Answered: Radial fields and zero circulation…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Radial fields and zero circulation Consider the radial vector
fields F = r/ | r | ^p, where p is a real number and r = ⟨x, y, z⟩ .
Let C be any circle in the xy-plane centered at the origin.
a. Evaluate a line integral to show that the field has zero circulation on C.
b. For what values of p does Stokes’ Theorem apply? For those values of p, use the surface integral in Stokes’ Theorem to show that the field has zero circulation on C.

Expert Solution

Step 1

Given the vector field $F = \frac{r}{{|r|}^{p}}$ where $p$ is a real number and $r = <x, y, z>$ .

The closed curve $C$ is any circle in the $x y$ -plane centered at the origin.

We have to evaluate a line integral to show that the field has zero circulation on $C$ and find the values of $p$ for which Stoke's theorem applies.

Step 2

Given that the closed curve $C$ is any circle in the $x y$ -plane centered at the origin.

The equation of the circle centered at the origin and has radius $a$ is $x^{2} + y^{2} = a^{2}$ .

The parametric equation of the circle $r (t) = <a \cos t, a \sin t, 0>$ for $0 \leq t \leq 2 π$ .

$\begin{array}{rcl} |r (t)| & = & \sqrt{{(a \cos t)}^{2} + {(a \sin t)}^{2} + 0^{2}} \\ = & \sqrt{a^{2} \cos^{2} t + a^{2} \sin^{2} t} \\ = & \sqrt{a^{2} (\cos^{2} t + \sin^{2} t)} \\ = & \sqrt{a^{2}} \\ |r (t)| & = & a \end{array}$

And,

$r' (t) = <\frac{d}{d t} (a \cos t), \frac{d}{d t} (a \sin t), \frac{d}{d t} (0)> r' (t) = <- a \sin t, a \cos t, 0>$

Here,

$\begin{array}{rcl} F & = & \frac{r}{{|r|}^{p}} \\ = & \frac{<a \cos t, a \sin t, 0>}{a^{p}} \\ F & = & \frac{1}{a^{p}} <a \cos t, a \sin t, 0> \end{array}$

The line integral,

$\begin{array}{rcl} \oint_{C} F . d r & = & \int_{0}^{2 π} F . r' (t) d t \\ = & \int_{0}^{2 π} \frac{1}{a^{p}} <a \cos t, a \sin t, 0> . <- a \sin t, a \cos t, 0> d t \\ = & \frac{1}{a^{p}} \int_{0}^{2 π} (- a^{2} \sin t \cos t + a^{2} \sin t \cos t) d t \\ = & \frac{1}{a^{p}} \int_{0}^{2 π} 0 . d t \\ \oint_{C} F . d r & = & 0 \end{array}$

Thus, the value of the line integral is $\begin{array}{rcl} \oint_{C} F . d r & = & 0 \end{array}$ .

Therefore, the vector field has zero circulation on $C$ .

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