Question

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

\({\rm{F}}(x,y,z) = \left( {{e^x}\sin y,{e^y}\sin z,{e^z}\sin x} \right)\)

Short answer

(a) The\({\mathop{\rm curl}\nolimits} {\bf{F}}\) is \(\left\langle { - {e^y}\cos z, - {e^z}\cos x, - {e^x}\cos y} \right\rangle \)

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

Step-by-step solution

Concept of Divergence of the vector field

Divergenceis avector operatorthat operates on avector field, producing ascalar fieldgiving the quantity of the vector field's source at each point.

\(\begin{array}{l}{\mathop{\rm div}\nolimits} {\bf{F}} = \nabla \cdot {\bf{F}}\\{\mathop{\rm div}\nolimits} {\bf{F}} = \frac{{\partial {F_x}}}{{\partial x}} + \frac{{\partial {F_y}}}{{\partial y}} + \frac{{\partial {F_z}}}{{\partial z}}\end{array}\)

Calculation of the\({\mathop{\rm curl}\nolimits} {\bf{F}}\)

(a)

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

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

We know that \({\rm{F}}(x,y,z) = \left\langle {{e^x}\sin y,{e^y}\sin z,{e^z}\sin x} \right\rangle \)

Substitute \({e^x}\sin y\) for \(P,{e^y}\sin z\) for \(Q\) and \({e^z}\sin x\) for \(R\) in equation \(\left( 1 \right),\)

\(\begin{array}{l}{\mathop{\rm curl}\nolimits} {\bf{F}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\{{e^x}\sin y}&{{e^y}\sin z}&{{e^z}\sin x}\end{array}} \right|\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left\{ {\begin{array}{*{20}{c}}{\left[ {\frac{\partial }{{\partial y}}\left( {{e^z}\sin x} \right) - \frac{\partial }{{\partial z}}\left( {{e^y}\sin z} \right)} \right]{\rm{i}} - \left[ {\frac{\partial }{{\partial x}}\left( {{e^z}\sin x} \right) - \frac{\partial }{{\partial z}}\left( {{e^x}\sin y} \right)} \right]{\rm{j}} + }\\{\left[ {\frac{\partial }{{\partial x}}\left( {{e^y}\sin z} \right) - \frac{\partial }{{\partial y}}\left( {{e^x}\sin y} \right)} \right]{\rm{k}}}\end{array}} \right\}\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left\{ {\begin{array}{*{20}{c}}{\left[ {\left( {{e^z}\sin x} \right)\frac{\partial }{{\partial y}}(1) - \left( {{e^y}} \right)\frac{\partial }{{\partial z}}(\sin z)} \right]{\rm{i}} - \left[ {\left( {{e^z}} \right)\frac{\partial }{{\partial x}}(\sin x) - \left( {{e^x}\sin y} \right)\frac{\partial }{{\partial z}}(1)} \right]{\rm{j}} + }\\{\left[ {\left( {{e^y}\sin z} \right)\frac{\partial }{{\partial x}}(1) - \left( {{e^x}} \right)\frac{\partial }{{\partial y}}(\sin y)} \right]{\rm{k}}}\end{array}} \right\}\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left[ {0 - \left( {{e^y}\cos z} \right)} \right]{\rm{i}} - \left[ {\left( {{e^z}\cos x} \right) - 0} \right]{\rm{j}} + \left[ {0 - \left( {{e^x}\cos y} \right)} \right]{\rm{k}}\end{array}\)

Modify the equation.

\(\begin{array}{l}{\mathop{\rm curl}\nolimits} {\bf{F}} = - \left( {{e^y}\cos z} \right){\rm{i}} - \left( {{e^z}\cos x} \right){\rm{j}} - \left( {{e^x}\cos y} \right){\rm{k}}\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left\langle { - {e^y}\cos z, - {e^z}\cos x, - {e^x}\cos y} \right\rangle \end{array}\)

Thus, the\({\mathop{\rm curl}\nolimits} {\bf{F}}\) is \(\left\langle { - {e^y}\cos z, - {e^z}\cos x, - {e^x}\cos y} \right\rangle \)

Calculation for the divergence of the vector field

(b)

Consider the standard equation of a divergence of vector field for \({\rm{F}} = P{\rm{i}} + Q{\rm{j}} + R{\rm{k}}\) \({\mathop{\rm div}\nolimits} {\rm{F}} = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\)

Substitute \({e^x}\sin y\) for \(P,{e^y}\sin z\) for \(Q\) and \({e^z}\sin x\) for \(R\) in equation \(\left( 2 \right),\)

\(\begin{array}{l}{\mathop{\rm div}\nolimits} {\rm{F}} = \frac{\partial }{{\partial x}}\left( {{e^x}\sin y} \right) + \frac{\partial }{{\partial y}}\left( {{e^y}\sin z} \right) + \frac{\partial }{{\partial z}}\left( {{e^z}\sin x} \right)\\{\mathop{\rm div}\nolimits} {\rm{F}} = (\sin y)\frac{\partial }{{\partial x}}\left( {{e^x}} \right) + (\sin z)\frac{\partial }{{\partial y}}\left( {{e^y}} \right) + (\sin x)\frac{\partial }{{\partial z}}\left( {{e^z}} \right)\\{\mathop{\rm div}\nolimits} {\rm{F}} = (\sin y)\left( {{e^x}} \right) + (\sin z)\left( {{e^y}} \right) + (\sin x)\left( {{e^z}} \right)\\{\mathop{\rm div}\nolimits} {\rm{F}} = {e^x}\sin y + {e^y}\sin z + {e^z}\sin x\end{array}\)

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