Answer
$\mathop \oint \limits_{\cal C} \sin x{\rm{d}}x + z\cos y{\rm{d}}y + \sin y{\rm{d}}z = 0$
Work Step by Step
1. Write ${\bf{F}} = \left( {\sin x,z\cos y,\sin y} \right)$.
2. Next, we check if ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\sin x,z\cos y,\sin y} \right)$ 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
$0=0$, ${\ \ \ }$ $\cos y = \cos y$, ${\ \ \ }$ $0=0$
These results imply that ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\sin x,z\cos y,\sin y} \right)$ 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
$f\left( {x,y,z} \right) = \smallint \sin x{\rm{d}}x = - \cos x + j\left( {y,z} \right)$
b. taking the integral of $\dfrac{{\partial f}}{{\partial y}}$ with respect to $y$ gives
$f\left( {x,y,z} \right) = \smallint z\cos y{\rm{d}}y = z\sin y + g\left( {x,z} \right)$
c. taking the integral of $\dfrac{{\partial f}}{{\partial z}}$ with respect to $z$ gives
$f\left( {x,y,z} \right) = \smallint \sin y{\rm{d}}z = z\sin y + h\left( {x,y} \right)$
Since the three ways of expressing $f\left( {x,y,z} \right)$ must be equal, we get
$ - \cos x + j\left( {y,z} \right) = z\sin y + g\left( {x,z} \right) = z\sin y + h\left( {x,y} \right)$
Thus, we conclude that the potential function is $f\left( {x,y,z} \right) = - \cos x + z\sin y + C$, where $C$ is a constant.
Since ${\bf{F}} = \nabla f$, so ${\bf{F}}$ is conservative. By Theorem 1, the circulation of ${\bf{F}}$ around a closed ellipse is zero, that is
$\mathop \oint \limits_{\cal C} \sin x{\rm{d}}x + z\cos y{\rm{d}}y + \sin y{\rm{d}}z = 0$
