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Section 17.1 : Curl and Divergence

4. Determine if the following vector field is conservative.

\[\vec F = 6x\,\vec i + \left( {2y - {y^2}} \right)\vec j + \left( {6z - {x^3}} \right)\vec k\]

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Start Solution

We know all we need to do here is compute the curl of the vector field.

\[\begin{align*}{\mathop{\rm curl}\nolimits} \vec F & = \nabla \times \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\displaystyle \frac{\partial }{{\partial x}}}&{\displaystyle \frac{\partial }{{\partial y}}}&{\displaystyle \frac{\partial }{{\partial z}}}\\{6x}&{2y - {y^2}}&{6z - {x^3}}\end{array}} \right|\\ & = \frac{\partial }{{\partial y}}\left( {6z - {x^3}} \right)\vec i + \frac{\partial }{{\partial z}}\left( {6x} \right)\vec j + \frac{\partial }{{\partial x}}\left( {2y - {y^2}} \right)\vec k\\ & \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{\partial }{{\partial y}}\left( {6x} \right)\vec k - \frac{\partial }{{\partial x}}\left( {6z - {x^3}} \right)\vec j - \frac{\partial }{{\partial z}}\left( {2y - {y^2}} \right)\vec i\\ & = \underline {3{x^2}\vec j} \end{align*}\] Show Step 2

So, we found that \({\mathop{\rm curl}\nolimits} \vec F = 3{x^2}\vec j \ne \vec 0\) for this vector field and so the vector field is NOT conservative.