The work done by the force vec{F} = x^2 hat{i | vedclass

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(B) The force is given by $\vec{F}(x, y) = x^2 \hat{i} + y^2 \hat{j}$.For a force to be conservative,the curl of the force must be zero,i.e.,$
abla \

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zero

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(B) The force is given by $\vec{F}(x, y) = x^2 \hat{i} + y^2 \hat{j}$.For a force to be conservative,the curl of the force must be zero,i.e.,$
abla 	imes \vec{F} = 0$.Here,$\frac{\partial F_y}{\partial x} = \frac{\partial}{\partial x}(y^2) = 0$ and $\frac{\partial F_x}{\partial y} = \frac{\partial}{\partial y}(x^2) = 0$.Since $\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 0$,the force $\vec{F}$ is a conservative force.The work done by a conservative force around any closed path is always zero.Therefore,the work done by the force $\vec{F}$ around the closed path $OABC$ is zero.

The work done by the force $\vec{F} = x^2 \hat{i} + y^2 \hat{j}$ around the closed path $OABC$ shown in the figure is:

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