Question

Use Stokes' Theorem to compute oint_C 1/2 z^2 dx + (xy)dy+2020dz, where C is the triangle with vertices at(1,0,0),(0,2,0), and (0,0,2) traversed in the order.

Use Stokes' Theorem to compute \(\displaystyle\oint_{{C}}\frac{{1}}{{2}}{z}^{{2}}{\left.{d}{x}\right.}+{\left({x}{y}\right)}{\left.{d}{y}\right.}+{2020}{\left.{d}{z}\right.}\), where C is the triangle with vertices at(1,0,0),(0,2,0), and (0,0,2) traversed in the order.

Answers (1)

2021-02-06

Step 1
The aim is to find \(\displaystyle\oint_{{C}}\frac{{1}}{{2}}{z}^{{2}}{\left.{d}{x}\right.}+{x}{y}{\left.{d}{y}\right.}+{2020}{\left.{d}{z}\right.}\) where C is the triangle.
(1, 0, 0), (0, 2, 0), (0, 0, 2).
Step 2
Given the stoke’s theorem,
\(\int \int_{S}\ curl\ F. dS = \int_C F.dr\)
Where S is \(\displaystyle\frac{{x}}{{1}}+\frac{{y}}{{2}}+\frac{{z}}{{2}}={1}\)
Then curl \(\displaystyle{F}={\left(\frac{{\partial{R}}}{{\partial{y}}}-\frac{{\partial{Q}}}{{\partial{z}}}\right)}{i}+{\left(\frac{{\partial{P}}}{{\partial{z}}}-\frac{{\partial{R}}}{{\partial{x}}}\right)}{j}+{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{k}\)
Here \(\displaystyle{F}=\frac{{1}}{{2}}{Z}^{{2}}{i}+{x}{y}{j}+{2020}{k}\)
\(\displaystyle{c}{u}{r}{l}{F}={\left(\frac{{\partial{R}}}{{\partial{y}}}-\frac{{\partial{Q}}}{{\partial{z}}}\right)}{i}+{\left(\frac{{\partial{P}}}{{\partial{z}}}-\frac{{\partial{R}}}{{\partial{x}}}\right)}+{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}={z}{j}+{y}{k}\)
If the surface S is,
\(\displaystyle\int\int_{{S}}{F}.{d}{S}=\int\int_{{D}}-{P}{\left(\frac{{\partial{g}}}{{\partial{x}}}\right)}-{Q}{\left(\frac{{\partial{g}}}{{\partial{y}}}\right)}+{R}{d}{A}\)
Where \(F = Pi + Qj + Rk\).
Where, \( z = 2-y-2x.\)
Then using the stoke's theorem,
\(\displaystyle\int\int_{{S}}\ {c}{u}{r}{l}\ {F}.{d}{S}=\int\int_{{D}}-{2}{z}^{{2}}={x}{y}+{2020}{d}{A}\)
Substitute \(z = 2-y-2x\),
\(\displaystyle\int\int{C}{u}{r}{l}{F}\int\int_{{D}}-{2}+{2}{x}+{2}{y}-{2}{x}-{2}{y}{d}{A}\)
\(\displaystyle=-{2}\int\int_{{D}}{d}{A}\)
Hence, the area of the triangle is \(\displaystyle\frac{{{b}{a}{s}{e}\times{h}{e}{i}{g}{h}{t}}}{{2}}=\frac{{1}}{{2}}\).

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