Answer
$\mathop \smallint \limits_{\cal C} 2xyz{\rm{d}}x + {x^2}z{\rm{d}}y + {x^2}y{\rm{d}}z = 16$
Work Step by Step
1. First, we write ${\bf{F}} = \left( {2xyz,{x^2}z,{x^2}y} \right)$.
2. Next, we check if ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {2xyz,{x^2}z,{x^2}y} \right)$ satisfies the cross-partials condition:
$\dfrac{{\partial {F_1}}}{{\partial y}} = \dfrac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ }$ $\dfrac{{\partial {F_2}}}{{\partial z}} = \dfrac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ }$ $\dfrac{{\partial {F_3}}}{{\partial x}} = \dfrac{{\partial {F_1}}}{{\partial z}}$
We get
$2xz = 2xz$, ${\ \ \ }$ ${x^2} = {x^2}$, ${\ \ \ }$ $2xy = 2xy$
These results imply that ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {2xyz,{x^2}z,{x^2}y} \right)$ satisfies the cross-partials condition. Therefore, by Theorem 4, there is a potential function for ${\bf{F}}$.
3. Now, we find a potential function for ${\bf{F}}$.
Let the potential function for ${\bf{F}}$ be $f\left( {x,y,z} \right)$ such that ${\bf{F}} = \nabla f = \left( {\dfrac{{\partial f}}{{\partial x}},\dfrac{{\partial f}}{{\partial y}},\dfrac{{\partial f}}{{\partial z}}} \right)$. So,
a. taking the integral of $\dfrac{{\partial f}}{{\partial x}}$ with respect to $x$ gives
$f\left( {x,y,z} \right) = \smallint 2xyz{\rm{d}}x = {x^2}yz + j\left( {y,z} \right)$
b. taking the integral of $\dfrac{{\partial f}}{{\partial y}}$ with respect to $y$ gives
$f\left( {x,y,z} \right) = \smallint {x^2}z{\rm{d}}y = {x^2}yz + g\left( {x,z} \right)$
c. taking the integral of $\dfrac{{\partial f}}{{\partial z}}$ with respect to $z$ gives
$f\left( {x,y,z} \right) = \smallint {x^2}y{\rm{d}}z = {x^2}yz + h\left( {x,y} \right)$
Since the three ways of expressing $f\left( {x,y,z} \right)$ must be equal, we get
${x^2}yz + j\left( {y,z} \right) = {x^2}yz + g\left( {x,z} \right) = {x^2}yz + h\left( {x,y} \right)$
Thus, we conclude that the potential function is $f\left( {x,y,z} \right) = {x^2}yz + C$, where $C$ is a constant.
Since ${\bf{F}} = \nabla f$, by definition ${\bf{F}}$ is conservative. By Theorem 1, the line integral of ${\bf{F}}$ over the path ${\bf{r}}\left( t \right) = \left( {{t^2},\sin \left( {\frac{{\pi t}}{4}} \right),{{\rm{e}}^{{t^2} - 2t}}} \right)$ for $0 \le t \le 2$
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{\bf{r}}\left( 2 \right)} \right) - f\left( {{\bf{r}}\left( 0 \right)} \right) = f\left( {4,1,1} \right) - f\left( {0,0,1} \right) = 16$
So, $\mathop \smallint \limits_{\cal C} 2xyz{\rm{d}}x + {x^2}z{\rm{d}}y + {x^2}y{\rm{d}}z = 16$.
