Chapter 3, Problem 31P

(a)

To determine

The value of $\left(\nabla \cdot r\right)T$.

Explanation of Solution

Given:

The vector $\left(T\right)$ is $2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z$.

The position vector $\left(r\right)$ is $x{a}_x+y{a}_y+z{a}_z$.

Calculation:

Calculate the value of $\left(\nabla \cdot r\right)T$ using the relation.

  $\left(\nabla \cdot r\right)T=\left(\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot r\right)T$

  $\begin{array}{c}\left(\nabla \cdot r\right)T=\left(\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(x{a}_x+y{a}_y+z{a}_z\right)\right)\left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)\\ =\left(\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(y\right)+\frac{\partial }{\partial z}\left(z\right)\right)\left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)\\ =\left(3\right)\left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)\\ =6zy{a}_x+3x{y}^2{a}_y+3x^{2}yz{a}_z\end{array}$

Thus, the value of $\left(\nabla \cdot r\right)T$ is $\underset{\_}{6zy{a}_x+3x{y}^2{a}_y+3x^{2}yz{a}_z}$.

(b)

To determine

The value of $\left(r\cdot \nabla \right)T$.

Explanation of Solution

Given:

The vector $\left(T\right)$ is $2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z$.

The position vector $\left(r\right)$ is $x{a}_x+y{a}_y+z{a}_z$.

Calculation:

Calculate the value of $\left(r\cdot \nabla \right)T$ using the relation.

  $\left(r\cdot \nabla \right)T=\left(r\cdot \left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\right)T$

  $\begin{array}{c}\left(r\cdot \nabla \right)T=\left(\left(x{a}_x+y{a}_y+z{a}_z\right)\cdot \left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\right)\left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)\\ =\left(\left(x\right)\frac{\partial }{\partial x}+\left(y\right)\frac{\partial }{\partial y}+\left(z\right)\frac{\partial }{\partial z}\right)\left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)\\ =\{\begin{array}{c}\left(x\right)\frac{\partial \left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)}{\partial x}+\left(y\right)\frac{\partial \left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)}{\partial y}\\ +\left(z\right)\frac{\partial \left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)}{\partial z}\end{array}\\ =\{\begin{array}{c}\left(x\right)\left(\left(0\right)a_x+\left(y^2\right)a_y+\left(2xyz\right)a_z\right)+\left(y\right)\left(\left(2z\right)a_x+\left(2xy\right)a_y+\left(x^{2}z\right)a_z\right)\\ +\left(z\right)\left(\left(2y\right)a_x+\left(0\right)a_y+\left(x^{2}y\right)a_z\right)\end{array}\end{array}$

  $\begin{array}{c}\left(r\cdot \nabla \right)T=\left(x{y}^2{a}_y+2x^{2}yz{a}_z\right)+\left(2yz{a}_x+2x{y}^2{a}_y+x^{2}yz{a}_z\right)+\left(2yz{a}_x+x^{2}yz{a}_z\right)\\ =4yz{a}_x+3x{y}^2{a}_y+4x^{2}yz{a}_z\end{array}$

Thus, the value of $\left(r\cdot \nabla \right)T$ is $\underset{\_}{4yz{a}_x+3x{y}^2{a}_y+4x^{2}yz{a}_z}$.

(c)

To determine

The value of $\nabla \cdot r\left(r\cdot T\right)$.

Explanation of Solution

Given:

The vector $\left(T\right)$ is $2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z$.

The position vector $\left(r\right)$ is $x{a}_x+y{a}_y+z{a}_z$.

Calculation:

Calculate the value of $\nabla \cdot r\left(r\cdot T\right)$ using the relation.

  $\nabla \cdot r\left(r\cdot T\right)=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot r\left(r\cdot T\right)$

  $\begin{array}{c}\nabla \cdot r\left(r\cdot T\right)=\{\begin{array}{c}\left(\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(x{a}_x+y{a}_y+z{a}_z\right)\right)\\ \left(\left(x{a}_x+y{a}_y+z{a}_z\right)\cdot \left(2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z\right)\right)\end{array}\\ =\left(\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(y\right)+\frac{\partial }{\partial z}\left(z\right)\right)\left(\left(x\right)\left(2zy\right)+\left(y\right)\left(x{y}^2\right)+\left(z\right)\left(x^{2}yz\right)\right)\\ =\left(3\right)\left(2xzy+x{y}^3+x^{2}y{z}^2\right)\\ =6xzy+3x{y}^3+3x^{2}y{z}^2\end{array}$

Thus, the value of $\nabla \cdot r\left(r\cdot T\right)$ is $\underset{\_}{6xzy+3x{y}^3+3x^{2}y{z}^2}$.

(d)

To determine

The value of $\left(r\cdot \nabla \right)r^2$.

Explanation of Solution

Given:

The vector $\left(T\right)$ is $2zy{a}_x+x{y}^2{a}_y+x^{2}yz{a}_z$.

The position vector $\left(r\right)$ is $x{a}_x+y{a}_y+z{a}_z$.

Calculation:

Calculate scalar of position vector $\left(r\right)$ using the relation.

  $\left|r\right|=\sqrt{x^2+y^2+z^2}$

Calculate the magnitude of square of position vector $\left(r^2\right)$ using the relation.

  $r^2={\left|r\right|}^2$

  $\begin{array}{c}{r}^2={\left(\sqrt{x^2+y^2+z^2}\right)}^2\\ =\left(x^2+y^2+z^2\right)\end{array}$

Calculate the value of $\left(r\cdot \nabla \right)r^2$ using the relation.

  $\left(r\cdot \nabla \right)r^2=\left(r\cdot \left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\right)r^2$

  $\begin{array}{c}\left(r\cdot \nabla \right)r^2=\left(\left(x{a}_x+y{a}_y+z{a}_z\right)\cdot \left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\right)\left(x^2+y^2+z^2\right)\\ =\left(\left(x\right)\frac{\partial }{\partial x}+\left(y\right)\frac{\partial }{\partial y}+\left(z\right)\frac{\partial }{\partial z}\right)\left(x^2+y^2+z^2\right)\\ =\left(x\right)\frac{\partial \left(x^2+y^2+z^2\right)}{\partial x}+\left(y\right)\frac{\partial \left(x^2+y^2+z^2\right)}{\partial y}+\left(z\right)\frac{\partial \left(x^2+y^2+z^2\right)}{\partial z}\\ =\left(x\right)\left(2x\right)+\left(y\right)\left(2y\right)+\left(z\right)\left(2z\right)\end{array}$

  $\left(r\cdot \nabla \right)r^2=2x^2+2y^2+2z^2$

Thus, the value of $\left(r\cdot \nabla \right)r^2$ is $\underset{\_}{2x^2+2y^2+2z^2}$.