Question

Find the curl and the divergence of the vector field f(x, y, z) = x2yz i + xy2z j + xyz2 k

Answer 1

$\mathbf f(x,y,z)=x^2yz\,\mathbf i+xy^2z\,\mathbf j+xyz^2\,\mathbf k$

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(x^2yz)}{\partial x}+\dfrac{\partial(xy^2z)}{\partial y}+\dfrac{\partial(xyz^2)}{\partial z}$
$=2xyz+2xyz+2xyz$
$=6xyz$

$\mathrm{curl}\mathbf f(x,y,z)=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\\\x^2yz&xy^2z&xyz^2\end{vmatrix}$
$=\left(\dfrac{\partial(xyz^2)}{\partial y}-\dfrac{\partial(xy^2z)}{\partial z}\right)\,\mathbf i-\left(\dfrac{\partial(xyz^2)}{\partial x}-\dfrac{\partial(x^2yz)}{\partial z}\right)\,\mathbf j+\left(\dfrac{\partial(xy^2z)}{\partial x}-\dfrac{\partial(x^2yz)}{\partial y}\right)\,\mathbf k$
$=\left(xz^2-xy^2\right)\,\mathbf i-\left(yz^2-x^2y\right)\,\mathbf j+\left(y^2z-x^2z\right)\,\mathbf k$