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Let \(\vec{F}\) be a conservative field and let \(C\) be a closed curve. Why are we able to conclude that \(\oint_{c} \vec{F} \cdot d \vec{r}=0 ?\)

Short Answer

Expert verified
In a conservative field, \\( \oint_{C} \vec{F} \cdot d\vec{r} = 0 \\\) because the integral depends only on endpoints, which are identical for closed curves.

Step by step solution

01

Understanding Conservative Fields

A conservative vector field is one where there exists a scalar potential function, \( V(x, y, z) \), such that \( \vec{F} = abla V \). This means that the field can be expressed as the gradient of some function. In a conservative field, the work done by the field along any path depends only on the endpoints, not on the path itself.
02

Applying the Fundamental Theorem for Line Integrals

The fundamental theorem for line integrals states that if \( \vec{F} = abla V \) for some potential function \( V \), then the line integral of \( \vec{F} \) along a curve \( C \) from point \( A \) to point \( B \) is equal to \( V(B) - V(A) \).
03

Evaluating the Integral Over a Closed Curve

Since \( C \) is a closed curve, the starting and ending points of the curve are the same. Therefore, in a conservative field, the integral of \( \vec{F} \) around the closed curve is \( V(B) - V(A) = V(A) - V(A) = 0 \). This means that for a closed curve in a conservative field, the line integral is zero.

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Key Concepts

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

Scalar Potential Function
In vector calculus, a scalar potential function is a key concept when dealing with conservative vector fields. If a vector field \( \vec{F} \) can be expressed as the gradient of a scalar potential function \( V(x, y, z) \), we call the vector field conservative. This is denoted by \( \vec{F} = abla V \). The potential function \( V \) is a scalar field, which means it assigns a single real value to each point in space. This value corresponds to the field's potential energy at that point.
In essence, knowing the potential function allows us to deduce the behavior of the vector field by applying the gradient operator, which gives us a vector field representing the rate and direction of change of \( V \). A critical attribute of these fields is that the line integral or the "work done" by the field over a path depends only on the start and end points, not the specific path taken. This characteristic significantly simplifies the calculations in various fields of physics and engineering.
Gradient
The gradient is a vector operation that plays an indispensable role in understanding scalar fields. When applied to a scalar potential function \( V \), it creates a vector field. This is represented as \( abla V \), which measures the rate of change of the function \( V \) in the space direction. The gradient points in the direction of the greatest increase of the function, and its magnitude gives the rate of the increase.
Let's break it down:
  • Suppose we have a potential function \( V = V(x, y, z) \). The gradient at any point is a vector composed of partial derivatives: \( abla V = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right) \).
  • Intuitively, you can think of the gradient as an arrow pointing uphill if the function represents a "landscape" of elevation.
The concept of gradient is forward-thinking because it helps to determine the direction of force or field in physical processes manifested by conservative fields. Thus, the gradient directly links the scalar potential function with the vector field that represents it.
Fundamental Theorem for Line Integrals
The Fundamental Theorem for Line Integrals is a crucial theorem to understand when working with conservative vector fields. It provides a method to calculate the line integral of a vector field over a path by using a scalar potential function.
Simply put, for a conservative vector field \( \vec{F} = abla V \) and any curve \( C \) connecting points \( A \) and \( B \), the line integral of \( \vec{F} \) over \( C \) is given by the difference in the potential function between these points:
  • \( \int_{C} \vec{F} \cdot d\vec{r} = V(B) - V(A) \).
This theorem highlights the elegance of conservative fields where the integral computation simplifies significantly. The theorem effectively tells us that when \( C \) is a closed curve (where \( A = B \)), the line integral equals zero because \( V(B) - V(A) = 0 \). This result is often leveraged in fields involving energy conservation and dynamics.
Closed Curve Line Integral
When dealing with line integrals over closed curves, especially in the context of conservative vector fields, an important result emerges: the integral is zero.
A closed curve implies that the same start and end points are used for the path traced by the curve. In the scenario of a conservative vector field, because the vector field can be expressed by a scalar potential function \( \vec{F} = abla V \), the line integral around a closed path is always zero. This is succinctly expressed as:
  • \( \oint_{C} \vec{F} \cdot d\vec{r} = 0 \)
Why does this happen? It's because the line integral's value is effectively determined by the difference in the potential at the start and end points of the path. Since these points are the same in a closed curve, their potential values cancel each other out. This result has profound implications in theoretical physics and engineering, where it's used to simplify complex systems, ensuring energy conservation and enabling sleek calculations.

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Most popular questions from this chapter

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 given by \(\vec{r}(t)=\) \(\langle\cos t, 3 \sin t\rangle\) and the \(z\) -values are determined by the function \(z=5-2 x-y ; \vec{F}=\left\langle-\frac{1}{3} y, 3 x, \frac{2}{3} y-3 x\right\rangle\)

In Exercises \(21-24,\) a vector field \(\vec{F}\) and a closed curve \(C,\) enclosing a region \(R,\) are given. Verify the Divergence Theorem by evaluating \(\oint_{C} \overrightarrow{\vec{F}} \cdot \vec{n} d s\) and \(\iint_{R} \operatorname{div} \vec{F} d A,\) 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).

Compute the flux of \(\vec{F}\) across \(\mathcal{S}\). (If \(\mathcal{S}\) is not a closed surface, choose \(\vec{n}\) so that it has a positive z-component, unless otherwise indicated.) \(\mathcal{S}\) is the closed surface composed of \(\mathcal{S}_{1},\) whose boundary is the ellipse in the \(x-y\) plane described by \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) and \(\mathcal{S}_{2},\) part of the elliptical paraboloid \(f(x, y)=1-\frac{x^{2}}{25}-\frac{y^{2}}{9}\) (see graph); \(\vec{F}=\langle 5,2,3\rangle .\)

In Exercises \(7-12,\) a vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{c} \vec{F} \cdot d \vec{r} .\) $$ \begin{aligned} &\vec{F}=\langle x, x+y\rangle ; C \text { is the portion of the parabola } y=x^{2} \text { from }\\\ &(0,0) \text { to }(1,1) \end{aligned} $$

Exercises \(21-24\) are designed to challenge your understanding and require no computation. Let \(\mathcal{S}\) be any closed surface enclosing a domain \(D\). Consider \(\vec{F}_{1}=\langle x, 0,0\rangle\) and \(\vec{F}_{2}=\left\langle y, y^{2}, z-2 y z\right\rangle\) These fields are clearly very different. Why is it that the total outward flux of each field across \(\mathcal{S}\) is the same?

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