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Section 6-1 : Curl and Divergence

1. Compute \({\mathop{\rm div}\nolimits} \vec F\) and \({\mathop{\rm curl}\nolimits} \vec F\) for \(\vec F = {x^2}y\,\vec i - \left( {{z^3} - 3x} \right)\vec j + 4{y^2}\vec k\).

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Let’s compute the divergence first and there isn’t much to do other than run through the formula.

\[{\mathop{\rm div}\nolimits} \vec F = \nabla \centerdot \vec F = \frac{\partial }{{\partial x}}\left( {{x^2}y} \right) + \frac{\partial }{{\partial y}}\left( {3x - {z^3}} \right) + \frac{\partial }{{\partial z}}\left( {4{y^2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{2xy}}\]

Be careful to watch for minus signs in front of any of the vector components (2nd component in this case!). It is easy to get in a hurry and miss them.

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The curl is a little more work but still just formula work so here is the curl.

\[\begin{align*}{\mathop{\rm curl}\nolimits} \vec F & = \nabla \times \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\displaystyle \frac{\partial }{{\partial x}}}&{\displaystyle \frac{\partial }{{\partial y}}}&{\displaystyle \frac{\partial }{{\partial z}}}\\{{x^2}y}&{3x - {z^3}}&{4{y^2}}\end{array}} \right|\\ & = \frac{\partial }{{\partial y}}\left( {4{y^2}} \right)\vec i + \frac{\partial }{{\partial z}}\left( {{x^2}y} \right)\vec j + \frac{\partial }{{\partial x}}\left( {3x - {z^3}} \right)\vec k - \frac{\partial }{{\partial y}}\left( {{x^2}y} \right)\vec k - \frac{\partial }{{\partial x}}\left( {4{y^2}} \right)\vec j - \frac{\partial }{{\partial z}}\left( {3x - {z^3}} \right)\vec i\\ & = 8y\vec i + 3\vec k - {x^2}\vec k + 3{z^2}\vec i\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\left( {8y + 3{z^2}} \right)\vec i + \left( {3 - {x^2}} \right)\vec k}}\end{align*}\]

Again, don’t forget the minus sign on the 2nd component.