Find the divergence of the field F: $$\vec{F} = \frac{yj - zk}{\sqrt{y^2+ z^2}}$$
Normally I find dot the gradient with F like this:
$$\langle 0,y,-z\rangle\cdot\langle F_x, F_y, F_z\rangle$$
And this simplifies to:
$$\langle 0, 1, -1\rangle$$
And then I tried dividing each component of the answer by:
$$\sqrt{y^2 +z^2},$$
but that isn't correct according to the answer in the back of my book.
How do I go about properly solving this? Thanks
On
$\nabla \cdot (0\mathbf i + \frac {y}{\sqrt{y^2+z^2}}\mathbf j - \frac {z}{\sqrt{y^2+z^2}}\mathbf k) = \frac {\partial}{\partial x}0+\frac {\partial}{\partial y}\frac {y}{\sqrt{y^2+z^2}} + \frac {\partial}{\partial z} \frac {-z}{\sqrt{y^2+z^2}}\\ 0+\frac {2y^2 + z^2}{(y^2+z^2)^\frac 32}-\frac {y^2 + 2z^2}{(y^2+z^2)^\frac 32} = \frac {y^2-z^2}{(y^2+z^2)^\frac 32}$
alternate
$\nabla \cdot \left((0\mathbf i + y\mathbf j - z\mathbf k) \left(\frac 1{(y^2+z^2)^\frac 12}\right)\right)\\ \left(\nabla\cdot(0,y,-z)\right)\left(\frac 1{(y^2+z^2)^\frac 12}\right) +(0,y,-z)\cdot \nabla \left(\frac 1{(y^2+z^2)^\frac 12}\right) \\ 0 + \left(\frac {(0,y,-z)\cdot(0,y,z)}{(y^2+z^2)^\frac 32}\right)\\ \frac {y^2-z^2}{(y^2+z^2)^\frac 32}$
$$\nabla \cdot \vec F=\frac {\partial F_x}{\partial x}+\frac {\partial F_y}{\partial y}+\frac {\partial F_z}{\partial z}=0+\frac{z^2}{(y^2+z^2)^{3/2}}+\frac{-y^2}{(y^2+z^2)^{3/2}}$$