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

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

=(0-0)i-(\cos z-\cos z)j+(1-1)k=0\,

Hence the given vector field F is conservative.

Getting the scalar potential \phi (x,y,z)\,

Here F=\nabla \phi =({\frac  {\partial \phi }{\partial x}})i+({\frac  {\partial \phi }{\partial y}})j+({\frac  {\partial \phi }{\partial z}})k\,

Therefore,F\cdot dr=[{\frac  {\partial }{\partial x}}i+{\frac  {\partial }{\partial y}}j+{\frac  {\partial }{\partial z}}k]\cdot (dxi+dyj+dzk)\,

or F\cdot dr={\frac  {\partial \phi }{\partial x}}dx+{\frac  {\partial \phi }{\partial y}}dy+{\frac  {\partial \phi }{\partial z}}dz=d\phi \,

Therefore,d\phi =F\cdot dr=[(y+\sin z)i+xj+x\cos zk]\cdot (dxi+dyj+dzk)\,

or d\phi =(y+\sin z)dx+xdy+x\cos zdz\,

or d\phi =(xdy+ydx)+(\sin zdx+x\cos zdz)\,

or d\phi =d(xy)+d(x\sin z)\,

This implies \phi =xy+x\sin z+C\,

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