(a) Let${\mathit{F}}_{\mathbf{1}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathit{i}$and${\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathit{x}\mathit{i}\mathbf{+}\mathit{y}\mathit{j}\mathbf{+}\mathit{z}\mathit{k}$Calculate the divergence and curl of${\mathit{F}}_{\mathbf{1}}$and${\mathit{F}}_{\mathbf{2}}$which one can be written as the gradient of a scalar? Find a scalar potential that does the job. Which one can be written as the curl of a vector? Find a suitable vector potential.

(b) Show thatlocalid="1654510098914" ${\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathit{y}\mathit{z}\mathbf{}\mathit{i}\mathbf{+}\mathit{z}\mathit{x}\mathbf{}\mathit{j}\mathbf{+}\mathit{x}\mathit{y}\mathbf{}\mathit{k}$can be written both as the gradient of a scalar and as the curl of a vector. Find scalar and vector potentials for this function.

Write the given vectors.

$$\begin{gathered} F_1=x^{2}k \\ {F}_2=xi+yj+zk \\ {F}_3=yzi+zxj+xyk \end{gathered}$$

The gradient of a scalar function$\mathit{F}$is defined as$\mathbf{\nabla }\mathit{F}$and curl of a vector function is defined as $\mathbf{\nabla }\mathbf{\times }\mathit{V}$ .

(a)

The dot product of vector R is obtained as follows:

$\nabla .R=[\frac{\partial r_1}{\partial x}+\frac{\partial r_2}{\partial y}+\frac{\partial r_3}{\partial z}]$

The curl of vector R is obtained as follows:

localid="1654604014780" $$\begin{gathered} \nabla \times R=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ r_1& r_2& r_3\end{array}\right| \\ =i\left(\frac{\partial r_1}{\partial y}-\frac{\partial r_3}{\partial z}\right)-j\left(\frac{\partial r_3}{\partial x}-\frac{\partial r_1}{\partial z}\right)+k\left(\frac{\partial r_2}{\partial x}-\frac{\partial r_1}{\partial y}\right) \end{gathered}$$

The Divergence of vectorlocalid="1654514915442" $F_1$is obtained as follows:

$$\begin{gathered} \nabla .F_1=\left[\frac{\partial \left(0\right)}{\partial x}+\frac{\partial \left(0\right)}{\partial y}+\frac{\partial \left(x^2\right)}{\partial z}\right] \\ =0 \end{gathered}$$

The Divergence of vectorlocalid="1654514923116" $F_2$is obtained as follows:

$$\begin{gathered} \nabla .F_2=\left[\frac{\partial \left(x\right)}{\partial x}+\frac{\partial \left(y\right)}{\partial y}+\frac{\partial \left(z\right)}{\partial z}\right] \\ =1+1+1 \\ =3 \end{gathered}$$

Thus, divergence of vectorslocalid="1654514937636" $F_1$andlocalid="1654514947237" $F_2$is obtained aslocalid="1654514955207" $\nabla .F_1=0$andlocalid="1654514967421" $\nabla .F_2=3$.

The Curl of vector is obtained as follows:

$$\begin{gathered} \nabla .F_1=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& x^2\end{array}\right| \\ =i(\frac{\partial \left(x^2\right)}{\partial y}-\frac{\partial \left(0\right)}{\partial z})-j(\frac{\partial \left(x^2\right)}{\partial x}-\frac{\partial \left(0\right)}{\partial z})+k\left(0\right) \\ =-2xj \end{gathered}$$

The curl of vectorlocalid="1654514978897" $F_2$is obtained as follows:

$$\begin{gathered} \nabla \times F_1=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& x^2\end{array}\right| \\ =i\left(\frac{\partial \left(z\right)}{\partial y}-\frac{\partial \left(y\right)}{\partial z}\right)-j\left(\frac{\partial \left(z\right)}{\partial x}-\frac{\partial \left(x\right)}{\partial z}\right)+k\left(\frac{\partial \left(y\right)}{\partial x}-\frac{\partial \left(x\right)}{\partial z}\right) \\ =0 \end{gathered}$$

Thus, curl of vectors$F_1$and$F_2$is obtained as$\nabla \times F_1=\left(-2x\right)j$and$\nabla \times F_2=0$.

