Question

Consider the given vector field. F(x, y, z) = x2yz i + xy2z j + xyz2 k (a) Find the curl of the vector field. (b) Find the divergence of the vector field.

Answer 1

Answer:  The required answers are

$(a)~curlF=\left(xz^2-xy^2,x^2y-yz^2,y^2z-x^2z\right),\\\\(b)~divF=2xyz+2xyz+2xyz=6xyz.$

Step-by-step explanation:  We are given to find the curl and divergence of the following vector field :

$F(x,y,z)=x^2yzi+xy^2zj+xyz^2k.$

We know that, for a vector field $F(x,y,z)=(F_1,F_2,F_3),$ we have

$curlF=\left(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z},\dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x},\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}\right),\\\\\\divF=\dfrac{\partial F_1}{\partial x}+\dfrac{\partial F_2}{\partial y}+\dfrac{\partial F_3}{\partial z}.$

The required partial derivatives are calculated as follows :

$\dfrac{\partial F_1}{\partial x}=\dfrac{\partial}{\partial x}(x^2yz)=2xyz,\\\\\dfrac{\partial F_2}{\partial x}=\dfrac{\partial}{\partial x}(xy^2z)=y^2z,\\\\\dfrac{\partial F_3}{\partial x}=\dfrac{\partial}{\partial x}(xyz^2)=yz^2,\\\\\dfrac{\partial F_1}{\partial y}=\dfrac{\partial}{\partial y}(x^2yz)=x^2z,\\\\\dfrac{\partial F_2}{\partial y}=\dfrac{\partial}{\partial y}(xy^2z)=2xyz,\\\\\dfrac{\partial F_3}{\partial y}=\dfrac{\partial}{\partial y}(xyz^2)=xz^2,\\\\\dfrac{\partial F_1}{\partial z}=\dfrac{\partial}{\partial z}(x^2yz)=x^2y,\\\\\dfrac{\partial F_2}{\partial z}=\dfrac{\partial}{\partial z}(xy^2z)=xy^2,\\\\\dfrac{\partial F_3}{\partial z}=\dfrac{\partial}{\partial z}(xyz^2)=2xyz.$

Therefore, we get

$curlF=\left(xz^2-xy^2,x^2y-yz^2,y^2z-x^2z\right),\\\\divF=2xyz+2xyz+2xyz=6xyz.$

Thus, the required answers are

$(a)~curlF=\left(xz^2-xy^2,x^2y-yz^2,y^2z-x^2z\right),\\\\(b)~divF=2xyz+2xyz+2xyz=6xyz.$