Solution:

The vector field is \(\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}\)

Find: \(\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}\)

Then, the double integral becomes \(\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}=\int_{{C}}{F}·{d}{r}\)

The surface of a cone is \(\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}\).

At \(\displaystyle{y}={2},{x}^{{2}}+{z}^{{2}}={4}\).

Let the parametric equation is \(\displaystyle{r}{\left({t}\right)}={<}{2}{\cos{{t}}},{2},{2}{\sin{{t}}}{>}\forall{0}≤{t}≤{2}\pi\) and its derivative is \(\displaystyle{r}'{\left({t}\right)}={<}−{2}{\sin{{t}}},{0},{2}{\cos{{t}}}{>}\).

Then, the vector field becomes

\(\displaystyle{F}{\left({r}{\left({t}\right)}\right)}=={\left({2}{\cos{{t}}}\right)}^{{2}}{\left({2}\right)}^{{3}}{\left({2}{\sin{{t}}}\right)}{i}+{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}}{j}+{8}{\sin{{t}}}{\cos{{t}}}{k}{\left({64}{{\cos}^{{2}}{t}}{\sin{{t}}}\right)}{i}+{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}}{j}+{8}{\sin{{t}}}{\cos{{t}}}{k}\)

Step 2

Obtain the integal:

\(\displaystyle\int_{{C}}{F}·{d}{r}={\int_{{0}}^{{{2}\pi}}}{0}{<}{64}{{\cos}^{{2}}{t}}{\sin{{t}}},{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}},{8}{\sin{{t}}}{\cos{{t}}}{>}{<}−{2}{\sin{{t}}},{0},{2}{\cos{{t}}}{>}{\int_{{0}}^{{{2}\pi}}}{\left(−{128}{{\cos}^{{2}}{t}}{{\sin}^{{2}}{t}}+{16}{\sin{{t}}}{{\cos}^{{2}}{t}}\right)}{\left.{d}{t}\right.}−{32}\pi\)

Therefore, \(\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}=−{32}\pi\).

The vector field is \(\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}\)

Find: \(\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}\)

Then, the double integral becomes \(\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}=\int_{{C}}{F}·{d}{r}\)

The surface of a cone is \(\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}\).

At \(\displaystyle{y}={2},{x}^{{2}}+{z}^{{2}}={4}\).

Let the parametric equation is \(\displaystyle{r}{\left({t}\right)}={<}{2}{\cos{{t}}},{2},{2}{\sin{{t}}}{>}\forall{0}≤{t}≤{2}\pi\) and its derivative is \(\displaystyle{r}'{\left({t}\right)}={<}−{2}{\sin{{t}}},{0},{2}{\cos{{t}}}{>}\).

Then, the vector field becomes

\(\displaystyle{F}{\left({r}{\left({t}\right)}\right)}=={\left({2}{\cos{{t}}}\right)}^{{2}}{\left({2}\right)}^{{3}}{\left({2}{\sin{{t}}}\right)}{i}+{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}}{j}+{8}{\sin{{t}}}{\cos{{t}}}{k}{\left({64}{{\cos}^{{2}}{t}}{\sin{{t}}}\right)}{i}+{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}}{j}+{8}{\sin{{t}}}{\cos{{t}}}{k}\)

Step 2

Obtain the integal:

\(\displaystyle\int_{{C}}{F}·{d}{r}={\int_{{0}}^{{{2}\pi}}}{0}{<}{64}{{\cos}^{{2}}{t}}{\sin{{t}}},{\sin{{\left({8}{\sin{{t}}}{\cos{{t}}}\right)}}},{8}{\sin{{t}}}{\cos{{t}}}{>}{<}−{2}{\sin{{t}}},{0},{2}{\cos{{t}}}{>}{\int_{{0}}^{{{2}\pi}}}{\left(−{128}{{\cos}^{{2}}{t}}{{\sin}^{{2}}{t}}+{16}{\sin{{t}}}{{\cos}^{{2}}{t}}\right)}{\left.{d}{t}\right.}−{32}\pi\)

Therefore, \(\displaystyle\int\int_{{S}}{c}{u}{r}{l}{F}·{d}{S}=−{32}\pi\).