VC5.70

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Here (2x\cos y+z\sin y)dx+(xz\cos y-x^{2}\sin y)dy+x\sin ydz=[(2x\cos y+z\sin y)i+(xz\cos y-x^{2}\sin y)j+x\sin yk]\cdot (dxi+dyj+dzk)=F\cdot dr\,,say --(1)

so that F=(2x\cos y+z\sin y)i+(xz\cos y-x^{2}\sin y)j+x\sin yk\, --(2)

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

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

=(x\cos y-x\sin y)i+(\sin y-\sin y)j+[z\cos y-2x\sin y-(-2x\sin y+\cos y)]k=0i+0j+0k=0\,

This shows that there exists a scalar function \phi (x,y,z)\, such that F=\nabla \phi \, or F\cdot dr=\nabla \phi \cdot dr\,,so using (1),we have (2x\cos y+z\sin y)dx+(xz\cos y-x^{2}\sin y)dy+x\sin ydz\,

=[{\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 \,

Hence the given differential equation reduce to d\phi =0\, so that \phi =c\, is the required solution.

Now,\nabla \phi =F\,

implies frac{\partial \phi }{\partial x}i+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k=(2x\cos y+z\sin y)i+(xz\cos y-x^{2}\sin y)j+x\sin yk\, implies

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

{\frac  {\partial \phi }{\partial y}}=xz\cos y-x^{2}\sin y\,

implies \phi =xz\sin y+x^{2}\cos y+f_{2}(z,x)\, --(4)

and {\frac  {\partial \phi }{\partial z}}=x\sin y\, whence \phi =xz\sin y+f_{3}(x,y)\, --(5)

(3),(4),(5) represent phi. These agree if we choose f_{(}y,z)=0,f_{2}(z,x)=0,f_{3}(x,y)=x^{2}\cos y\,

Hence \phi =x^{2}\cos y+c\,

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