Question

Given the vector field F= ∆(x3 + y3 + z3 + 3xyz), find div F and curl F.​

Answer 1

Since F is apparently a vector field, I assume you mean

`\vec F = \\abla(x^3+y^3+z^3+3xyz)`

with ∇ = gradient, whereas ∆ is often used to denote the Laplacian, ∆ = ∂²/∂x² + ∂²/∂y² + ∂²/∂z².

Let
`f(x,y,z)=x^3+y^3+z^3+3xyz`. Compute the gradient of f :

`\vec F = \\abla f(x,y,z) = (\partial f)/(\partial x) \, \vec\imath + (\partial f)/(\partial y) \, \vec\jmath + (\partial f)/(\partial z) \, \vec k`

`\vec F = (3x^2+3yz) \,\vec\imath + (3y^2 + 3xz) \,\vec\jmath + (3z^2+3xy) \,\vec k`

Now compute the divergence of F (incidentally, divergence of a gradient field is the Laplacian of the f):

`\mathrm{div} \, \vec F = (\partial(3x^2+3yz))/(\partial x) + (\partial(3y^2+3xz))/(\partial y) + (\partial(3z^2+3xy))/(\partial z)`

`\boxed{\mathrm{div} \, \vec F = 6x + 6y + 6z}`

and the curl: (the following is overkill, since any gradient field has curl zero, but it doesn't hurt to verify that)

`\mathrm{curl}\, \vec F = \left((\partial(3z^2+3xy))/(\partial y) - (\partial(3y^2+3xz))/(\partial z)\right) \,\vec\imath - \left((\partial(3z^2+3xy))/(\partial x) - (\partial(3x^2+3yz))/(\partial z)\right) \, \vec\jmath \\ ~~~~~~~~~~~~ + \left((\partial(3y^2+3xz))/(\partial x) - (\partial(3x^2+3yz))/(\partial y)\right) \,\vec k`

`\boxed{\mathrm{curl} \,\vec F = \vec 0}`