \geq R_{0}=2 d\).]

If \(\mathbf{u}=\operatorname{grad} F\) in a domain \(D\), then show that a) \(\int_{\left(x_{1}, y_{1}, z_{1}\right)}^{\left(x_{2}, z_{2}\right)} u_{T} d s=F\left(x_{2}, y_{2}, z_{2}\right)-F\left(x_{1}, y_{1}, z_{1}\right)\), where the integral is along any path in \(D\) joining the two points; b) \(\int_{C} u_{T} d s=0\) on any closed path in \(D\).

Let \(D\) be a domain that has a finite number of "holes" at points \(A_{1}, A_{2}, \ldots, A_{k}\), so that \(D\) is ( \(k+1)\)-tuply connected; cf. Fig. 5.23. Let \(P\) and \(Q\) be continuous and have continuous derivatives in \(D\), with \(\partial P / \partial y=\partial Q / \partial x\) in \(D\). Let \(C_{1}\) denote a circle about \(A_{1}\) in \(D\), enclosing none of the other \(A^{\text {'s }}\). Let \(C_{2}\) be chosen similarly for \(A_{2}\), and so on. Let $$ \oint_{C_{1}} P d x+Q d y=\alpha_{1}, \oint_{C_{2}} P d x+Q d y=\alpha_{2} \ldots, \oint_{C_{k}} P d x+Q d y=\alpha_{k} $$ a) Show that if \(C\) is an arbitrary simple closed path in \(D\) enclosing \(A_{1}, A_{2} \ldots, A_{k}\), then $$ \oint_{C} P d x+Q d y=\alpha_{1}+\alpha_{2}+\cdots+\dot{u}_{k} $$ b) Determine all possible values of the integral $$ \int_{\left(x_{1}, y_{1}\right)}^{\left(x_{2}, y_{2}\right)} P d x+Q d y $$ between two fixed points of \(D\), if it is known that this integral has the value \(K\) for one particular path. 9\. Let \(P\) and \(Q\) be continuous and have continuous derivatives, with \(\partial P / \partial y=\partial Q / \partial x\), except at the points \((4,0),(0,0),(-4,0)\). Let \(C_{1}\) denote the circle \((x-2)^{2}+y^{2}=9\); let \(C_{2}\) denote the circle \((x+2)^{2}+y^{2}=9\); let \(C_{3}\) denote the circle \(x^{2}+y^{2}=25\). Given that $$ \oint_{C_{1}} P d x+Q d y=11, \quad \oint_{C_{2}} P d x+Q d y=9, \quad \oint_{C_{3}} P d x+Q d y=13 $$ find $$ \int_{C_{4}} P d x+Q d y $$ «s where \(C_{4}\) is the circle \(x^{2}+y^{2}=1\). [Hint: Use the result of Problem \(8(\) a).]

Evaluate by the divergence theorem: a) \(\iint_{S} x d y d z+y d z d x+z d x d y\), where \(S\) is the sphere \(x^{2}+y^{2}+z^{2}=1\) and \(\mathbf{n}\) is the outer normal; b) \(\iint_{S} v_{n} d \sigma\), where \(\mathbf{v}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}, \mathbf{n}\) is the outer normal and \(S\) is the surface of the cube \(0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\); c) \(\iint_{S} e^{y} \cos z d y d z+e^{x} \sin z d z d x+e^{x} \cos y d x d y\), with \(S\) and \(\mathbf{n}\) as in (a); d) \(\iint_{S} \nabla F \cdot \mathbf{n} d \sigma\) if \(F=x^{2}+y^{2}+z^{2}, \mathbf{n}\) is the exterior normal, and \(S\) bounds a solid region \(R\); e) \(\iint_{S} \nabla F \cdot \mathbf{n} d \sigma\) if \(F=2 x^{2}-y^{2}-z^{2}\), with \(\mathbf{n}\) and \(S\) as in (d); f) \(\iint_{S} \nabla F \cdot \mathbf{n} d \sigma\) if \(F=\left[(x-2)^{2}+y^{2}+z^{2}\right]^{-1 / 2}\) and \(S\) and \(\mathbf{n}\) are as in (a).

Let a path (5.1) be given and let a change of parameter be made by an equation \(t=\) \(g(\tau), \alpha \leq \tau \leq \beta\), where \(g^{\prime}(\tau)\) is continuous and positive in the interval and \(g(\alpha)=\) \(h, g(\beta)=k \cdot\) As in \((5.4)\) the line integral \(\int f(x, y) d x\) on the path \(x=\phi(g(\tau)), y=\psi(g(\tau))\) is given by $$ \int_{\alpha}^{\beta} f[\phi(g(\tau)), \psi(g(\tau))] \frac{d}{d \tau} \phi(g(\tau)) d \tau . $$ Show that this equals the integral in (5.4), so that such a change of parameter does not affect the value of the line integral.

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