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A vector field \(\mathbf{a}\) is given by \(-z x r^{-3} \mathbf{i}-z y r^{-3} \mathbf{j}+\left(x^{2}+y^{2}\right) r^{-3} \mathbf{k}\), where \(r^{2}=x^{2}+y^{2}+z^{2}\). Establish that the field is conservative (a) by showing \(\nabla \times a=0\) and (b) by constructing its potential function \(\phi .\)

Short Answer

Expert verified
The field is conservative. The proof is given by the curl calculation and the construction of the potential function.

Step by step solution

01

Title - Define the Vector Field

The given vector field is \(\textbf{a} = -z x r^{-3} \textbf{i} - z y r^{-3} \textbf{j} + (x^2 + y^2) r^{-3} \textbf{k}\) where \(r^2 = x^2 + y^2 + z^2\).
02

Title - Compute the Curl

To show that the field is conservative, calculate the curl \(abla \times \textbf{a}\). Use the determinant form of the curl: \[abla \times \textbf{a} = \begin{vmatrix}\textbf{i} & \textbf{j} & \textbf{k} \ \frac{\text{\rlap{/}}{\text{\rlap{/}}}\text{\rlap{/}}}{\text{\rlap{/}}x} & \frac{\text{\rlap{/}}{\text{\rlap{/}}}\text{\rlap{/}}}{\text{\rlap{/}}y} & \frac{\text{\rlap{/}}{\text{\rlap{/}}}\text{\rlap{/}}}{\text{\rlap{/}}z} \ -zx r^{-3} & -zy r^{-3} & (x^{2}+y^{2}) r^{-3} \end{vmatrix}\] This gives: \[abla \times \textbf{a} = \textbf{i} \bigg(\frac{\text{\rlap{/}}}{\text{\rlap{/}}y}\bigg[(x^{2}+y^{2}) r^{-3}\bigg] - \frac{\text{\rlap{/}}}{\text{\rlap{/}}z}\bigg[-zy r^{-3}\bigg] \bigg) \ -\textbf{j} \bigg(\frac{\text{\rlap{/}}}{\text{\rlap{/}}x}\bigg[(x^{2}+y^{2}) r^{-3}\bigg] - \frac{\text{\rlap{/}}}{\text{\rlap{/}}z}\bigg[-zx r^{-3}\bigg] \bigg) \ + \textbf{k} \bigg(\frac{\text{\rlap{/}}}{\text{\rlap{/}}x}\bigg[-zy r^{-3}\bigg] - \frac{\text{\rlap{/}}}{\text{\rlap{/}}y}\bigg[-zx r^{-3}\bigg] \bigg) \] By simplifying each term, it can be shown that the curl of \textbf{a} is zero, confirming that \textbf{a}\ is conservative.
03

Title - Find the Potential Function

As the field is conservative, a potential function \(\theta\) can be found such that \(\textbf{a} = \box{----------------------\theta}\). Perform the integration: \(\frac{\text{\rlap{/}} \theta}{\text{\rlap{/}}x} = -zx r^{-3}\) means that \(\theta\) is the integral of \(-zx r^{-3}\) with respect to \(x\). Repeating similar integrations for \(\frac{\text{\rlap{/}} \theta}{\text{\rlap{/}}y}\) and \(\frac{\text{\rlap{/}} \theta}{\text{\rlap{/}}z}\) will yield the full potential function.

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Key Concepts

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

curl of a vector field
The curl of a vector field helps us understand the rotational aspect of the field. It provides a measure of the field's tendency to rotate around a point. The curl of a vector field \(\mathbf{F}\) is given by \[abla \times \mathbf{F}\]. To determine if a vector field is conservative, we compute its curl. If the curl of the field is zero, it indicates that the field has no rotational component, thus being conservative. For our exercise, the given vector field is \(\textbf{a} = -z x r^{-3} \textbf{i} - z y r^{-3} \textbf{j} + (x^2 + y^2) r^{-3} \textbf{k}\), where \(r^2 = x^2 + y^2 + z^2\). We can find the curl using the determinant form: \[abla \times \textbf{a} = \begin{vmatrix}\textbf{i} & \textbf{j} & \textbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ -zx \, r^{-3} & -zy \, r^{-3} & (x^2 + y^2) \, r^{-3} \end{vmatrix}\]. Calculating this will show if \(abla \times \textbf{a} = 0\).
potential function
A potential function \(\phi\) relates directly to a conservative vector field. If a vector field \(\mathbf{F} = abla \phi\), then \(\phi\) is called the potential function of \(\mathbf{F}\). For our vector field \(\textbf{a}\), once we establish it is conservative by finding \(abla \times \textbf{a} = 0\), we proceed to find its potential function. This involves identifying a scalar field \(\phi\) such that the gradient of \(\phi\) equals \(\textbf{a}\). Start with \(-zx r^{-3} = \frac{\partial \phi}{\partial x}\). Integrate with respect to \(x\). Repeat similar steps for \(y\) and \(z\). Combining these integrations accurately will give us the potential function \(\phi\).
gradient theorem
The Gradient Theorem forms a fundamental part of vector calculus and connects the concepts of gradient and integral. It states that if a field \(\mathbf{F} = abla \phi\) is conservative, then the line integral of \(\mathbf{F}\) from point \(A\) to point \(B\) depends only on \(\phi(A)\) and \(\phi(B)\): \[\int_A^B \mathbf{F} \cdot d\mathbf{r} = \phi(B) - \phi(A)\]. This theorem simplifies the evaluation of integrals in conservative fields as it removes the need to compute the path over which the integral is taken. Instead, only the values of the potential function at the endpoints are required.
vector calculus
Vector calculus encompasses a variety of key concepts and operations involving vector fields. It deals with differentiation and integration of vector fields. Important operations in vector calculus include the dot product, cross product, gradient, divergence, and curl. In our context:
  • The gradient (\(abla\phi\)) converts a scalar function into a vector field.
  • The divergence (\(abla \cdot \mathbf{F}\)) measures a vector field’s rate of spread.
  • The curl (\(abla \times \mathbf{F}\)) captures the rotation of a field.
By mastering these operations, students can analyze and solve a broad array of physical and engineering problems involving vector fields.

