Question

Verify that force fields is conservative. Then find a scalar potential $\mathit{\phi }$ such that $\mathit{F}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathit{\phi }\mathbf{,}\mathbf{}\mathit{F}\mathbf{=}\left(3x^{2}yz-3y\right)\mathit{i}\mathbf{+}\left(x^{3}z-3x\right)\mathit{j}\mathbf{+}\left(x^{3}y+2z\right)\mathit{k}$ .

Short answer

The force field is conservative and the scalar potential $\phi =-z^2-y\left(x^{3}z-3x\right)$

Step-by-step solution

Given Information

The given force vector is $F=\left(3x^{2}yz-3y\right)i+\left(x^{3}z-3x\right)j+\left(x^{3}y+2z\right)k$.

Definition

Force field is said to be conservative if curl of force becomes zero.

Concept and formula

The formula to find $\nabla \times F$ in order to show that field is conservative is mentioned below.

$\nabla \times F=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& y& z\end{array}\right|$

where $F=xi+yj+zk$

In order to find the scalar potential$\phi$ ,where$F=-\nabla \phi$ , the formula for work done is given below.

$$\begin{gathered} W=\int F.dr \\ F=\nabla W \\ \phi =-W \end{gathered}$$

Conservative force field

Use the formula $\nabla \times F=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& y& z\end{array}\right|$.

Put .$F=\left(3x^{2}yz-3y\right)i+\left(x^{3}z-3x\right)j+\left(x^{3}y+2z\right)k$

$$\begin{gathered} \nabla \times F=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 3x^{2}yz-3y& x^{3}z-3x& x^{3}y+2z\end{array}\right| \\ \nabla \times F=\left(x^3-x^3\right)i-\left(3x^{2}y-3x^{2}y\right)j+\left(3x^{2}z-3-3x^{2}z+3\right)k \\ \nabla \times F=0 \end{gathered}$$

This implies that force field is conservative.

Calculate Work done

Use the formula $W=\int F.dr$

$W=\int \left(3x^{2}yz-3y\right)dx+\left(x^{3}z-3x\right)dy+\left(x^{3}y+2z\right)dz$

Take path from (0,0,0) to (x,y,z)

(i) From (0,0,0) to (x,0,0)

y = 0

z = 0

dy = 0

dz = 0

$$\begin{gathered} W_1=\underset{0}{\overset{x}{\int }}0dx \\ {W}_1=0 \end{gathered}$$

(ii) From (x,0,0) to (x,0,z)

y = 0

dy = 0

dx = 0

$$\begin{gathered} W_2=\underset{0}{\overset{z}{\int }}2zdz \\ {W}_2=z^2 \end{gathered}$$

(iii) From (x,0,z) to (x,y,z)

dx = 0

dz = 0

$$\begin{gathered} W_3=\underset{0}{\overset{y}{\int }}\left(x^{3}z-3x\right)dy \\ {W}_3=x^{3}yz-3xy \end{gathered}$$

The total work done is given below.

$$\begin{gathered} W=W_1+W_2+W_3 \\ W=z^2+x^{3}yz-3xy \\ W=z^2+y(x^{3}z-3x) \end{gathered}$$

Calculate scalar potential

Use the formula $F=\nabla W$

Also, use the given relation between force and scalar potential which is given below.

$F=-\nabla \phi$

From both the above formulas we get that $\phi =-W$

$\Rightarrow \phi =-z^2-y\left(x^{3}z-3x\right)$

The force field is conservative and the scalar potential $\phi =-z^2-y\left(x^{3}z-3x\right)$.