Question

Find (a) the curl and (b) the divergence of the vector field.

\({\rm{F}}(x,y,z) = xy{e^z}{\rm{i}} + yz{e^x}{\rm{k}}\)

Short answer

(a) The curl \({\bf{F}}\) is \(z{e^x}{\rm{i}} + \left( {xy{e^z} - yz{e^x}} \right){\rm{j}} - x{e^z}{\rm{k}}\)

(b) The divergence of the vector field is \(y\left( {{e^z} + {e^x}} \right)\).

Step-by-step solution

Given information

Given data:

\({\rm{F}}(x,y,z) = xy{e^z}{\rm{i}} + yz{e^x}{\rm{k}}\).

(a)

standard equation of a curl \({\bf{F}}\)

Consider the standard equation of a curl\({\bf{F}}\)for\({\rm{F}} = P{\rm{i}} + Q{\rm{j}} + R{\rm{k}}\).

\({\mathop{\rm curl}\nolimits} {\bf{F}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\P&Q&R\end{array}} \right|...............(1)\)

the curl \({\bf{F}}\)

Substitute \(xy{e^z}\) for $P, 0$ for \(Q\) and \(yz{e^x}\) for \(R\) in equation (1),

\(\begin{aligned}{\mathop{\rm curl}\nolimits} {\bf{F}} & = \left| {\begin{aligned}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\{xy{e^z}}&0&{yz{e^x}}\end{aligned}} \right|\\ & = \left\{ {\begin{aligned}{*{20}{c}}{\left( {\frac{\partial }{{\partial y}}\left( {yz{e^x}} \right) - \frac{\partial }{{\partial z}}(0)} \right){\rm{i}} - \left( {\frac{\partial }{{\partial z}}\left( {xy{e^z}} \right) - \frac{\partial }{{\partial x}}\left( {yz{e^x}} \right)} \right){\rm{j}} + }\\{\left( {\frac{\partial }{{\partial x}}(0) - \frac{\partial }{{\partial y}}\left( {xy{e^z}} \right)} \right){\rm{k}}}\end{aligned}} \right\}\\ & = \left\{ {\begin{aligned}{*{20}{c}}{\left( {\left( {z{e^x}} \right)\frac{\partial }{{\partial y}}(y) - (0)\frac{\partial }{{\partial z}}(1)} \right){\rm{i}} - \left( {(xy)\frac{\partial }{{\partial z}}\left( {{e^z}} \right) - (yz)\frac{\partial }{{\partial x}}\left( {{e^x}} \right)} \right){\rm{j}} + }\\{\left( {(0)\frac{\partial }{{\partial x}}(1) - \left( {x{e^z}} \right)\frac{\partial }{{\partial y}}(y)} \right){\rm{k}}}\end{aligned}} \right\}\\ & = \left( {\left( {z{e^x}} \right)(1) - (0)} \right){\rm{i}} - \left( {(xy)\left( {{e^z}} \right) - (yz)\left( {{e^x}} \right)} \right){\rm{j}} + \left( {(0) - \left( {x{e^z}} \right)(1)} \right){\rm{k}}\end{aligned}\)

Solve further

\({\mathop{\rm curl}\nolimits} {\bf{F}} = \left( {z{e^x}} \right){\rm{i}} - \left( {xy{e^z} - yz{e^x}} \right){\rm{j}} - x{e^z}{\rm{k}}\)

Thus, the curl \({\bf{F}}\) is \(z{e^x}{\rm{i}} + \left( {xy{e^z} - yz{e^x}} \right){\rm{j}} - x{e^z}{\rm{k}}\).

(b)

divergence of the vector field

Consider the standard equation of a divergence of vector field for\({\rm{F}} = P{\rm{i}} + Q{\rm{j}} + R{\rm{k}}\)

\({\mathop{\rm div}\nolimits} {\rm{F}} = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}..........(2)\)

divergence of the vector field

Substitute \(xy{e^z}\) for P, 0 for \(Q\) and \(yz{e^x}\) for \(R\) in equation (2),

\(\begin{aligned}{c}{\mathop{\rm div}\nolimits} {\rm{F}} & = \frac{\partial }{{\partial x}}\left( {xy{e^z}} \right) + \frac{\partial }{{\partial y}}(0) + \frac{\partial }{{\partial z}}\left( {yz{e^x}} \right)\\ & = \left( {y{e^z}} \right)\frac{\partial }{{\partial x}}(x) + (0)\frac{\partial }{{\partial y}}(1) + \left( {y{e^x}} \right)\frac{\partial }{{\partial z}}(z)\\ & = \left( {y{e^z}} \right)(1) + (0)(0) + \left( {y{e^x}} \right)(1)\\ & = \left( {y{e^z}} \right) + 0 + \left( {y{e^x}} \right)\end{aligned}\)

Modify the equation,

\(\begin{aligned}{\mathop{\rm div}\nolimits} {\rm{F}} & = y{e^z} + y{e^x}\\ & = y\left( {{e^z} + {e^x}} \right)\end{aligned}\)

Thus, the divergence of the vector field is \(y\left( {{e^z} + {e^x}} \right)\).