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{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\\cos y&-x\sin y&0\end{vmatrix}}\,

=0i+0j+(-\sin y+\sin y)k=0\, showing that the given line integral is independent of path in space and there exists a function \phi (x,y,z)\, such that F=\nabla \phi \, so that

\cos yi-x\sin yj={\frac  {\partial \phi }{\partial x}}i+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k\,

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

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

and {\frac  {\partial \phi }{\partial z}}=0\, implies \phi =f_{3}(x,y)\,

Now (1),(2),(3) each represent phi. These agree if we choose f_{1}(y,z)=0,f_{2}(z,x)=0,f_{3}(x,y)=x\cos y\,

Therefore \phi (x,y,z)=x\cos y\,

Therefore,the required integral =\int F\cdot dr=\int \nabla \phi \cdot dr=\int d\phi =[x\cos y]_{{1,0}}^{{0,1}}=-1\,

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