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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2. IrishBoy123

$$\int \ xy^2 \ dx + x^2y \ dy$$ $$= \int \ <xy^2, x^2y> \bullet <dx,dy>$$ 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

$d\left( \frac{ x^2 y^2}{ 2} \right)=xy^2dx+x^2ydy$

4. anonymous

nice @surjithayer

5. 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

6. 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: $\Large \left\{ \begin{gathered} \frac{{\partial U}}{{\partial x}} = {F_x} = x{y^2} \hfill \\ \hfill \\ \frac{{\partial U}}{{\partial y}} = {F_y} = {x^2}y \hfill \\ \hfill \\ \frac{{\partial U}}{{\partial z}} = {F_z} = 0 \hfill \\ \end{gathered} \right.$

7. 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: $\Large U\left( {2,3,z} \right) - U\left( {0,0,z} \right) = ...?$