Electrostatics and integration

we multiply by (idx+jdy+kdz) then we partial integrate from 0 to 2 on x axis and from 0 to 2 on y axis

Too easy: [trick required is]

$\displaystyle{6xydx+3x^2dy=3d(x^2y)}$

@Tanishq Varshney Please do not use text in the latex code , and don't randomly use Huge words . It looks ugly and difficult to read. I have edited several of your questions . It will be better and helpful for us if you take care it in future.
Thanks .

i believe that there is no need of doing $\displaystyle{6xydx+3x^2dy=3d(x^2y)}$ there's a more easy way to do it!! just integrate E with (idx+jdy+kdz) to get $\displaystyle{6xydx+3(x^2-y^2)dy)}$ and put x=y to get $\displaystyle{6x^2dx)}$ and integrate with limits from 0 to 2!!

This question can be solved without a complex integral of the sorts given below.

Just think about the journey from $(0,0,0)$ to $(2,2,2)$ in three steps:

• Moving along $X-axis$ from $(0,0,0)$ to $(2,0,0)$ .

• Moving from $(2,0,0)$ to $(2,2,0)$ along the $X-Y$ plane.

• Moving from $(2,2,0)$ to $(2,2,2)$ through space.

Now, for the first step, the work will be done only but the force along $X-axis$ , but since force along this axis is given by $6xy$ and $y=0$ for the $X-axis$ , hence the force is zero, or in other words, no work is done.

For the second step, the work is done only by the force in the upward direction, i.e. $3({x}^{2}-{y}^{2})$ . Since, for this step, the X-coordinate is fixed at $x=2$ , the work is done by the force $12-3{y}^{2}$ , which can be written as

$\int _{ 0 }^{ 2 }{ (12-3{ y }^{ 2 }) } dy$

which when simplified gives $16 V$ of potential difference.

For the third step, there is again no force acting along the $Z-axis$ and hence no work is done yet again.

Therefore, total potential developed is only due to the second step i.e. $16J$

Note

You can choose any path. Since, potential difference is a state function, choosing the path I suggested simplifies the scenario.