Question

# Use Stokes' Theorem to evaluate int_C F * dr where C is oriented counterclockwise as viewed from above. F(x,y,z)=(x+y^2)i+(y+z^2)j+(z+x^2)k, C is the triangle with vertices (3,0,0),(0,3,0), and (0,0,3).

Use Stokes' Theorem to evaluate $$\displaystyle\int_{{C}}{F}\cdot{d}{r}$$ where C is oriented counterclockwise as viewed from above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}+{y}^{{2}}\right)}{i}+{\left({y}+{z}^{{2}}\right)}{j}+{\left({z}+{x}^{{2}}\right)}{k}$$,
C is the triangle with vertices (3,0,0),(0,3,0), and (0,0,3).

2021-02-26

Solution:
The vector field is $$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}+{y}^{{2}}\right)}{i}+{\left({y}+{z}^{{2}}\right)}{j}+{\left({z}+{x}^{{2}}\right)}{k}$$.
The equation of the plane is $$x+y+z=3$$.
Consider $$z=g(x,y)=3−x−y.$$
Use Stokes’ Theorem and get the surface integral set up.
$$\int_{C} F \times dr = \int \int_{S} curl\ F \times dS = \int \int_{S} curlF \times grad\ fdA$$
Obtain the curl of F:
$$\displaystyle{c}{u}{r}{l}{F}={\left[\begin{array}{ccc} {i}&{j}&{k}\\\frac{\partial}{{\partial{x}}}&\frac{\partial}{{\partial{y}}}&\frac{\partial}{{\partial{z}}}\\{x}+{y}^{{2}}&{y}+{z}^{{2}}&{z}+{x}^{{2}}\end{array}\right]}={<}-{2}{z},-{2}{x},-{2}{y}{>}$$
Conclusion:
Then, $$\displaystyle{f{{\left({x},{y},{z}\right)}}}={z}-{3}+{x}+{y}.\nabla{f}={<}{1},{1},{1}{>}$$
Plug $$\displaystyle{z}={0},{x}+{y}={3}\Rightarrow{y}={3}-{x}$$
$$\displaystyle\int_{{C}}{F}\cdot{d}{r}=\int\int_{{S}}{\left\langle-{2}{z},-{2}{x},-{2}{y}\right\rangle}\cdot{<}{1},{1},{1}{>}{d}{A}$$
$$\displaystyle={\int_{{0}}^{{3}}}{\int_{{0}}^{{{3}-{x}}}}-{2}{\left({z}+{x}+{y}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle=-{6}{\int_{{0}}^{{3}}}{\int_{{0}}^{{{3}-{x}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\because{z}+{x}+{y}={3}$$
$$\displaystyle=-{6}{\int_{{0}}^{{3}}}{\left[{3}-{x}\right]}{\left.{d}{x}\right.}$$
That is, $$\displaystyle\int_{{C}}{F}\cdot{d}{r}=-{6}{\left[{3}{x}-\frac{{x}^{{2}}}{{2}}\right]}=-{27}$$.