The divergence of vector$F_1$is 0. So,$F_1$can be expressed as curl of some vector. The divergence of vector$F_2$is 0. So, $F_2$it can be expressed as gradient of a scalar.

Out of the all the given functions, the curl of vector$F_2$is 0. So, it can be expressed as gradient of scalar function.

Let us assume $F_2=\nabla V$. Expand $F_2=\nabla V$as,

$xi+yj+zk=\frac{\partial V_x}{\partial X}i+\frac{\partial V_y}{\partial y}j+\frac{\partial V_z}{\partial Z}k$

On comparing the right and left side of the above equation, we obtain,

$$\begin{gathered} \frac{\partial V_X}{\partial X}=X \\ \frac{\partial V_y}{\partial y}=y \\ \frac{\partial V_z}{\partial Z}=Z \end{gathered}$$

Integrate the above functions as,

$$\begin{gathered} \int \partial V_x=\int x\partial x \\ {v}_x=\frac{x^2}{2}+C_1 \\ \int \partial V_y=\int y\partial y \\ {v}_y=\frac{y^2}{2}+C_2 \\ \int \partial V_Z=\int Z\partial Z \\ {v}_z=\frac{z^2}{2}+C_3 \end{gathered}$$

Thus, the scalar function v can be written as$v=\frac{1}{2}\left(x^2+y^2+z^2\right)+c$. Thus the scalar function is obtained as $v=\frac{1}{2}\left(x^2+y^2+z^2\right)+c$.

Out of the all the given functions, the dot product of vector $F_1$is 0. So, it can be expressed as curl of a vector, as divergence of curl is always 0.

Let us assume $\nabla \times F_1=A$. Expand $\nabla \times F_1=A$as,

$\nabla \times A=F_1$

$$\begin{gathered} \left|\begin{array}{ccc}i& i& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ A_X& A_Y& A_Z\end{array}\right|=x^2 \\ \left(\frac{\partial A_Z}{\partial y}-\frac{\partial Ay}{\partial y}\right)i-\left(\frac{\partial A_Z}{\partial x}-\frac{\partial A_x}{\partial z}\right)j+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)k=x^{2}k \end{gathered}$$

On comparing the right and left side of the above equation, we obtain,

$$\begin{gathered} \left(\frac{\partial A_Z}{\partial y}-\frac{\partial Ay}{\partial z}\right)=0................\left(1\right) \\ \left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)=0................\left(2\right) \\ \left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)=x^2................\left(3\right) \end{gathered}$$

On comparing above result, we obtain,localid="1654515009362" $A_x=0$andlocalid="1654515027821" $A_y=0$. Comparing

equation (1),

$$\begin{gathered} \left(\frac{\partial A_y}{\partial x}-\frac{\partial Ax}{\partial y}\right)=x^2 \\ \frac{\partial A_y}{\partial x}=x^2 \end{gathered}$$

Integrate the resulting (1),

$$\begin{gathered} \int \frac{\partial A_y}{\partial x}\partial x=\int x^2\partial x \\ \int \partial A_y=\int x^2\partial x \\ {A}_y=\frac{x^3}{3}+C \end{gathered}$$

Thus, the vector potential can be written as$A=\left(\frac{x^3}{3}+c\right)j$.

(b)

The curl of vector $F_3$is obtained as follows:

$$\begin{gathered} \nabla \times F_3=\left|\begin{array}{ccc}i& i& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ yz& zx& xy\end{array}\right| \\ =i\left(\frac{\partial \left(xy\right)}{\partial y}-\frac{\partial \left(zx\right)}{\partial z}\right)-j\left(\frac{\partial \left(xy\right)}{\partial y}-\frac{\partial \left(yz\right)}{\partial z}\right)+k\left(\frac{\partial \left(zx\right)}{\partial x}-\frac{\partial \left(yz\right)}{\partial z}\right) \\ =0 \end{gathered}$$

Thus, curl of vector $F_3$is obtained as $\nabla \times F_3=0$. Thus, can also be written as gradient of scalar function.

