VC5.73

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Here yzdx+(xz+1)dy+xydz=[yzi+(xz+1)j+xyk]\cdot (dxi+dyj+dzk)=F\cdot dr\, (say)

--(1)

so that F=yzi+(xz+1)j+xyk\, --(2)

Therefore {\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\yz&xz+1&xy\end{vmatrix}}\,

=[{\frac  {\partial }{\partial y}}(xy)-{\frac  {\partial }{\partial z}}(xz+1)]i-[{\frac  {\partial }{\partial x}}(xy)-{\frac  {\partial }{\partial z}}(yz)]j+[{\frac  {\partial }{\partial x}}(xz+1)-{\frac  {\partial }{\partial y}}(yz)]k\,

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

Hence the form under the integral sign is exact and so the given integral is independent of the path C joining (1,0,0)\, to (2,1,4)\,

Therefore \int _{C}[yzdx+(xz+1)dy+xydz]=\int _{C}[d(xyz)+dy]=\int _{C}d(xyz+y)=\int _{{1,0,0}}^{{2,1,4}}d(xyz+y)=[xyz+y]_{{1,0,0}}^{{2,1,4}}=(8+1)-(0+0)=9\,

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