Step 1

Given the force field,

\(\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}\)

Also, C is the surface boundary of S:\(\displaystyle{z}={4}−{4}{x}^{{2}}−{y}^{{2}},{z}\ge{0}\)

Then, C is an ellipse

\(\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}\)

Then, using the Stoke's Theorem, work done W is given by,

\(\displaystyle{W}=\int_{{C}}{F}\cdot{d}{r}\)

\(\displaystyle=\int\int_{{S}}{\left(\nabla\times{F}\right)}\cdot{d}{S}\)...(1)

Here, \(\displaystyle\nabla\times{F}={\left({0},{0},{2}\right)}\)

Also, note that the orientation of the surface is clockwise,

Hence,

\(n=(0,0,-1)\)

Step 2

Then, from equation(1),

\(\displaystyle{W}=\int\int_{{S}}{\left({0},{0},{2}\right)}\cdot{\left({0},{0},-{1}\right)}{d}{S}\)

\(\displaystyle=-{2}\int\int_{{S}}{d}{S}\)

\(\displaystyle=-{2}\times{\left(\pi\times{1}\times{2}\right)}\)

\(=-12.566\ \text{units}\)

Inverting the orientation of the surface S,

\(\displaystyle{W}=\int\int_{{S}}{\left({0},{0},{2}\right)}\cdot{\left({0},{0},{1}\right)}{d}{S}\)

\(\displaystyle=-{2}\int\int_{{S}}{d}{S}\)

\(\displaystyle=-{2}\times{\left(\pi\times{1}\times{2}\right)}\)

\(=12.566\ \text{units}\)

Thus, the correct option is,

b)Inverting the orientation of the surface S, we can apply Stoke's Theorem and conclude that \(W=12.56\) units.