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.