Answer
The vector field is conservative.
The potential function is $f\left( {x,y,z} \right) = 2x + 4y + {{\rm{e}}^z} + C$, where $C$ is a constant.
Work Step by Step
We have ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {2,4,{{\rm{e}}^z}} \right)$.
1. We check if ${\bf{F}}$ satisfies the cross-partials condition:
$\frac{{\partial {F_1}}}{{\partial y}} = \frac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ \ }$ $\frac{{\partial {F_2}}}{{\partial z}} = \frac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ \ }$ $\frac{{\partial {F_3}}}{{\partial x}} = \frac{{\partial {F_1}}}{{\partial z}}$
We get
$0=0$, ${\ \ \ \ }$ $0=0$, ${\ \ \ \ }$ $0=0$
From these results, we conclude that ${\bf{F}} = \left( {2,4,{{\rm{e}}^z}} \right)$ satisfies the cross-partials condition. Therefore, by Theorem 4 in Section 17.3, ${\bf{F}}$ is conservative. Thus, there is a potential function for ${\bf{F}}$.
2. Find a potential function for ${\bf{F}}$.
Let the potential function for ${\bf{F}}$ be $f\left( {x,y,z} \right)$ such that ${\bf{F}} = \nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$. So,
a. taking the integral of $\frac{{\partial f}}{{\partial x}}$ with respect to $x$ gives
$f\left( {x,y,z} \right) = \smallint 2{\rm{d}}x = 2x + m\left( {y,z} \right)$
b. taking the integral of $\frac{{\partial f}}{{\partial y}}$ with respect to $y$ gives
$f\left( {x,y,z} \right) = \smallint 4{\rm{d}}y = 4y + n\left( {x,z} \right)$
c. taking the integral of $\frac{{\partial f}}{{\partial z}}$ with respect to $z$ gives
$f\left( {x,y,z} \right) = \smallint {{\rm{e}}^z}{\rm{d}}z = {{\rm{e}}^z} + o\left( {x,y} \right)$
Since the three ways of expressing $f\left( {x,y,z} \right)$ must be equal, we get
$2x + m\left( {y,z} \right) = 4y + n\left( {x,z} \right) = {{\rm{e}}^z} + o\left( {x,y} \right)$
From here, we conclude that the potential function is $f\left( {x,y,z} \right) = 2x + 4y + {{\rm{e}}^z} + C$, where $C$ is a constant.
