Find the divergence of the fields shown on the attached image.

a) Divergence of a vector field is a scalar quantity that represents how the field spreads out, or "diverges", in different directions. It is usually denoted as

`vecgrad*vecF` and is calculated as

`vecgrad * vecF = (dF_x)/(dx) + (dF_y)/(dy) + (dF_z)/(dz)` .

In the given vector field, the components are

`F_x = x` , so `(dF_x)/(dx) = 1`

`F_y = y^3z^2` , so `(dF_y)/(dy) = 3y^2z^2`

`F_z = xz^3` , so `(dF_z)/(dz) = 3xz^2` .

Thus, the divergence of the given vector field is

`vec grad * vecF = 1 + 3y^2z^2 + 3xz^2` .

b) The divergence of this vector field can be calculated the same way. Here,

`(dF_x)/(dx) = cosy`

`(dF_y)/(dy) = 2xy`

and `(dF_z)/(dz) = 0`

So the divergence is

`vec grad * vecF = cosy + 2xy` .