Chapter 3, Problem 30P

(a)

To determine

To prove the expression: $\nabla \cdot \left(VA\right)=V\nabla \cdot A+A\cdot \nabla V$.

Explanation of Solution

Given:

The vector field is $A$.

The scalar field is $V$.

Calculation:

Consider the vector field $\left(A\right)$,

  $A=A_x{a}_x+A_y{a}_y+A_z{a}_z$

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

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

  $\begin{array}{c}\nabla \cdot \left(VA\right)=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(V\left(A_x{a}_x+A_y{a}_y+A_z{a}_z\right)\right)\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(V{A}_x{a}_x+V{A}_y{a}_y+V{A}_z{a}_z\right)\\ =\frac{\partial \left(V{A}_x\right)}{\partial x}+\frac{\partial \left(V{A}_y\right)}{\partial y}+\frac{\partial \left(V{A}_z\right)}{\partial z}\\ =A_x\frac{\partial V}{\partial x}+V\frac{\partial A_x}{\partial x}+A_y\frac{\partial V}{\partial y}+V\frac{\partial A_y}{\partial y}+A_z\frac{\partial V}{\partial z}+V\frac{\partial A_z}{\partial z}\end{array}$

  $\nabla \cdot \left(VA\right)=V\left(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)+A_x\frac{\partial V}{\partial x}+A_y\frac{\partial V}{\partial y}+A_z\frac{\partial V}{\partial z}$        (I)

Calculate the value of $V\nabla \cdot A+A\cdot \nabla AV$ using the relation.

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

  $\begin{array}{c}V\nabla \cdot A+A\cdot \nabla V=\{\begin{array}{c}V\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(A_x{a}_x+A_y{a}_y+A_z{a}_z\right)\\ +\left(A_x{a}_x+A_y{a}_y+A_z{a}_z\right)\cdot \left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)V\end{array}\\ =\{\begin{array}{c}V\left(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)\\ +\left(A_x{a}_x+A_y{a}_y+A_z{a}_z\right)\cdot \left(\frac{\partial V}{\partial x}{a}_x+\frac{\partial V}{\partial y}{a}_y+\frac{\partial V}{\partial z}{a}_z\right)\end{array}\\ =V\left(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)+A_x\frac{\partial V}{\partial x}+A_y\frac{\partial V}{\partial y}+A_z\frac{\partial V}{\partial z}\end{array}$

  $V\nabla \cdot A+A\cdot \nabla V=V\left(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)+A_x\frac{\partial V}{\partial x}+A_y\frac{\partial V}{\partial y}+A_z\frac{\partial V}{\partial z}$        (II)

The equation (I) and equation (II) are equal.

Thus, it is proved that the expression: $\nabla \cdot \left(VA\right)=V\nabla \cdot A+A\cdot \nabla V$.

(b)

To determine

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

Explanation of Solution

Given:

The vector field $\left(A\right)$ is $2x{a}_x+3y{a}_y-4z{a}_z$.

The scalar field $\left(V\right)$ is $xyz$.

Calculation:

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

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

  $\begin{array}{c}\nabla \cdot \left(VA\right)=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(\left(xyz\right)\left(2x{a}_x+3y{a}_y-4z{a}_z\right)\right)\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(2x^{2}yz{a}_x+3x{y}^{2}z{a}_y-4xy{z}^2{a}_z\right)\\ =\frac{\partial \left(2x^{2}yz\right)}{\partial x}+\frac{\partial \left(3x{y}^{2}z\right)}{\partial y}+\frac{\partial \left(-4xy{z}^2\right)}{\partial z}\\ =4xyz+6xyz-8xyz\end{array}$

  $\nabla \cdot \left(VA\right)=2xyz$

Thus, the value of $\nabla \cdot \left(VA\right)$ is $\underset{\_}{2xyz}$.