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Here (y^{2}z^{3}\cos x-4x^{3}z)dx+2z^{3}y\sin xdy+(3y^{2}z^{2}\sin x-x^{4})dz=[(y^{2}z^{3}\cos x-4x^{3}z)i+2z^{3}y\sin xj+(3y^{2}z^{2}\sin x-x^{4})k]\cdot (dxi+dyj+dzk)=F\cdot dr\, (say) --(1)

so that (y^{2}z^{3}\cos x-4x^{3}z)i+2z^{3}y\sin xj+(3y^{2}z^{2}\sin x-x^{4})k\, --(2)

Therefore,{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\y^{2}z^{3}\cos x-4x^{3}z&2z^{3}y\sin x&3y^{2}z^{2}\sin x-x^{4}\end{vmatrix}}\,

=[{\frac  {\partial }{\partial x}}(3y^{2}z^{2}\sin x-x^{4})-{\frac  {\partial }{\partial z}}(2z^{3}y\sin x)]i+[{\frac  {\partial }{\partial x}}(3y^{2}z^{2}\sin x-x^{4})-{\frac  {\partial }{\partial z}}(y^{2}z^{3}\cos x-4x^{3}z)]j+[{\frac  {\partial }{\partial x}}(2z^{3}y\sin x)-{\frac  {\partial }{\partial y}}(y^{2}z^{3}\cos x-4x^{3}z)]k\,

=(6yz^{2}\sin x-6z^{2}y\sin x)i-[3y^{2}z^{3}\cos x-4x^{3}-(3y^{2}z^{2}\cos x-4x^{3})]j+(2z^{3}y\cos x-2yz^{3}\cos x)k=0i-0j+0k=0\,

Hence there exists a scalar function \phi (x,y,z)\, such that F=\nabla \phi \, so that F\cdot dr=\nabla \phi \cdot dr\,. So using (1),we get

(y^{2}z^{3}\cos x-4x^{3}z)dx+2z^{3}y\sin xdy+(3y^{2}z^{2}\sin x-x^{4})dz=[{\frac  {\partial \phi }{\partial x}}i+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k]\cdot (dxi+dyj+dzk)={\frac  {\partial \phi }{\partial x}}dx+{\frac  {\partial \phi }{\partial y}}dy+{\frac  {\partial \phi }{\partial z}}dz=d\phi \,

This shows that the given expression is exact differential of some function \phi (x,y,z)\,

Now \nabla \phi =F\,

={\frac  {\partial \phi }{\partial x}}i+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k=(y^{2}z^{3}\cos x-4x^{3}z)i+2z^{3}y\sin xj+(3y^{2}z^{2}\sin x-x^{4})k\,

implies {\frac  {\partial \phi }{\partial x}}=y^{2}z^{3}\cos x-4x^{3}z\, where \phi =y^{2}z^{3}\sin x-x^{4}z+f_{1}(y,z)\, --(1)

{\frac  {\partial \phi }{\partial y}}=2z^{3}y\sin x\, where \phi =y^{2}z^{3}\sin x+f_{2}(x,z)\, --(2)

and {\frac  {\partial \phi }{\partial z}}=3y^{2}z^{2}\sin x-x^{4}\, where \phi =y^{2}z^{3}\sin x-x^{4}z+f_{3}(x,y)\, --(3)

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

Therefore \phi =y^{2}z^{3}\sin x-x^{4}z\, to which may be added any constant.

Thus \phi =y^{2}z^{3}\sin x-x^{4}z+c\,

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