Radial fields in R are conservative Prove that the radial fieldwhere r = (x, y, z) and p is a real number, is conser-vative on any region not containing the origin. For what values ofp is F conservative on a region that contains the origin?F

The given radial field is: F=r⇀r⇀p where, r⇀=x,y,z So, r⇀=x2+y2+z2 Therefore, the radial field is:…

Answered: Radial fields in R are conservative…

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 in R are conservative Prove that the radial field
where r = (x, y, z) and p is a real number, is conser-
vative on any region not containing the origin. For what values of
p is F conservative on a region that contains the origin?
F

Expert Solution

Step 1

The given radial field is:

$F = \frac{\overset{⇀}{r}}{{|\overset{⇀}{r}|}^{p}}$

where, $\overset{⇀}{r} = <x, y, z>$

So, $|\overset{⇀}{r}| = \sqrt{x^{2} + y^{2} + z^{2}}$

Therefore, the radial field is:

$F = \frac{x i + y j + z k}{{(x^{2} + y^{2} + z^{2})}^{p / 2}}$

So,

$\begin{array}{rcl} P & = & \frac{x}{{(x^{2} + y^{2} + z^{2})}^{p / 2}} \\ Q & = & \frac{y}{{(x^{2} + y^{2} + z^{2})}^{p / 2}} \\ R & = & \frac{z}{{(x^{2} + y^{2} + z^{2})}^{p / 2}} \end{array}$

Step 2

So, the partial derivatives are:

$\begin{array}{rcl} P_{y} & = & - \frac{x (\frac{p}{2}) {(x^{2} + y^{2} + z^{2})}^{\frac{p - 2}{2}} (2 y)}{{(x^{2} + y^{2} + z^{2})}^{p}} \\ P_{y} & = & - \frac{p x y}{{(x^{2} + y^{2} + z^{2})}^{\frac{^{p + 2}}{2}}} \end{array}$

Similarly,

$\begin{array}{rcl} Q_{x} & = & - \frac{p x y}{{(x^{2} + y^{2} + z^{2})}^{\frac{p + 2}{2}}} \end{array}$

$\begin{array}{rcl} P_{z} & = & - \frac{p x z}{{(x^{2} + y^{2} + z^{2})}^{\frac{p + 2}{2}}} \\ R_{x} & = & - \frac{p x z}{{(x^{2} + y^{2} + z^{2})}^{\frac{p + 2}{2}}} \end{array}$