VC3.10

Given, $\phi (x,y,z)=(x^2+y^2+z^2)e^{-(x^2+y^2+z^2)^{\frac{1}{2}}}\,$

Let this be equation 1.

Let $r=xi+yj+zk,r^2=|r|^2=x^2+y^2+z^2\,$ [Equation 2]

Using (2),(1)= $\phi=r^2e^{-r}\,$ [Equation 3]

From (2), $\frac{\partial r}{\partial x}=\frac{x}{r}\,$ [Equation 4]

Now $\nabla\phi=\frac{\partial\phi}{\partial x}i+\frac{\partial\phi}{\partial y}j+\frac{\partial\phi}{\partial z}k\,$ [Equation 5]

Here, $\frac{\partial\phi}{\partial x}=\frac{\partial}{\partial x}(r^2e^{-r})\,$

= $\frac{\partial (r^2e^{-r})}{\partial r}\cdot\frac{\partial r}{\partial x}\,$

= $(2re^{-r}-r^2e^{-r})(\frac{x}{r})\,$ by (4)

= $(2-r)e^{-r}x\,$ [Equation 6]

Similarly, $\frac{\partial\phi}{\partial y}=(2-r)e^{-r}y,\frac{\partial\phi}{\partial z}=(2-r)e^{-r}z\,$ [Equation 7]

Using (6)and (7),(5)reduces to

$\nabla\phi=(2-r)e^{-r}(xi+yj+zk)=(2-r)e^{-r} r\,$ by (2)

Therefore, $|\nabla\phi|=|(2-r)e^{-r}r|=(2-r)e^{-r}r\,$