The work done by the force F=(x^2-y^2)i-(2xy+y)j displacing a particle in the xy plane from (0, 0) to (1, 1) along the parabola  is

Question : The work done by the force F=(x^2-y^2)i-(2xy+y)j displacing a particle in the xy plane from (0, 0) to (1, 1) along the parabola  isSolution :Given force \[\vec{F}=\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\] Then curl of F is \[ curl\left(\vec{F}\right)=\nabla\times\vec{F}=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)\times\left(\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\right)\] \[=\left(\hat{i}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial x}-\left(\hat{i}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial x}+\left(\hat{j}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial…

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