Question

To determine

(a)To find the\({\mathop{\rm curl}\nolimits} {\bf{F}}\).

(b) To find the divergence of the vector field.

Short answer

(a) The\({\mathop{\rm curl}\nolimits} {\bf{F}}\)is\(x(\cos xy - \cos zx){\rm{i}} + y(\cos yz - \cos xy){\rm{j}} - z(\cos zx - \cos yz){\rm{k}}\).

(b) The divergence of the vector field is\(0\).

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\({\rm{F}}(x,y,z) = \sin yz{\rm{i}} + \sin zx{\rm{j}} + \sin xy{\rm{k}}\)

Now substitute \(\sin yz\) for \(P,\sin zx\) for \(Q\) and \(\sin xy\) for \(R\) in equation \(\left( 1 \right),\)

\(\begin{array}{l}{\mathop{\rm curl}\nolimits} {\bf{F}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\{\sin yz}&{\sin zx}&{\sin xy}\end{array}} \right|\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left\{ {\begin{array}{*{20}{c}}{\left( {\frac{\partial }{{\partial y}}(\sin xy) - \frac{\partial }{{\partial z}}(\sin zx)} \right){\rm{i}} - \left( {\frac{\partial }{{\partial z}}(\sin yz) - \frac{\partial }{{\partial x}}(\sin xy)} \right){\rm{j}} + \left( {\frac{\partial }{{\partial x}}(\sin zx) - \frac{\partial }{{\partial y}}(\sin yz)} \right){\rm{k}}}\\{}\end{array}} \right\}\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left\{ {\begin{array}{*{20}{c}}{\left( {(x)\frac{\partial }{{\partial y}}(\sin xy) - (x)\frac{\partial }{{\partial z}}(\sin zx)} \right){\rm{i}} - \left( {(y)\frac{\partial }{{\partial z}}(\sin yz) - (y)\frac{\partial }{{\partial x}}(\sin xy)} \right){\rm{j}} + \left( {(z)\frac{\partial }{{\partial x}}(\sin zx) - (z)\frac{\partial }{{\partial y}}(\sin yz)} \right){\rm{k}}}\\{}\end{array}} \right\}\\{\mathop{\rm curl}\nolimits} {\bf{F}} = \left\{ {\begin{array}{*{20}{c}}{((x)(\cos xy) - (x)(\cos zx)){\rm{i}} - ((y)(\cos yz) - (y)(\cos xy)){\rm{j}} + ((z)(\cos zx) - (z)(\cos yz)){\rm{k}}}\\{}\end{array}} \right\}\end{array}\)

On further simplification

\({\mathop{\rm curl}\nolimits} {\bf{F}} = x(\cos xy - \cos zx){\rm{i}} - y(\cos yz - \cos xy){\rm{j}} + z(\cos zx - \cos yz){\rm{k}}\)

Thus, the \({\mathop{\rm curl}\nolimits} {\bf{F}}\) is \(x(\cos xy - \cos zx){\rm{i}} + y(\cos yz - \cos xy){\rm{j}} - z(\cos zx - \cos yz){\rm{k}}\).

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 \(\sin yz\) for \(P,\sin zx\) for \(Q\) and \(\sin xy\) for \(R\) in equation \(\left( 2 \right),\)

\(\begin{array}{l}{\mathop{\rm div}\nolimits} {\rm{F}} = \frac{\partial }{{\partial x}}(\sin yz) + \frac{\partial }{{\partial y}}(\sin zx) + \frac{\partial }{{\partial z}}(\sin xy)\\{\mathop{\rm div}\nolimits} {\rm{F}} = (\sin yz)\frac{\partial }{{\partial x}}(1) + (\sin zx)\frac{\partial }{{\partial y}}(1) + (\sin xy)\frac{\partial }{{\partial z}}(1)\\{\mathop{\rm div}\nolimits} {\rm{F}} = (\sin yz)(0) + (\sin zx)(0) + (\sin xy)(0)\\{\mathop{\rm div}\nolimits} {\rm{F}} = 0 + 0 + 0\\{\mathop{\rm div}\nolimits} {\rm{F}} = 0\end{array}\)

Thus, the divergence of the vector field is\(0\).