Evaluate the line integral oint_{C} frac{y}{(x-1)^{2}+4y^{2}} dx+frac{-(x-1)}{(x-1)^{2}+4y^{2}} dy where C is circle defined by x^{2}+y^{2}=27

Evaluate the line integral oint_{C} frac{y}{(x-1)^{2}+4y^{2}} dx+frac{-(x-1)}{(x-1)^{2}+4y^{2}} dy where C is circle defined by x^{2}+y^{2}=27

asked 2021-01-31
Evaluate the line integral \(\displaystyle\oint_{{{C}}}\ {\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\ {\left.{d}{x}\right.}+{\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\ {\left.{d}{y}\right.}\) where C is circle defined by \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={27}\)

Answers (1)

Our aim is to apply Green's Theorem. Let us take \(\displaystyle{L}{\left({x},{y}\right)}={\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\) and \(\displaystyle{M}{\left({x},{y}\right)}={\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\). It follows that \(\displaystyle{\frac{{\partial{L}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\right)}\)
\(\displaystyle={\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\) Let D be the interior of C, that is, D is a circular disk given by \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}\leq{27}\). By Green's Theorem, the given line integral can be obtained as \(\displaystyle\oint_{{{C}}}{L}{\left.{d}{x}\right.}+{M}{\left.{d}{y}\right.}=\int\int_{{{D}}}{\left({\frac{{\partial{M}}}{{\partial{x}}}}-{\frac{{\partial{L}}}{{\partial{y}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle\int\int_{{{D}}}{0}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={0}\) Therefore, the value of the line integrals is \(\displaystyle\oint_{{{C}}}{\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{x}\right.}+{\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{y}\right.}={0}\)

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