VC5.70

From Exampleproblems

Here $(2x\cos y+z\sin y)dx+(xz\cos y-x^2\sin y)dy+x\sin y dz=[(2x\cos y+z\sin y)i+(xz\cos y-x^2\sin y)j+x\sin y k]\cdot(dx i+dy j+dz k)=F\cdot dr\,$ ,say --(1)

so that $F=(2x\cos y+z\sin y)i+(xz\cos y-x^2\sin y)j+x\sin y k\,$ --(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 $(2x\cos y+z\sin y)dx+(xz\cos y-x^2\sin y)dy+x\sin y dz\,$

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

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

Now, $\nabla\phi=F\,$

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