# A particle moves along line segments from the origin to the points(3,0,0),(3,3,1),(0,3,1), and back to the origin under the influence of the force field F(x,y,z)=z^2i+3xyj+4y^2k. Use Stokes' Theorem to find the work done.

A particle moves along line segments from the origin to the points(3,0,0),(3,3,1),(0,3,1), and back to the origin under the influence of the force field $$\displaystyle{F}{\left({x},{y},{z}\right)}={z}^{{2}}{i}+{3}{x}{y}{j}+{4}{y}^{{2}}{k}$$.
Use Stokes' Theorem to find the work done.

2021-02-10

Step 1
From stokes theorem $$\displaystyle\oint{F}.{d}{r}=\oint\oint\nabla\times{F}{d}{S}$$.
Thus, calculate the curl of $$\displaystyle{F}={z}^{{2}}{i}+{3}{x}{y}{j}+{4}{y}^{{2}}{k}$$:
$$\displaystyle\nabla\times{F}={\left[\begin{array}{ccc} {i}&{j}&{k}\\\frac{\partial}{{\partial{x}}}&\frac{\partial}{{\partial{y}}}&\frac{\partial}{{\partial{z}}}\\{z}^{{2}}&{3}{x}{y}&{4}{y}^{{2}}\end{array}\right]}$$
$$=8yi+2zj+3yk$$
Step 2
Thus, $$\displaystyle\oint{F}.{d}{r}=\oint\oint{\left({8}{y}{\left.{d}{y}\right.}{\left.{d}{z}\right.}+{2}{z}{\left.{d}{x}\right.}{\left.{d}{z}\right.}+{3}{y}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\right)}$$. Now as per the given information $$\displaystyle{0}\le{x}\le{3},{0}\le{y}\le{3}{\quad\text{and}\quad}{0}\le{z}\le{1}$$.
Therefore the integral becomes:
$$\displaystyle\oint{F}.{d}{r}={\int_{{{z}={0}}}^{{1}}}{\int_{{{y}={0}}}^{{3}}}{8}{y}{\left.{d}{y}\right.}{\left.{d}{z}\right.}+{\int_{{{z}={0}}}^{{1}}}{\int_{{{x}={0}}}^{{3}}}{2}{z}{\left.{d}{x}\right.}{\left.{d}{z}\right.}+{\int_{{{x}={0}}}^{{3}}}{\int_{{{y}={0}}}^{{3}}}{3}{y}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=\frac{{1}}{{2}}{\left({8}\times{3}^{{2}}+{2}\times{3}+{3}\times{3}\times{3}^{{2}}\right)}$$
$$\displaystyle=\frac{{159}}{{2}}$$
$$=79.5$$ unit
Step 3
Thus work done is 79.5 unit.