Question

Consider the vector field. f(x, y, z) = xy2z2i x2yz2j x2y2zk (a) find the curl of the vector field?

Answer 1

Observe that the given vector field is a gradient field:

Let
`f(x,y,z)=\\abla g(x,y,z)`, so that

`(\partial g)/(\partial x) = x y^2 z^2`

`(\partial g)/(\partial y) = x^2 y z^2`

`(\partial g)/(\partial z) = x^2 y^2 z`

Integrating the first equation with respect to
`x`, we get

`g(x,y,z) = \frac12 x^2 y^2 z^2 + h(y,z)`

Differentiating this with respect to
`y` gives

`(\partial g)/(\partial y) = x^2 y z^2 + (\partial h)/(\partial y) = x^2 y z^2 \\\\ \implies (\partial h)/(\partial y) = 0 \implies h(y,z) = i(z)`

Now differentiating
`g` with respect to
`z` gives

`(\partial g)/(\partial z) = x^2 y^2 z + (di)/(dz) = x^2 y^2 z \\\\ \implies (di)/(dz) = 0 \implies i(z) = C`

Putting everything together, we find a scalar potential function whose gradient is
`f`,

`f(x,y,z) = \\abla \left(\frac12 x^2 y^2 z^2 + C\right)`

It follows that the curl of
`f` is 0 (i.e. the zero vector).