Green's theorem

Tuesday, January 18, 2011

Green's theorem

Hello!

5b. Green's theorem gives us a possibility to compute the area of a plane region integrating along its boundary. Actually, it can help for more complex tasks then computing area. There are two (and more) forms of that integral,

$\displaystyle{\left|\oint_{{C}}{x}{\left.{d}{y}\right.}\right|}$ and $\displaystyle{\left|\oint_{{C}}{y}{\left.{d}{x}\right.}\right|},$ where $\displaystyle{C}$ is the bounding curve.

In our case $\displaystyle{y}$ better suits as the independent variable, so compute $\displaystyle\oint_{{C}}{x}{\left.{d}{y}\right.}.$ The curve consists of two parts, and the integral is the sum of two integrals, $\displaystyle\int_{{{C}_{{1}}}}$ and $\displaystyle\int_{{{C}_{{2}}}},$ where $\displaystyle{C}_{{1}}$ is the segment and $\displaystyle{C}_{{2}}$ is the semi-circumference. Note that to get round the boundary in the right direction, we have to integrate over $\displaystyle{C}_{{1}}$ from the larger $\displaystyle{y}$ to the smaller.

The rest is simple,

$\displaystyle\int_{{{C}_{{1}}}}={\int_{{1}}^{{-{1}}}}{x}{\left.{d}{y}\right.}={\int_{{1}}^{{-{1}}}}{1}{\left.{d}{y}\right.}=-{2},$ and $\displaystyle\int_{{{C}_{{2}}}}={\int_{{-{1}}}^{{{1}}}}{x}{\left.{d}{y}\right.}={\int_{{-{1}}}^{{{1}}}}{{y}}^{{2}}{\left.{d}{y}\right.}=\frac{{2}}{{3}}.$

So the area is $\displaystyle{\left|-{2}+\frac{{2}}{{3}}\right|}$ = 4/3.

The question 5a should be asked separately. If you would ask it again, please make clear what symbol is after $\displaystyle{\left({x}+{2}{{y}}^{{2}}\right)}.$ $\displaystyle{\left(-{j}\right)}$ which means unit vector of the y-axis?

Ninenotes

Tuesday, January 18, 2011

Green's theorem

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