Question

Is there a vector field \({\rm{G}}\) on \({R^3}\)such that \(curl\,G = \left( {x\sin y,\cos y,z - xy} \right)\)? Explain.

Short answer

There is no vector field \({\rm{G}}\) on \({\mathbb{R}^3}\).

Step-by-step solution

Concept of Curl of the vector field

The curl is a vector operator that describes the infinitesimal circulation of a vector field in three-dimensional Euclidean space.

The curl of a vector field\({\bf{F}}(x,y,z)\)is the vector field

\({\mathop{\rm curl}\nolimits} {\bf{F}} = \nabla \times {\bf{F}} = \left( {\frac{{\partial {F_3}}}{{\partial y}} - \frac{{\partial {F_2}}}{{\partial z}}} \right)\widehat \imath - \left( {\frac{{\partial {F_3}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial z}}} \right)\widehat \jmath + \left( {\frac{{\partial {F_2}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial y}}} \right)\widehat {\bf{k}}\)

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

The given equation is\({\mathop{\rm curlG}\nolimits} = \langle x\sin y,\cos y,z - xy\rangle \).

Consider the standard equation of a divergence of vector field for \({\rm{F}} = P{\rm{i}} + Q{\rm{j}} + R{\rm{k}}\). \({\mathop{\rm div}\nolimits} {\rm{F}} = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}\;\;\;\;\;\;\;\;\;(1)\)

For \({\mathop{\rm div}\nolimits} ({\mathop{\rm curl}\nolimits} {\rm{F}})\),

Substitute \(x\sin y\) for \(P,\cos y\) for \(Q\) and \(z - xy\) for \(R\) in equation \(\left( 1 \right),\)

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

Since\({\mathop{\rm div}\nolimits} ({\mathop{\rm curl}\nolimits} {\rm{F}}) \ne 0\), there is no vector field \({\rm{G}}\) on \({R^3}\).

Thus, there is no vector field \({\rm{G}}\) on \({R^3}\).