Step 1

Stock's Theorem:-

\(\displaystyle\int_{{C}}{F}.{d}{r}\int\int{c}{u}{r}{l}{\left({F}\right)}{d}{s}\)

Given that a curve say D (as name not mention in problem) of intersection of the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={1}\) and the plane z=x and \(F=y\ i+z\ j+ 2x\ k\)

Also, S be inside of this curve, oriented with upward-pointing normal, parameterize of this curve is

\(\displaystyle{r}={<}{x},{y},{z}\ge{<}{x},{y},{x}{>}\) as z = z

\(\displaystyle{r}={\cos{{0}}}{i}+{\sin{{0}}}{j}+{\cos{{0}}}{k}\)

since, we know \(\displaystyle{x}={1}.{\cos{{0}}}={\cos{{0}}},{y}={1}.{\sin{{0}}}={\sin{{0}}},{z}={x}={\cos{{0}}}\) as radius =1

Here, \(\displaystyle{0}\le{0}\le{2}\pi{\left({a}{s}{x}^{{2}}+{y}^{{2}}={1}^{{2}}\right)}\)

then,

\(\displaystyle{r}={<}{\cos{{0}}},{\sin{{0}}},{\cos{{0}}}{>}\)

\(\displaystyle\frac{{{d}{r}}}{{{d}{0}}}={r}'={<}-{\sin{{0}}},{\cos{{0}}},-{\sin{{0}}}{>}\)

\(F(x,y,z)=yi+zj+2xk\)

\(\displaystyle{F}{\left({r}{\left({0}\right)}\right)}={\sin{{0}}}{i}+{\cos{{0}}}{j}+{2}{\cos{{0}}}{k}\)

Step 2

Now,

\(\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{F}{\left({r}{\left({0}\right)}\right)}\frac{{{d}{r}}}{{{d}{0}}}{d}{0}\)

\(\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{<}{\sin{{0}}},{\cos{{0}}},{2}{\cos{{0}}}{>}{<}-{\sin{{0}}},{\cos{{0}}},-{\sin{{0}}}{>}{d}{0}\)

\(\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{\left(-{{\sin}^{{2}}{0}}+{{\cos}^{{2}}{0}}-{2}{\sin{{0}}}{\cos{{0}}}\right)}{d}{0}\)

Note: \(\displaystyle-{{\cos}^{{2}}{0}}-{{\sin}^{{2}}{0}}={\cos{{20}}},{\sin{{20}}}={2}{\sin{{0}}}{\cos{{0}}}\)

\(\displaystyle\int_{{D}}{F}{d}{r}={\int_{{0}}^{{{2}\pi}}}{\left({\cos{{20}}}-{\sin{{20}}}\right)}{d}{0}={{\left[\frac{{{\sin{{20}}}}}{{2}}+\frac{{{\cos{{20}}}}}{{2}}\right]}_{{0}}^{{{2}\pi}}}\)

Note: \(\displaystyle-{\sin{{\left({2}\pi\right)}}}={\sin{{\left({0}\right)}}}={0},{\cos{{\left({2}\pi\right)}}}-{\cos{{\left({0}\right)}}}={1}-{1}={0}\)

\(\displaystyle\int_{{D}}{F}{d}{r}={0}\)

By, using Stock's Theorem, we have

\(\displaystyle\int\int{c}{u}{r}{l}{\left({F}\right)}{d}{s}=\int_{{D}}{F}.{d}{r}={0}\)

\(\displaystyle\int\int{c}{u}{r}{l}{\left({F}\right)}{d}{s}={0}\)