Step 1

Here use the stokes theorem to find the integral of the function.

And given that the function \(\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}\)

The curve of intersection of the plane \(x + z =10\) and the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={9}\)

Step 2

The Stokes theorem is given as:

\(\displaystyle\int_{{C}}{F}\cdot{d}{r}=\int\int_{{S}}{c}{u}{r}{l}{F}\cdot{d}{s}\)

if \(\displaystyle{F}={P}\hat{{{i}}}+{Q}\hat{{{j}}}+{R}\hat{{{k}}}\)

Then \(\displaystyle{C}{u}{r}{l}{F}={\left({R}_{{y}}-{Q}_{{z}}\right)}\hat{{{i}}}+{\left({P}_{{z}}-{R}_{{x}}\right)}\hat{{{j}}}+{\left({Q}_{{x}}-{P}_{{y}}\right)}\hat{{{k}}}\)

Given \(\displaystyle{F}={x}{y}\hat{{{i}}}+{3}{z}\hat{{{j}}}+{5}{y}\hat{{{k}}}\)

Therefore,

Curl \(\displaystyle{F}={\left({5}-{3}\right)}\hat{{{i}}}+{\left({0}-{0}\right)}\hat{{{j}}}+{\left({0}-{x}\right)}\hat{{{k}}}={2}\hat{{{i}}}-{x}\hat{{{k}}}\)

Step 3

Let S be the path of the plane \(x + z =10\) inside the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={9}\)

If the surface S is of the form \(z = g(x, y)\) and is oriented upwards,

So, by the formula:

\(\displaystyle\int\int_{{S}}{F}\cdot{d}{S}=\int\int_{{D}}-{P}{\left(\frac{{\partial{g}}}{{\partial{x}}}-{Q}\frac{{\partial{g}}}{{\partial{y}}}\right)}+{R}{d}{A}\)

Here S is the part of the plane \(z = 10-x\)

And D is region inside the circle \(\displaystyle{x}^{{2}}+{y}^{{2}}={0},{z}={0}\)

\(\displaystyle\int\int_{{S}}{C}{u}{r}{l}{F}\cdot{d}{s}=\int\int_{{D}}{\left({2}-{x}\right)}{d}{A}\)

Step 4

Solve further:

\(\displaystyle\int\int_{{S}}{F}\cdot{d}{S}=={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{3}}}{\left({2}-{r}^{{2}}{\cos{{0}}}\right)}{r}{d}{r}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\left[{\int_{{0}}^{{3}}}{2}{r}{d}{r}-{\int_{{0}}^{{3}}}{r}^{{3}}{\cos{{\left({0}\right)}}}{d}{r}\right]}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{{\left[\frac{{r}^{{2}}}{{2}}+\frac{{r}^{{4}}}{{4}}{\cos{{0}}}\right]}_{{0}}^{{3}}}{d}{0}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\left[{9}-{\cos{{\left({0}\right)}}}\frac{{81}}{{4}}\right]}{d}{0}\)

\(\displaystyle={18}\pi-{0}={56.54}\)