VC5.66

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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, $F\cdot dr=[\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k]\cdot(dx i+dy j+dz k)\,$

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, $d\phi=F\cdot dr=[(y+\sin z)i+xj+x\cos z k]\cdot(dx i+dy j+dz k)\,$

or $d\phi=(y+\sin z)dx+x dy+x\cos z dz\,$

or $d\phi=(xdy+ydx)+(\sin z dx+x\cos z dz)\,$

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

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

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