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anonymous

  • one year ago

Evaluate : IntegralC xy2dx + x2ydy if "c" is a piecewise smooth curve from (0,0) to (2,3).

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  1. rsst123
    • one year ago
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  2. IrishBoy123
    • one year ago
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    \(\int \ xy^2 \ dx + x^2y \ dy\) \( = \int \ <xy^2, x^2y> \bullet <dx,dy>\) you haven't specified the actual path but we can check that \(curl [<xy^2, x^2y>] = 2xy - 2yx = 0 \) so the path doesn't matter. is that the point for you? you are looking for the potential function?

  3. anonymous
    • one year ago
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    \[d\left( \frac{ x^2 y^2}{ 2} \right)=xy^2dx+x^2ydy\]

  4. anonymous
    • one year ago
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    nice @surjithayer

  5. Michele_Laino
    • one year ago
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    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

  6. Michele_Laino
    • one year ago
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    hint: since our field is conservative, then there exists a function, say U, such that: \[\Large {\mathbf{F}} = \nabla U\] or using cartesian coordinates: \[\Large \left\{ \begin{gathered} \frac{{\partial U}}{{\partial x}} = {F_x} = x{y^2} \hfill \\ \hfill \\ \frac{{\partial U}}{{\partial y}} = {F_y} = {x^2}y \hfill \\ \hfill \\ \frac{{\partial U}}{{\partial z}} = {F_z} = 0 \hfill \\ \end{gathered} \right.\]

  7. Michele_Laino
    • one year ago
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    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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