anonymous
  • anonymous
Evaluate : IntegralC xy2dx + x2ydy if "c" is a piecewise smooth curve from (0,0) to (2,3).
Mathematics
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katieb
  • katieb
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IrishBoy123
  • IrishBoy123
\(\int \ xy^2 \ dx + x^2y \ dy\) \( = \int \ \bullet \) you haven't specified the actual path but we can check that \(curl [] = 2xy - 2yx = 0 \) so the path doesn't matter. is that the point for you? you are looking for the potential function?
anonymous
  • anonymous
\[d\left( \frac{ x^2 y^2}{ 2} \right)=xy^2dx+x^2ydy\]

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anonymous
  • anonymous
nice @surjithayer
Michele_Laino
  • Michele_Laino
if we write your field as follows: \[\Large {\mathbf{F}} = \left( {x{y^2},{x^2}y,0} \right)\] then we have: \[\Large \nabla \times {\mathbf{F}} = 0\] so, good job @IrishBoy123
Michele_Laino
  • Michele_Laino
hint: since our field is conservative, then there exists a function, say U, such that: \[\Large {\mathbf{F}} = \nabla U\] or using cartesian coordinates: \[\Large \left\{ \begin{gathered} \frac{{\partial U}}{{\partial x}} = {F_x} = x{y^2} \hfill \\ \hfill \\ \frac{{\partial U}}{{\partial y}} = {F_y} = {x^2}y \hfill \\ \hfill \\ \frac{{\partial U}}{{\partial z}} = {F_z} = 0 \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
hint: after a simple integration of that system I got this function: \[\Large U\left( {x,y,z} \right) = \frac{{{x^2}{y^2}}}{2} + k\] where k is an arbitrary real constant so your integral is given by this subsequent computation: \[\Large U\left( {2,3,z} \right) - U\left( {0,0,z} \right) = ...?\]

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