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Chapter 5: Problem 8

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: \(\int_{C} f \mathbf{T} d s = \iint_{S} \mathbf{n} \times \nabla f d \sigma\)

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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.

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Most popular questions from this chapter

Evaluate the following line integrals: a) \(\oint_{C} y^{2} d x+x y d y\), where \(C\) is the square with vertices \((1,1),(-1,1),(-1,-1)\), \((1,-1)\) b) \(\oint_{C} y d x-x d y\), where \(C\) is the circle \(x^{2}+y^{2}=1(\) cf. (5.24)); c) \(\oint_{C} x^{2} y^{2} d x-x y^{3} d y\), where \(C\) is the triangle with vertices \((0,0),(1,0),(1,1)\).

Let \(F(x, y)=x^{2}-y^{2}\). Evaluate a) \(\int_{(0,0)}^{(2,8)} \nabla F \cdot d \mathbf{r}\) on the curve \(y=x^{3}\); b) \(\oint \frac{\partial F}{\partial n} d s\) on the circle \(x^{2}+y^{2}=1\), if \(\mathbf{n}\) is the outer normal and \(\frac{\partial F}{\partial n}=\nabla F \cdot \mathbf{n}\) is the directional derivative of \(F\) in the direction of \(\mathbf{n}\) (Section 2.14).

Transform the integrals, using the substitution given: a) \(\int_{0}^{1} \int_{0}^{y}\left(x^{2}+y^{2}\right) d x d y, u=y, v=x\); b) \(\iint_{R_{T y}}(x-y) d x d y\), where \(R_{x y}\) is the region \(x^{2}+y^{2} \leq 1\), and \(x=u+\left(1-u^{2}-v^{2}\right)\), \(y=v+\left(1-u^{2}-v^{2}\right)\); (Hint: Use as \(R_{u v}\) the region \(u^{2}+v^{2} \leq 1\).) c) \(\iint_{R_{x y}} x y d x d y\), where \(R_{x y}\) is the region \(x^{2}+y^{2} \leq 1\) and \(x=u^{2}-v^{2}, y=2 u v\).

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 \(R \geq R_{0}=2 d\).]

(Degree of mapping of one surface into another) Let \(S_{w v w}\) and \(S_{x y z}\) be surfaces forming the boundaries of regions \(R_{u v w}\) and \(R_{x y z}\) respectively: it is assumed that \(R_{u v w}\) and \(R_{x y}\). are bounded and closed and that \(R_{x y z}\) is simply connected. Let \(S_{u v w}\) and \(S_{x y z}\) be oriented by the outer normal. Let \(s, t\) be parameters for \(S_{u v w}\) : $$ u=u(s, t), \quad v=v(s, t), \quad w=w(s, t), $$ the normal having the direction of $$ \left(u_{s} \mathbf{i}+v_{s} \mathbf{j}+w_{s} \mathbf{k}\right) \times\left(u_{\mathbf{i}} \mathbf{i}+v_{t} \mathbf{j}+w_{t} \mathbf{k}\right) $$ Let $$ x=x(u, v, w), \quad y=y(u, v, w), \quad z=z(u, v, w) $$ be functions defined and having continuous derivatives in a domain containing \(S_{u v w}\), an let these equations define a mapping of \(S_{u v w}\) into \(S_{x y z}\). The degree \(\delta\) of this mapping i defined as \(1 / 4 \pi\) times the solid angle \(\Omega(O, S)\) of the image \(S\) of \(S_{u v w}\) with respect t a point \(O\) interior to \(S_{x y z}\). If \(O\) is the origin, the degree is hence given by the integra (Kronecker integral) $$ \delta=\frac{1}{4 \pi} \iint_{R_{u t}}\left|\begin{array}{ccc} x & y & z \\ \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s} & \frac{\partial z}{\partial s} \\ \frac{\partial x}{\partial t} & \frac{\partial y}{\partial t} & \frac{\partial z}{\partial t} \end{array}\right| \frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}} d s d t $$ where \(x, y, z\) are expressed in terms of \(s, t\) by (a) and (b). It can be shown that \(\delta\), as thu defined, is independent of the choice of the interior point \(O\), that \(\delta\) is a positive or negativ integer or zero, and that \(\delta\) does measure the effective number of times that \(S_{x y z}\) is covered Let \(S_{k v w}\) be the sphere \(u=\sin s \cos t, v=\sin s \sin t, w=\cos s, 0 \leq s \leq \pi, 0\) \(t \leq 2 \pi\). Let \(S_{x y z}\) be the sphere \(x^{2}+y^{2}+z^{2}=1\). Evaluate the degree for the followin; mappings of \(S_{u v w}\) into \(S_{x y z}\); a) \(x=v, y=-w, z=u\) b) \(x=u^{2}-v^{2}, y=2 u v, z=w \sqrt{2-w^{2}}\)
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