91Ó°ÊÓ

The divergence of curl of a unction is defined as$∇.(∇×v)$ . The vector is defined as $v=v_{x}i+v_{y}j+v_{z}k.$ the $∇$operator is defined as

$∇=\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k.$

The curl of vector $\stackrel{→}{v}$is computed as follows:

$$\begin{gathered} ∇xv=\left|\begin{array}{ccc}i& j& y\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ v_x& v_y& v_z\end{array}\right| \\ =\left(\frac{∂v_z}{∂y}-\frac{∂v_y}{∂z}\right)i-\left(\frac{∂v_z}{∂x}-\frac{∂v_x}{∂z}\right)j+\left(\frac{∂v_y}{∂x}-\frac{∂v_x}{∂y}\right)k \end{gathered}$$

Now compute divergence of curl of vector $\stackrel{→}{v}$,as :

$$\begin{gathered} ∇.(∇×v)=\left(\frac{∂}{∂x}i+\frac{∂}{∂y}+j\frac{∂}{∂z}k\right).\left(\left(\frac{∂v_z}{∂y}-\frac{∂v_y}{∂z}\right)i-\left(\frac{∂v_z}{∂x}-\frac{∂v_x}{∂z}\right)j+\left(\frac{∂v_y}{∂x}-\frac{∂v_x}{∂y}\right)k\right) \\ =\frac{∂}{∂x}\left(\frac{∂v_z}{∂y}-\frac{∂v_y}{∂z}\right)+\frac{∂}{∂y}\left(\frac{∂v_z}{∂x}-\frac{∂v_x}{∂z}\right)+\frac{∂}{∂z}\left(\frac{∂v_y}{∂x}-\frac{∂v_x}{∂y}\right) \\ =0 \end{gathered}$$

Thus the value of divergence of cutl of a function is 0.

To compute an expression substitute the vectors and other required expression and then simplify.

The vector $\stackrel{→}{v}$ is defined as $x^{2}i+3x{z}^{2}j-2xzk$. The ldivergence of curl of the vector $\stackrel{→}{v}$is computed as follows:

$$\begin{gathered} ∇.(∇×v_a)=\frac{∂}{∂x}\left(\frac{∂v_z}{∂y}-\frac{∂v_y}{∂z}\right)+\frac{∂}{∂y}\left(\frac{∂v_z}{∂x}-\frac{∂v_x}{∂z}\right)+\frac{∂}{∂z}\left(\frac{∂v_y}{∂x}-\frac{∂v_x}{∂y}\right) \\ =\frac{∂}{∂x}\left(\frac{∂(-2xz)}{∂y}-\frac{∂\left(3x{z}^2\right)}{∂z}\right)+\frac{∂}{∂y}\left(\frac{∂(-2xz)}{∂x}-\frac{∂\left(x^2\right)}{∂z}\right)+\frac{∂}{∂z}\left(\frac{∂\left(3x{z}^2\right)}{∂x}-\frac{∂\left(x^2\right)}{∂y}\right) \\ =\frac{∂}{∂x}(0-6xz)+\frac{∂}{∂y}(0-(-2z\left)\right)+\frac{∂}{∂z}(3z^2-0) \\ =-6z-0+6z \\ =0 \end{gathered}$$

Thus the value of $∇.(∇×v_a)$is 0.