$\overline {F}=(y \sin z-\sin x)\widehat{i}+ (x\sin z+2yz)\widehat{j}+ (xy\cos z+y^2)\widehat{k}$ ***$\nabla \times F

\=\begin{vmatrix} \ i & j & k \ \dfrac{\partial }{\partial x}&\dfrac{\partial }{\partial y}&\dfrac{\partial }{\partial z} \ y \sin z - \sin x & x \sin z+2yz & xy\cos z +y^2 \end{vmatrix}$*** $= i(\dfrac{\partial}{\partial y}[xy\cos z +y^2] -\dfrac{\partial}{\partial z}[x \sin z +2yz] ) - j(\dfrac{\partial}{\partial x}[x \sin z +2yz] -\dfrac{\partial}{\partial z}[y\sin z-\sin x]) + k(\dfrac{\partial}{\partial x}[x \sin z +2yz]- \dfrac{\partial}{\partial y}[y \sin z-\sin x])$ $ \= i(x\cos z +2y -x\ cos z -2y ) -j(\sin z +0-\sin z +0) +k(\sin z+0 - \sin z -0) \ = 0$ $ \ As \: \nabla \times \overline {F} = 0, \: \overline {F} \text{ is irrotational.} \ \therefore\text{, there exist a scalar potential of F such that } \overline F =\nabla \phi $ $r = xi +yj + zk \ \overline F = \nabla \phi $ $(y \sin z-\sin x)\widehat{i}+ (x\sin z+2yz)\widehat{j}+ (xy\cos z+y^2)\widehat{k} = \widehat{i} \dfrac{\partial \phi }{\partial x}+ \widehat{j} \dfrac{\partial \phi }{\partial y}+ \widehat{k} \dfrac{\partial \phi }{\partial z} $ $\ \text{Comparing both sides, we get, } \ \dfrac{\partial \phi }{\partial x} = (y \sin z-\sin x) , \quad \dfrac{\partial \phi }{\partial y} = (x\sin z+2yz), \quad \dfrac{\partial \phi }{\partial z} =(xy\cos z+y^2) $ $ \ Now, d\phi = \dfrac{\partial \phi }{\partial x}dx +\dfrac{\partial \phi }{\partial y}dy +\dfrac{\partial \phi }{\partial z}dz \ \therefore, d\phi = (y \sin z-\sin x)dx + (x\sin z+2yz)dy +(xy\cos z+y^2) dz$ $ \ \text{Integrating both sides from (0,0,0 to (x,y,z)),} \ \int d\phi = \int \limits_0^x (0 -\sin x)dx + \int \limits_0^y (0+0)dy + \int \limits_0^z (xy \cos z +y^2 ) dz \ ...From \: \: 0 \:to\: x, y =0, z=0 , dy= 0, dz =0, From\:\: 0 \:to\: y, z= 0, dz =0$ $\phi = \cos x -1+0+xy[\sin z]_0^z+zy^2 \ \bbox[3pt, yellow]{\phi = \cos x -1 + xy \sin z +zy^2 } $