# Vector Calculus: Finding out divergence and curl of vector field

Vector Calculus – Finding out divergence and curl of vector field:

In case of fluid flow,

Divergence:  Relates to the Way in which fluid flows away
or, toward from a point.
It has Scaler values.

Curl:        Rotational properties of a fluid at a point.
It has vector values.


Vector field in 3 space with xyz co-ordinate system,
$\overrightarrow{F}(x,y,z) = < f(x,y,z), g(x,y,z), h(x,y,z) >$

Then divergence of function $\overrightarrow{F} (x, y, z)$ is,
$div \vec{F} = \nabla . \vec{F} = < \frac{\delta}{\delta x}, \frac{\delta}{\delta y}, \frac{\delta}{\delta z} > . < f, g, h >$
$div \vec{F} = \frac{\delta f}{\delta x} + \frac{\delta g}{\delta y} + \frac{\delta h}{\delta z}$

The curl of the function $\vec{F} (x, y, z)$ is,
$curl \vec{F} = \nabla \times \vec{F}$

$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\delta}{\delta x} & \frac{\delta}{\delta y} & \frac{\delta}{\delta z} \\ f & g & h \end{vmatrix}$

$curl \vec{F} = < \frac{\delta h}{\delta y} - \frac{\delta g}{\delta z}, \frac{\delta f}{\delta z} - \frac{\delta h}{\delta x}, \frac{\delta g}{\delta x} - \frac{\delta f}{\delta y} >$

#### Example:

$curl \vec{F} = < x^2y, 2y^3z, 3z >$

Here,
$f(x,y,z) = x^2y \\ f(x,y,z) = 2y^3z \\ f(x,y,z) = 3z$

We know divergence function,
$div \vec{F} = \frac{\delta f}{\delta x} + \frac{\delta g}{\delta y} + \frac{\delta h}{\delta z} \ \ \ \ \ (1)$

Now,
$\frac{\delta f}{\delta x} = \frac{\delta}{\delta x} x^2y = 2xy \\ \\ \frac{\delta g}{\delta y} = \frac{\delta}{\delta y} 2y^3z = 6y^2z \\ \\ \frac{\delta h}{\delta z} = \frac{\delta}{\delta z} 3z = 3$

So, from $eqn(1)$ we get,
$div \vec{F} = 2xy + 6y^2z + 3$

Also, The curl of the function,
$curl \vec{F} = \nabla \times \vec{F}$
$= \begin{vmatrix} \hat{k} & \hat{k} & \hat{k} \\ \frac{\delta}{\delta x} & \frac{\delta}{\delta y} & \frac{\delta}{\delta z} \\ f & g & h \end{vmatrix}$
$curl \vec{F} = < \frac{\delta h}{\delta y} - \frac{\delta g}{\delta z}, \frac{\delta f}{\delta z} - \frac{\delta h}{\delta x}, \frac{\delta g}{\delta x} - \frac{\delta f}{\delta y} > \ \ \ \ \ (2)$

calculating partial derivatives,
$\frac{\delta h}{\delta y} = \frac{\delta}{\delta y} 3z = 0 \\ \\ \frac{\delta g}{\delta z} = \frac{\delta}{\delta z} 2y^3z = 2y^3 \\ \\ \frac{\delta f}{\delta z} = \frac{\delta}{\delta z} x^2y = 0 \\ \\ \frac{\delta h}{\delta x} = \frac{\delta}{\delta x} 3z = 0 \\ \\ \frac{\delta g}{\delta x} = \frac{\delta}{\delta x} 2y^3z = 0 \\ \\ \frac{\delta f}{\delta y} = \frac{\delta}{\delta y} x^2y = x^2$

So, from $eqn(2)$ we write,
$curl \vec{F} = < 0 - 2y^3, 0 - 0, 0 - x^2 > \\ \\ curl \vec{F} = < -2y^3, 0, - x^2 > \\ or, \\ curl \vec{F} = -2y^3 \hat{i} - x^2 \hat{k};$