Question

Evaluate :
IntegralC xy2dx + x2ydy

if "c" is a piecewise smooth curve from
(0,0) to (2,3).

Answer 1

\(\int \ xy^2 \ dx + x^2y \ dy\)

\( = \int \ <xy^2, x^2y> \bullet <dx,dy>\)

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?

Answer 2

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

Answer 3

nice @surjithayer

Answer 4

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

Answer 5

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

Answer 6

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