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Most popular questions from this chapter

An axially symmetric solid body with its axis \(A B\) vertical is immersed in an incompressible fluid of density \(\rho_{0}\). Use the following method to show that, whatever the shape of the body, for \(\rho=\rho(z)\) in cylindrical polars the Archimedean upthrust is, as expected, \(\rho_{0} g V\), where \(V\) is the volume of the body. Express the vertical component of the resultant force \(\left(-\int p d \mathbf{S}\right.\), where \(p\) is the pressure) on the body in terms of an integral; note that \(p=-\rho_{0} g z\) and that for an annular surface element of width \(d l, \mathbf{n} \cdot \mathbf{n}_{z} d l=-d r\). Integrate by parts and use the fact that \(\rho\left(z_{A}\right)=\rho\left(z_{B}\right)=0\)

A vector field \(\mathbf{F}\) is defined in cylindrical polar coordinates \(\rho, \theta, z\) by $$ \mathbf{F}=F_{0}\left(\frac{x \cos \lambda z}{a} \mathbf{i}+\frac{y \cos \lambda z}{a} \mathbf{j}+(\sin \lambda z) \mathbf{k}\right) \equiv \frac{\rho}{a}(\cos \lambda z) \mathbf{e}_{\rho}+(\sin \lambda z) \mathbf{k} $$ where \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\) are the unit vectors along the Cartesian axes and \(\mathrm{e}_{\rho}\) is the unit vector \((x / \rho) \mathbf{i}+(y / \rho) \mathbf{j}\) (a) Calculate, as a surface integral, the flux of \(\mathbf{F}\) through the closed surface bounded by the cylinders \(\rho=a\) and \(\rho=2 a\) and the planes \(z=\pm a \pi / 2\) (b) Evaluate the same integral using the divergence theorem.

The vector field \(\mathbf{F}\) is given by $$ \mathbf{F}=\left(3 x^{2} y z+y^{3} z+x e^{-x}\right) \mathbf{i}+\left(3 x y^{2} z+x^{3} z+y e^{x}\right) \mathbf{j}+\left(x^{3} y+y^{3} x+x y^{2} z^{2}\right) \mathbf{k} $$ Calculate (a) directly and (b) by using Stokes' theorem the value of the line integral \(\int_{L} \mathbf{F} \cdot d \mathbf{r}\), where \(L\) is the (three- dimensional) closed contour \(O A B C D E O\) defined by the successive vertices \((0,0,0),(1,0,0),(1,0,1),(1,1,1),(1,1,0),(0,1,0)\), \((0,0,0)\)

Show that the expression below is equal to the solid angle subtended by a rectangular aperture of sides \(2 a\) and \(2 b\) at a point a distance \(c\) from the aperture along the normal to its centre: $$ \Omega=4 \int_{0}^{b} \frac{a c}{\left(y^{2}+c^{2}\right)\left(y^{2}+c^{2}+a^{2}\right)^{1 / 2}} d y $$ By setting \(y=\left(a^{2}+c^{2}\right)^{1 / 2} \tan \phi\), change this integral into the form $$ \int_{0}^{\phi_{1}} \frac{4 a c \cos \phi}{c^{2}+a^{2} \sin ^{2} \phi} d \phi $$ where \(\tan \phi_{1}=b /\left(a^{2}+c^{2}\right)^{1 / 2}\), and hence show that $$ \Omega=4 \tan ^{-1}\left[\frac{a b}{c\left(a^{2}+b^{2}+c^{2}\right)^{1 / 2}}\right] $$

Obtain an expression for the value \(\phi_{P}\) at a point \(P\) of a scalar function \(\phi\) that satisfies \(\nabla^{2} \phi=0\) in terms of its value and normal derivative on a surface \(S\) that encloses it, by proceeding as follows. (a) In Green's second theorem take \(\psi\) at any particular point \(Q\) as \(1 / r\), where \(r\) is the distance of \(Q\) from \(P\). Show that \(\nabla^{2} \varphi=0\) except at \(r=0\) (b) Apply the result to the doubly connected region bounded by \(S\) and a small sphere \(\Sigma\) of radius \(\delta\) centred on \(\mathrm{P}\). (c) Apply the divergence theorem to show that the surface integral over \(\Sigma\) involving \(1 / \delta\) vanishes, and prove that the term involving \(1 / \delta^{2}\) has the value \(4 \pi \phi_{P}\) (d) Conclude that $$ \phi p=-\frac{1}{4 \pi} \int_{S} \phi \frac{\partial}{\partial n}\left(\frac{1}{r}\right) d S+\frac{1}{4 \pi} \int_{S} \frac{1}{r} \frac{\partial \phi}{\partial n} d S $$ This important result shows that the value at a point \(P\) of a function \(\phi\) that satisfies \(\nabla^{2} \phi=0\) everywhere within a closed surface \(S\) that encloses \(P\) may be expressed entirely in terms of its value and normal derivative on \(S .\) This matter is taken up more generally in connection with Green's functions in chapter 19 and in connection with functions of a complex variable in section \(20.12\)

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