# VC5.70

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

$\frac{\partial\phi}{\partial x}=2x\cos y+z\sin y\,$ 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 $\phi=x^2\cos y+c\,$

##### Toolbox

 Get A Wifi Network Switcher Widget for Android