A body is displaced from origin to ( 1m,1m) by a force $F=\left(2y\hat{i}+3x^2\hat{j}\right)$ along the path x=y. Find the work done in Joules along the path.

$\begin{array}{l}F=\left(2y\widehat{i}+3x^2\widehat{j}\right)\\ dx=(dx\widehat{i}+dy\widehat{j})\\ Fdx=\left(2ydx+3x^{2}dy\right)\end{array}$

we cannot integrate F.dr or $\left(2ydx+3x^{2}dy\right)$ as such to find the work done.

But we can change this expression.

Along the path x = y 

$\begin{array}{l}\left(2ydx+3x^{2}dy\right)=\left(2xdx+3y^{2}dy\right)\\ W_1={\int }_{(0,0)}^{(1m,1m)} Fdr={\int }_{(0,0)}^{(1m,1m)} \left(2xdx+3y^{2}dy\right)\\ ={\left[x^2+y^3\right]}_{(0,0)}^{(1,1)}\\ =2J\end{array}$