Curl and Divergence

Assume that \(\overrightarrow F=\langle P,Q,R\rangle\) where \(P\), \(Q\), and \(R\) are functions defined on \(D\). Then we have \begin{eqnarray*}\nabla \cdot (f\overrightarrow F)&=&\frac{\partial}{\partial x}(fP)+ \frac{\partial}{\partial y}(fQ)+ \frac{\partial}{\partial z}(fR)\\ &=& \frac{\partial f}{\partial x}f\cdot P+f\cdot\frac{\partial P}{\partial x}+ \frac{\partial f}{\partial y}f\cdot Q+f\cdot\frac{\partial Q}{\partial y}+ \frac{\partial f}{\partial z}f\cdot R+f\cdot\frac{\partial R}{\partial z}\\ &=&\langle f_x,f_y,f_z\rangle\cdot\langle P,Q,R\rangle+ f \cdot (P_x+Q_y+R_z)\\ &=& \nabla f\cdot \overrightarrow F+f\nabla\cdot \overrightarrow F. \end{eqnarray*}

Assume that \(\overrightarrow F=\langle P,Q,R\rangle\) where \(P\), \(Q\), and \(R\) are functions defined on \(D\). Then we have \begin{eqnarray*}\nabla \cdot (\nabla \times \overrightarrow F)&=& \nabla\cdot \left\langle R_y-Q_z, P_z-R_x, Q_x-P_y\right\rangle =R_{yx}-Q_{zx}+P_{zy}-R_{xy}+Q_{xz}-P_{yz}=0. \end{eqnarray*}

\begin{eqnarray*}\nabla \cdot \left(\nabla (fg)\right))&=& \nabla\cdot \left\langle \frac{\partial}{\partial x}(fg), \frac{\partial}{\partial y}(fg), \frac{\partial}{\partial z}(fg)\right\rangle\\&=& \frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}(fg)\right)+ \frac{\partial}{\partial y}\left(\frac{\partial}{\partial y}(fg)\right)+ \frac{\partial}{\partial z}\left(\frac{\partial}{\partial z}(fg)\right) =f_{xx}g+2f_xg_x+fg_{xx}+f_{yy}g+2f_yg_y+fg_{yy}+f_{zz}+2f_zg_z+fg_{zz}\\ &=&g\left(f_{xx}+f_{yy}+f_{zz}\right)+ f\left(g_{xx}+g_{yy}+g_{zz}\right)+2\langle f_x,f_y,f_z\rangle\cdot \langle g_x,g_y,g_z\rangle\\ &=& g \nabla\cdot \langle f_x,f_y,f_z\rangle+ f\nabla\cdot \langle g_x,g_y,g_z\rangle+ 2\nabla f\cdot \nabla g\\ &=&f \nabla \cdot (\nabla g)+ g\nabla\cdot (\nabla f)+2\nabla f\cdot \nabla g. \end{eqnarray*}

Let us first evaluate \(\nabla f\times \nabla g\). \begin{eqnarray*}\nabla f\times\nabla g=\mbox{det }\left|\begin{array}{ccc} \overrightarrow i& \overrightarrow j& \overrightarrow k\\ f_x &f_y& f_z\\ g_x& g_y& g_z\\ \end{array}\right|=\langle f_yg_z-f_zg_y,f_zg_x-f_xg_z,f_xg_y-f_yg_x\rangle.\end{eqnarray*} Therefore \begin{eqnarray*} \nabla\cdot \left(\nabla f\times \nabla g\right)&=& \frac{\partial}{\partial x}\left(f_yg_z-f_zg_y\right)+ \frac{\partial}{\partial y}\left(f_zg_x-f_xg_z\right)+ \frac{\partial}{\partial z}\left(f_xg_y-f_yg_x\right)\\ &=& f_{xy}g_z+f_yg_{xz}-f_{xz}g_y-f_zg_{xy} + f_{yz}g_x+f_zg_{xy}-f_{xy}g_z-f_xg_{yz} + f_{xz}g_y+f_xg_{yz}-f_{yz}g_x-f_yg_{xz} \\&=& 0. \end{eqnarray*} Therefore, the correct answer is A.

Using the definition of divergence we obtain \begin{eqnarray*} \nabla\cdot \langle x^2y^2e^z, xe^{xy},-z\cos(xy)\rangle&=& 2xy^2e^z+x^2e^{xy}-\cos(xy). \end{eqnarray*}

The curl of the vector field \(\overrightarrow F\) is \begin{eqnarray*} \nabla\times \langle x^2y, ye^z, -z\cos x\rangle&=&\mbox{det }\left|\begin{array}{ccc} \overrightarrow i& \overrightarrow j& \overrightarrow k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ x^2y& ye^z& -z\cos x\\ \end{array}\right| &=& -ye^z\overrightarrow i -\left(z\sin x -0\right) j-x^2\overrightarrow k=-\langle ye^z,z\sin x,x^2\rangle. \end{eqnarray*}

Using the definition of the curl, we obtain \begin{eqnarray*} \nabla\times \langle 3x,y,z\rangle&=&\mbox{det }\left|\begin{array}{ccc} \overrightarrow i& \overrightarrow j& \overrightarrow k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ 3x& y& z\\ \end{array}\right| &=& 0. \end{eqnarray*}

Using the definition of the curl, we obtain \begin{eqnarray*} \nabla\times \langle z,x,2x\rangle&=&\mbox{det }\left|\begin{array}{ccc} \overrightarrow i& \overrightarrow j& \overrightarrow k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ z& x& 2x\\ \end{array}\right| &=& 0\overrightarrow i -\overrightarrow j+\overrightarrow k=\langle 0,-1,1\rangle. \end{eqnarray*}