Respuesta :
Answer: The required answers are
[tex](a)~curlF=\left(xz^2-xy^2,x^2y-yz^2,y^2z-x^2z\right),\\\\(b)~divF=2xyz+2xyz+2xyz=6xyz.[/tex]
Step-by-step explanation: We are given to find the curl and divergence of the following vector field :
[tex]F(x,y,z)=x^2yzi+xy^2zj+xyz^2k.[/tex]
We know that, for a vector field [tex]F(x,y,z)=(F_1,F_2,F_3),[/tex] we have
[tex]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}.[/tex]
The required partial derivatives are calculated as follows :
[tex]\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.[/tex]
Therefore, we get
[tex]curlF=\left(xz^2-xy^2,x^2y-yz^2,y^2z-x^2z\right),\\\\divF=2xyz+2xyz+2xyz=6xyz.[/tex]
Thus, the required answers are
[tex](a)~curlF=\left(xz^2-xy^2,x^2y-yz^2,y^2z-x^2z\right),\\\\(b)~divF=2xyz+2xyz+2xyz=6xyz.[/tex]
