Question

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

\({\rm{F}}(x,y,z) = x{e^{xy}}\sin z\,j + y{\tan ^{ - 1}}\left( {x/z} \right)k\)

Short answer

(a) The \({\mathop{\rm curl}\nolimits} {\bf{F}}\) is \(\left[ {{{\tan }^{ - 1}}\left( {\frac{x}{z}} \right) - {e^{xy}}\cos z} \right]{\rm{i}} - \left( {\frac{{yz}}{{1 + {x^2}}}} \right){\rm{j}} + \left( {y{e^{xy}}\sin z} \right){\rm{k}}\).

(b) The divergence of the vector field is \(x{e^{xy}}\sin z - \frac{{xy}}{{{x^2} + {z^2}}}\).

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 \({\rm{F}}(x,y,z) = {e^{xy}}\sin z{\rm{j}} + y{\tan ^{ - 1}}\left( {\frac{x}{z}} \right){\rm{k}}\)

Substitute \(0\) for \(P,{e^{xy}}\sin z\) for \(Q\) and \(y{\tan ^{ - 1}}\left( {\frac{x}{z}} \right)\) for \(R\) in equation \(\left( 1 \right),\)

Thus, the\({\mathop{\rm curl}\nolimits} {\bf{F}}\)is\(\left[ {{{\tan }^{ - 1}}\left( {\frac{x}{z}} \right) - {e^{xy}}\cos z} \right]{\rm{i}} - \left( {\frac{{yz}}{{1 + {x^2}}}} \right){\rm{j}} + \left( {y{e^{xy}}\sin z} \right){\rm{k}}\).

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 \(0\) for \(P,{e^{xy}}\sin z\) for \(Q\) and \(y{\tan ^{ - 1}}\left( {\frac{x}{z}} \right)\) for \(R\) in equation \(\left( 2 \right),\)

Thus, the divergence of the vector field is \(x{e^{xy}}\sin z - \frac{{xy}}{{{x^2} + {z^2}}}\).