# Use Stokes' Theorem to evaluate int_C F*dr where C is oriented counterclockwise as viewed above.F(x,y,z)=xy i + 3zj+5yk, C is the curve of intersection of the plane x+z=10 and the cylinder x^2+y^2=9.

Use Stokes' Theorem to evaluate $$\int_C F \cdot dr$$ where C is oriented counterclockwise as viewed above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}$$, C is the curve of intersection of the plane $$x+z=10$$ and the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={9}$$.

2021-01-07

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:
$$\displaystyle\int\int_{{S}}{F}\cdot{d}{S}=={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{3}}}{\left({2}-{r}^{{2}}{\cos{{0}}}\right)}{r}{d}{r}{d}{0}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{\left[{\int_{{0}}^{{3}}}{2}{r}{d}{r}-{\int_{{0}}^{{3}}}{r}^{{3}}{\cos{{\left({0}\right)}}}{d}{r}\right]}{d}{0}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{{\left[\frac{{r}^{{2}}}{{2}}+\frac{{r}^{{4}}}{{4}}{\cos{{0}}}\right]}_{{0}}^{{3}}}{d}{0}$$
$$\displaystyle={\int_{{0}}^{{{2}\pi}}}{\left[{9}-{\cos{{\left({0}\right)}}}\frac{{81}}{{4}}\right]}{d}{0}$$
$$\displaystyle={18}\pi-{0}={56.54}$$