Q. 21 - Calculus

Question

In Exercises 17–24, find a potential function for the given vector field.

$\mathbf{F}(x,y,z)=yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}$

Step-by-Step Solution

Answer

A potential function for the given vector field is $f(x,y,z)=xyz$. 

Step 1. Given Information

In given exercises we have to find a potential function for the given vector field. 

$\mathbf{F}(x,y,z)=yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}$

Step 2. Since F ( x , y , z ) = y z i + x z j + x y k

$$\begin{gathered} \mathbf{F}(x,y,z)=\int \left(yz\right)dx+\mathrm{B}+\mathrm{C} \\ \mathbf{F}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathrm{yz}\int d\mathrm{x}+\mathrm{B}+\mathrm{C} \\ \mathbf{F}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathrm{yzx}+\mathrm{\alpha }+\mathrm{B}+\mathrm{C} \\ \mathbf{F}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathrm{xyz}+\mathrm{\alpha }+\mathrm{B}+\mathrm{C} \end{gathered}$$

where $\alpha$ is an arbitrary constant and B is the integral with respect to y of the terms in $F_2(x,y,z)$ in which the factor x does not appear.

Step 3. In this case, that is all of F 2 ( x , y , z ) , so

$$\begin{gathered} B=\int xzdy \\ B=xz\int dy \\ B=xzy+\beta \\ B=xyz+\beta \end{gathered}$$

where β is an arbitrary constant.

Step 4. Now finding C = ∫ x y d z

$$\begin{gathered} C=xy\int dz \\ C=xyz+\gamma \end{gathered}$$

where $\gamma$ is an arbitrary constant. 

Setting the constants equal to zero since they do not affect the gradient of $f(x,y,z)$

We have,

$f(x,y,z)=xyz$