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).

Answers (1)

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, \(\displaystyle\int_{{C}}{F}\cdot{d}{r}=-{6}{\left[{3}{x}-\frac{{x}^{{2}}}{{2}}\right]}=-{27}\).

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