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Section 5-6 : Conservative Vector Fields

For problems 1 – 3 determine if the vector field is conservative.

  1. \(\vec F = \left( {{x^3} - 4x{y^2} + 2} \right)\vec i + \left( {6x - 7y + {x^3}{y^3}} \right)\vec j\) Solution
  2. \(\vec F = \left( {2x\sin \left( {2y} \right) - 3{y^2}} \right)\vec i + \left( {2 - 6xy + 2{x^2}\cos \left( {2y} \right)} \right)\vec j\) Solution
  3. \(\vec F = \left( {6 - 2xy + {y^3}} \right)\vec i + \left( {{x^2} - 8y + 3x{y^2}} \right)\vec j\) Solution

For problems 4 – 8 find the potential function for the vector field.

  1. \(\displaystyle \vec F = \left( {6{x^2} - 2x{y^2} + \frac{y}{{2\sqrt x }}} \right)\vec i - \left( {2{x^2}y - 4 - \sqrt x } \right)\vec j\) Solution
  2. \(\vec F = {y^2}\left( {1 + \cos \left( {x + y} \right)} \right)\vec i + \left( {2xy - 2y + {y^2}\cos \left( {x + y} \right) + 2y\sin \left( {x + y} \right)} \right)\vec j\) Solution
  3. \(\vec F = \left( {2{z^4} - 2y - {y^3}} \right)\vec i + \left( {z - 2x - 3x{y^2}} \right)\vec j + \left( {6 + y + 8x{z^3}} \right)\vec k\) Solution
  4. \(\displaystyle \vec F = \frac{{2xy}}{{{z^3}}}\vec i + \left( {2y - {z^2} + \frac{{{x^2}}}{{{z^3}}}} \right)\vec j - \left( {4{z^3} + 2yz + \frac{{3{x^2}y}}{{{z^4}}}} \right)\vec k\) Solution
  5. Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) where C is the portion of the circle centered at the origin with radius 2 in the 1st quadrant with counter clockwise rotation and \(\displaystyle \vec F\left( {x,y} \right) = \left( {2xy - 4 - \frac{1}{2}\sin \left( {\frac{1}{2}x} \right)\sin \left( {\frac{1}{2}y} \right)} \right)\,\vec i + \left( {{x^2} + \frac{1}{2}\cos \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right)} \right)\vec j\). Solution
  6. Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) where \(\vec F\left( {x,y} \right) = \left( {2y{{\bf{e}}^{x\,y}} + 2x{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}} \right)\,\vec i + \left( {2x{{\bf{e}}^{x\,y}} - 2y{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}} \right)\vec j\) and C is the curve shown below.
    Solution