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 $f(x,y,z)=\nabla g(x,y,z)$, so that

$\dfrac{\partial g}{\partial x} = x y^2 z^2$

$\dfrac{\partial g}{\partial y} = x^2 y z^2$

$\dfrac{\partial g}{\partial z} = x^2 y^2 z$

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

$g(x,y,z) = \dfrac12 x^2 y^2 z^2 + h(y,z)$

Differentiating this with respect to $y$ gives

$\dfrac{\partial g}{\partial y} = x^2 y z^2 + \dfrac{\partial h}{\partial y} = x^2 y z^2 \\\\ \implies \dfrac{\partial h}{\partial y} = 0 \implies h(y,z) = i(z)$

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

$\dfrac{\partial g}{\partial z} = x^2 y^2 z + \dfrac{di}{dz} = x^2 y^2 z \\\\ \implies \dfrac{di}{dz} = 0 \implies i(z) = C$

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

$f(x,y,z) = \nabla \left(\dfrac12 x^2 y^2 z^2 + C\right)$

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