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Exercises \(21-24\) are designed to challenge your understanding and require no computation. (a) Green's Theorem can be used to find the area of a region enclosed by a curve by evaluating a line integral with the appropriate choice of vector field \(\vec{F}\). What condition on \(\vec{F}\) makes this possible? (b) Likewise, Stokes' Theorem can be used to find the surface area of a region enclosed by a curve in space by evaluating a line integral with the appropriate choice of vector field \(\vec{F}\). What condition on \(\vec{F}\) makes this possible?

Short Answer

Expert verified
(a) Green's: \( \vec{F} = (-y, x) \). (b) Stokes': \( \nabla \times \vec{F} = \hat{n} \).

Step by step solution

01

Understanding Green's Theorem

Green's Theorem relates a line integral around a simple, closed curve in a plane to a double integral over the plane region it encloses. It states that for the vector field \( \vec{F} = (P, Q) \), the integral \( \oint_C \vec{F} \cdot d\vec{r} = \iint_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \; dA \) holds, where \( R \) is the region enclosed by \( C \).
02

Condition for Green's Theorem to Find Area

To use Green's Theorem to find the area of the region \( R \), we set the vector field \( \vec{F} = (-y, x) \). This choice satisfies the condition that \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 \), leading the integral to equal twice the area of \( R \). Thus, the area can be found by \( \frac{1}{2} \oint_C (-y, x) \cdot d\vec{r} \).
03

Understanding Stokes' Theorem

Stokes' Theorem extends Green's Theorem to three dimensions and relates a surface integral over the surface \( S \) to a line integral over its boundary curve \( C \). It states \( \oint_C \vec{F} \cdot d\vec{r} = \iint_S (abla \times \vec{F}) \cdot d\vec{S} \), where \( \vec{F} \) is a vector field and \( abla \times \vec{F} \) is the curl of \( \vec{F} \).
04

Condition for Stokes' Theorem to Find Surface Area

To use Stokes' Theorem for finding the surface area of \( S \), we choose \( \vec{F} \) such that \( abla \times \vec{F} = \hat{n} \), where \( \hat{n} \) is the unit normal to the surface \( S \). This ensures that the integral becomes the surface area of \( S \), allowing it to be evaluated using the line integral over \( C \).

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Key Concepts

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

Green's Theorem
Green's Theorem provides a fascinating connection between a line integral around a closed curve in the plane and a double integral over the region it encloses. This theorem is particularly helpful in converting a usually complex line integral into a potentially simpler double integral. Given a vector field \( \vec{F} = (P, Q) \), Green's Theorem can be expressed as: \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] Here, \( C \) is the positively oriented, simple, closed curve, and \( R \) is the region bounded by \( C \). The beauty of this theorem lies in its application to find the area of the region \( R \). By choosing the vector field \( \vec{F} = (-y, x) \), the term \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \) equals 2, which makes the integral directly twice the area of \( R \). Thus, the area can be calculated as: \[ \text{Area of } R = \frac{1}{2} \oint_C (-y, x) \cdot d\vec{r} \] This clever use of Green's Theorem simplifies the computation of areas significantly.
Stokes' Theorem
Stokes' Theorem elegantly extends Green's Theorem to three-dimensional spaces. It relates a surface integral over a surface \( S \) to a line integral around its boundary \( C \). The theorem states: \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S (abla \times \vec{F}) \cdot d\vec{S} \] where \( \vec{F} \) is a vector field, and \( abla \times \vec{F} \) is the curl of \( \vec{F} \). This intricate relationship simplifies the task of evaluating complex surface integrals. In the context of finding surface areas, Stokes' Theorem can be applied with a strategic choice of \( \vec{F} \) so that \( abla \times \vec{F} = \hat{n} \), where \( \hat{n} \) is the unit normal to the surface \( S \). When this condition is satisfied, the surface integral computes directly as the surface area, and the line integral around the boundary \( C \) provides the required result. This application showcases the theorem's utility in breaking down three-dimensional geometry into manageable line integrals.
Line Integrals
Line integrals are an essential concept in calculus, especially in the study of vector fields. They allow for the integration of functions or vector fields along a curvilinear path or a loop. When dealing with a line integral in the context of vector fields, it can be represented as: \[ \oint_C \vec{F} \cdot d\vec{r} \] where \( \vec{F} \) is the vector field through which the line integral is evaluated, \( \vec{r} \) is the parameterization of the curve \( C \), and \( d\vec{r} \) represents the differential element along the curve. Line integrals are crucial for understanding how a vector field behaves around and within a given path. In concepts like Green's and Stokes' Theorems, line integrals facilitate the transformation of different types of integrals (like surface or double integrals) into more convenient forms. This transformation often unravels complex spatial relationships and simplifies calculations in diverse applications like fluid flow and electromagnetism.
Vector Fields
Vector fields are fundamental elements in multivariable calculus, representing a function that assigns each point in space a vector. They are essential for describing velocities, forces, and directions, making them pivotal in physics and engineering. Mathematically, a vector field \( \vec{F} \) in three dimensions is expressed as: \[ \vec{F}(x, y, z) = P(x, y, z) \hat{i} + Q(x, y, z) \hat{j} + R(x, y, z) \hat{k} \] where \( P, Q, \) and \( R \) are functions describing the components of the vector field in the \( x, y, \) and \( z \) directions, respectively. Understanding vector fields helps in visualizing how they change across a region or space, influencing phenomena such as fluid dynamics and magnetic or electric fields. In calculus, analyzing concepts such as divergence, curl, and understanding theorems like Green's and Stokes' requires a firm grasp of vector fields. These concepts enable us to comprehend how vector quantities move, rotate, or diverge across space, providing insights into a variety of real-world conditions and systems.

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Most popular questions from this chapter

A domain \(D\) in space is given. Parametrize each of the bounding surfaces of \(D\). \(D\) is the domain bounded by \(y=1-z^{2}, y=1-x^{2}, x=0\), \(y=0\) and \(z=0\).

A domain \(D\) in space is given. Parametrize each of the bounding surfaces of \(D\). \(D\) is the domain bounded by the cone \(x^{2}+y^{2}=(z-1)^{2}\) and the plane \(z=0\).

In Exercises \(13-16,\) a vector field \(\vec{F}\) and a closed curve \(C,\) enclosing a region \(R,\) are given. Verify Green's Theorem by evaluating \(\oint_{C} \vec{F} \cdot d \vec{r}\) and \(\iint_{R}\) curl \(\vec{F} d A,\) showing they are equal. \(\vec{F}=\langle x-y, x+y\rangle ; C\) is the closed curve composed of the parabola \(y=x^{2}\) on \(0 \leq x \leq 2\) followed by the line segment from (2,4) to (0,0).

Exercises \(21-24\) are designed to challenge your understanding and require no computation. Stokes' Theorem establishes equality between a particular line integral and a particular double integral. What types of circumstances would lead one to choose to evaluate the double integral over the line integral?

In Exercises \(9-12,\) a closed curve \(C\) that is the boundary of a surface \(S\) is given along with a vector field \(\vec{F}\). Verify Stokes' Theorem on \(C ;\) that is, show \(\oint_{c} \vec{F} \cdot d \vec{r}=\iint_{S}(\operatorname{curl} \vec{F}) \cdot \vec{n} d S\). $$ \begin{aligned} &C \text { is the curve parametrized by } \vec{r}(t)=\langle\cos t, \sin t, 1\rangle \text { and } \mathcal{S}\\\ &\text { is the portion of } z=x^{2}+y^{2} \text { enclosed by } c ; \vec{F}=(z,-x, y) \text { . } \end{aligned} $$

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