Chapter MA, Problem 1.15MA

(i)

To determine

To obtain: The value of $\nabla \cdot A$.

Answer to Problem 1.15MA

The value of $\nabla \cdot A$ is $\underset{\_}{2y{a}_x+3z{a}_y+5a_z}$.

Explanation of Solution

Definition used:

The vector differential operator is basically called the del operator $(\nabla )$, which in cartesian coordinate, is formulated as $\nabla =\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z$.

Given:

The vectors are given as $A=2xy{a}_x+3zy{a}_y+5z{a}_z$ and $B=\sin x{a}_x+2y{a}_y+5y{a}_z$.

Obtain $\nabla \cdot A$ as follows.

$\begin{array}{c}\nabla \cdot A=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)\\ =\frac{\partial }{\partial x}\left(2xy\right)a_x+\frac{\partial }{\partial y}\left(3zy\right)a_y+\frac{\partial }{\partial z}\left(5z\right)a_z\\ =2y{a}_x+3z{a}_y+5a_z\end{array}$

Therefore, the value of $\nabla \cdot A$ is $\underset{\_}{2y{a}_x+3z{a}_y+5a_z}$.

(ii)

To determine

To obtain: The value of $\nabla \times A$.

Answer to Problem 1.15MA

The value of $\nabla \times A$ is $\underset{\_}{-3y{a}_x-2x{a}_z}$.

Explanation of Solution

The vectors are given as $A=2xy{a}_x+3zy{a}_y+5z{a}_z$ and $B=\sin x{a}_x+2y{a}_y+5y{a}_z$.

Obtain $\nabla \times A$ as follows.

$\begin{array}{c}\nabla \times A=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 2xy& 3zy& 5z\end{array}\right|\\ =\left[\frac{\partial }{\partial y}\left(5z\right)-\frac{\partial }{\partial z}\left(3zy\right)\right]a_x-\left[\frac{\partial }{\partial x}\left(5z\right)-\frac{\partial }{\partial z}\left(2xy\right)\right]a_y+\left[\frac{\partial }{\partial x}\left(3zy\right)-\frac{\partial }{\partial y}\left(2xy\right)\right]a_z\\ =\left[0-3y\right]a_x-\left[0-0\right]a_y+\left[0-2x\right]a_z\\ =-3y{a}_x-2x{a}_z\end{array}$

Therefore, the value of $\nabla \times A$ is $\underset{\_}{-3y{a}_x-2x{a}_z}$.

(iii)

To determine

To obtain: The value of $\nabla \cdot \nabla \times A$.

Answer to Problem 1.15MA

The value of $\nabla \cdot \nabla \times A$ is $\underset{\_}{0}$.

Explanation of Solution

From part (b), the value of $\nabla \times A$ is $-3y{a}_x-2x{a}_z$.

Obtain $\nabla \cdot \nabla \times A$ as follows.

$\begin{array}{c}\nabla \cdot \nabla \times A=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(-3y{a}_x-2x{a}_z\right)\\ =\frac{\partial }{\partial x}\left(-3y\right)a_x-\frac{\partial }{\partial z}\left(2x\right)a_z\\ =\left(0\right)a_x-\left(0\right)a_z\\ =0\end{array}$

Therefore, the value of $\nabla \cdot \nabla \times A$ is $\underset{\_}{0}$.

(iv)

To determine

To obtain: The value of $\nabla \cdot A\times B$.

Answer to Problem 1.15MA

The value of $\nabla \times A$ is $\underset{\_}{-3y{a}_x-2x{a}_z}$.

Explanation of Solution

The vectors are given as $A=2xy{a}_x+3zy{a}_y+5z{a}_z$ and $B=\sin x{a}_x+2y{a}_y+5y{a}_z$.

Obtain $A\times B$ as follows.

$\begin{array}{c}A\times B=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 2xy& 3zy& 5z\\ \sin x& 2y& 5y\end{array}\right|\\ =\left[3zy\left(5z\right)-5z\left(2y\right)\right]a_x-\left[2xy\left(5z\right)-5z\left(\sin x\right)\right]a_y+\left[2xy\left(2y\right)-3zy\left(\sin x\right)\right]a_z\\ =\left[15y{z}^2-10yz\right]a_x-\left[10xyz-5z\sin x\right]a_y+\left[4x{y}^2-3zy\sin x\right]a_z\end{array}$

Obtain $\nabla \cdot A\times B$ as follows.

$\begin{array}{c}\nabla \cdot A\times B=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(\left[15y{z}^2-10yz\right]a_x-\left[10xyz-5z\sin x\right]a_y+\left[4x{y}^2-3zy\sin x\right]a_z\right)\\ =\frac{\partial }{\partial x}\left(15y{z}^2-10yz\right)a_x-\frac{\partial }{\partial y}\left(10xyz-5z\sin x\right)a_y+\frac{\partial }{\partial z}\left(4x{y}^2-3zy\sin x\right)a_z\\ =0a_x-\left(10xz-0\right)a_y+\left(0-3y\sin x\right)a_z\\ =-10xz{a}_y-3y\sin x{a}_z\end{array}$

Therefore, the value of $\nabla \cdot A\times B$ is $\underset{\_}{-10xz{a}_y-3y\sin x{a}_z}$.