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{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\\sin y+z\cos x&x\cos y+\sin z&y\cos z+\sin x\end{vmatrix}}\,

=[{\frac  {\partial }{\partial y}}(y\cos z+\sin x)-{\frac  {\partial }{\partial z}}(x\cos y+\sin z)]i-[{\frac  {\partial }{\partial x}}(y\cos z+\sin x-{\frac  {\partial }{\partial z}}(\sin y+z\cos x)]j+[{\frac  {\partial }{\partial x}}(x\cos y+\sin z)-{\frac  {\partial }{\partial y}}(\sin y+z\cos x)]k\,

=(\cos z-\cos z)i-(\cos x-\cos x)j+(\cos y-\cos y)k=0\,

Hence the given vector field is irrotational. Let F=\nabla \phi \,

(\sin y+z\cos x)i+(x\cos y+\sin z)j+(y\cos z+\sin x)k={\frac  {\partial \phi }{\partial x}}i+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k\,

Therefore {\frac  {\partial \phi }{\partial x}}=\sin y+z\cos x\, whence \phi =x\sin y+z\sin x+f_{1}(y,z)\, --(1)

{\frac  {\partial \phi }{\partial y}}=x\cos y+\sin z\, whence \phi =x\sin y+y\sin z+f_{2}(z,x)\, --(2)

and {\frac  {\partial \phi }{\partial z}}=y\cos z+\sin x\, whence \phi =y\sin z+z\sin x+f_{3}(x,y)\, --(3)

(1),(2) and(3) each represent phi.These agree if we choose f_{1}(y,z)=y\sin z,f_{2}(z,x)=z\sin x,f_{3}(x,y)=x\sin y\,

Therefore \phi =x\sin y+z\sin x+y\sin z+c\,,c being constant.

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