Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 17.1 : Curl and Divergence
3. Determine if the following vector field is conservative.
\[\vec F = \left( {4{y^2} + \frac{{3{x^2}y}}{{{z^2}}}} \right)\,\vec i + \left( {8xy + \frac{{{x^3}}}{{{z^2}}}} \right)\vec j + \left( {11 - \frac{{2{x^3}y}}{{{z^3}}}} \right)\vec k\]Show All Steps Hide All Steps
Start SolutionWe know all we need to do here is compute the curl of the vector field.
\[\begin{align*}{\mathop{\rm curl}\nolimits} \vec F = \nabla \times \vec F &= \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\displaystyle \frac{\partial }{{\partial x}}}&{\displaystyle \frac{\partial }{{\partial y}}}&{\displaystyle \frac{\partial }{{\partial z}}}\\{\displaystyle 4{y^2} + \frac{{3{x^2}y}}{{{z^2}}}}&{\displaystyle 8xy + \frac{{{x^3}}}{{{z^2}}}}&{\displaystyle 11 - \frac{{2{x^3}y}}{{{z^3}}}}\end{array}} \right|\\ & = \frac{\partial }{{\partial y}}\left( {11 - \frac{{2{x^3}y}}{{{z^3}}}} \right)\vec i + \frac{\partial }{{\partial z}}\left( {4{y^2} + \frac{{3{x^2}y}}{{{z^2}}}} \right)\vec j + \frac{\partial }{{\partial x}}\left( {8xy + \frac{{{x^3}}}{{{z^2}}}} \right)\vec k\\ & \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\,\,\,\,\,\, - \frac{\partial }{{\partial y}}\left( {4{y^2} + \frac{{3{x^2}y}}{{{z^2}}}} \right)\vec k - \frac{\partial }{{\partial x}}\left( {11 - \frac{{2{x^3}y}}{{{z^3}}}} \right)\vec j - \frac{\partial }{{\partial z}}\left( {8xy + \frac{{{x^3}}}{{{z^2}}}} \right)\vec i\\ & = - \frac{{2{x^3}}}{{{z^3}}}\vec i - \frac{{6{x^2}y}}{{{z^3}}}\vec j + \left( {8y + \frac{{3{x^2}}}{{{z^2}}}} \right)\vec k - \left( {8y + \frac{{3{x^2}}}{{{z^2}}}} \right)\vec k + \frac{{6{x^2}y}}{{{z^3}}}\vec j + \frac{{2{x^3}}}{{{z^3}}}\vec i\\ & = \underline {\vec 0} \end{align*}\] Show Step 2So, we found that \({\mathop{\rm curl}\nolimits} \vec F = \vec 0\) for this vector field and so the vector field is conservative.