# 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

Question
Modeling
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 $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={27}$$

2021-02-01
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 $$\displaystyle{\frac{{\partial{L}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\right)}$$
$$\displaystyle{\frac{{{\frac{{\partial}}{{\partial{y}}}}{\left({y}\right)}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}-{\frac{{\partial}}{{\partial{y}}}}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}{y}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}$$
$$\displaystyle={\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right.}}}}$$
$$\displaystyle{\frac{{\partial{M}}}{{\partial{x}}}}={\frac{{\partial}}{{\partial{x}}}}{\left({\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\right)}$$
$$\displaystyle{\frac{{{\frac{{\partial}}{{\partial{x}}}}{\left({x}-{1}\right)}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}-{\frac{{\partial}}{{\partial{x}}}}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}{\left({x}-{1}\right)}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}$$
$$\displaystyle={\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}$$ 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 $$\displaystyle\oint_{{{C}}}{L}{\left.{d}{x}\right.}+{M}{\left.{d}{y}\right.}=\int\int_{{{D}}}{\left({\frac{{\partial{M}}}{{\partial{x}}}}-{\frac{{\partial{L}}}{{\partial{y}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle\int\int_{{{D}}}{\left({\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}-{\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle\int\int_{{{D}}}{0}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={0}$$ Therefore, the value of the line integrals is $$\displaystyle\oint_{{{C}}}{\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{x}\right.}+{\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{y}\right.}={0}$$

### Relevant Questions

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