The vector $F_3$can be expressed as curl of a vector, as divergence of curl is always 0.

Let us assume $F_3=\nabla \times G$. Expand $F_3=\nabla \times G$as,

$F_3=\nabla \times G$

$\left(\frac{\partial G_x}{\partial y}-\frac{\partial G_y}{\partial z}\right)i-\left(\frac{\partial G_z}{\partial x}-\frac{\partial G_x}{\partial z}\right)j+\left(\frac{\partial G_y}{\partial x}-\frac{\partial G_x}{\partial y}\right)k=yzi+zxj+xyk$

On comparing the right and left side of the above equation, we obtain,

$$\begin{gathered} \left(\frac{\partial G_z}{\partial y}-\frac{\partial G_y}{\partial z}\right)=yz.............\left(1\right) \\ \left(\frac{\partial G_x}{\partial z}-\frac{\partial G_z}{\partial x}\right)=xz.............\left(2\right) \\ \left(\frac{\partial G_y}{\partial x}-\frac{\partial G_x}{\partial y}\right)=xy.............\left(3\right) \end{gathered}$$

Integrate $\left(\frac{\partial G_z}{\partial y}-\frac{\partial G_y}{\partial z}\right)=yz$

$$\begin{gathered} G_Z=\frac{y^{2}z}{4}+f\left(x,z\right)..........\left(1\right) \\ {G}_Y=\frac{y{z}^2}{4}+g\left(x,y\right).........\left(2\right) \end{gathered}$$

Integrate $\left(\frac{\partial G_x}{\partial z}-\frac{\partial G_z}{\partial x}\right)=xz$and $\left(\frac{\partial G_y}{\partial x}-\frac{\partial G_x}{\partial y}\right)=xy$

$$\begin{gathered} G_x=\frac{z^{2}x}{4}+h\left(x,y\right)..........\left(3\right) \\ {G}_y=\frac{z^{2}x}{4}+j\left(x,y\right).........\left(4\right) \end{gathered}$$

Integrate $\left(\frac{\partial G_y}{\partial x}-\frac{\partial G_x}{\partial y}\right)$=xy

role="math" localid="1656148241993" $$\begin{gathered} G_y=\frac{x^{2}y}{4}+f\left(x,y\right)..........\left(5\right) \\ {G}_x=\frac{x^{2}y}{4}+I\left(x,y\right).........\left(6\right) \end{gathered}$$

From the equations, (1), (2), (3), (4), (5), (6), the vector $G=G_{x}i+G_{y}j+G_{z}k$

can be written as $G=\frac{1}{4}\left\{x\left(z^2-y^2\right)i+y\left(x^2-z^2\right)j+z\left(y^2-x^2\right)k\right\}$.

Let the function $F_3$is written in terms of scalar potential as $F_3=\nabla \varphi$. The divergence of scalar potential $\varphi$is written as,

$\nabla \varphi =\frac{\partial \varphi }{\partial x}\stackrel{\wedge }{x}+\frac{\partial \varphi }{\partial y}\stackrel{\wedge }{y}+\frac{\partial \varphi }{\partial z}\stackrel{\wedge }{z}$

On comparing equation $F_3=yzi+zxj+xyk,$with $\nabla \varphi =\frac{\partial \varphi }{\partial x}\stackrel{\wedge }{x}+\frac{\partial \varphi }{\partial y}\stackrel{\wedge }{y}+\frac{\partial \varphi }{\partial z}\stackrel{\wedge }{z}$, we

obtain,

$$\begin{gathered} \frac{\partial \varphi }{\partial x}=yz \\ \frac{\partial \varphi }{\partial y}=zx \\ \frac{\partial \varphi }{\partial z}=xy \end{gathered}$$

On integrating any one equation of above three equations, we obtain,

$$\begin{gathered} \int \partial \varphi =\int yz\partial \times \\ \varphi =xyz+c \end{gathered}$$

Thus, the scalar potential for$F_{3}is\varphi =xyz+c$