# 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 where C is circle defined by ${x}^{2}+{y}^{2}=27$
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mhalmantus
Our aim is to apply Green's Theorem. Let us take $L\left(x,y\right)=\frac{y}{{\left(x-1\right)}^{2}+4{y}^{2}}$ and$M\left(x,y\right)=\frac{-\left(x-1\right)}{{\left(x-1\right)}^{2}+4{y}^{2}}$. It follows that$\frac{\partial L}{\partial y}=\frac{\partial }{\partial y}\left(\frac{y}{{\left(x-1\right)}^{2}+4{y}^{2}}\right)$
$\frac{\frac{\partial }{\partial y}\left(y\right)\left({\left(x-1\right)}^{2}+4{y}^{2}\right)-\frac{\partial }{\partial y}\left({\left(x-1\right)}^{2}+4{y}^{2}\right)y}{{\left({\left(x-1\right)}^{2}+4{y}^{2}\right)}^{2}}$
$=\frac{{\left(x-1\right)}^{2}-4{y}^{2}}{\left({\left(x-1\right)}^{2}+4{y}^{2}}$
$\frac{\partial M}{\partial x}=\frac{\partial }{\partial x}\left(\frac{-\left(x-1\right)}{{\left(x-1\right)}^{2}+4{y}^{2}}\right)$
$\frac{\frac{\partial }{\partial x}\left(x-1\right)\left({\left(x-1\right)}^{2}+4{y}^{2}\right)-\frac{\partial }{\partial x}\left({\left(x-1\right)}^{2}+4{y}^{2}\right)\left(x-1\right)}{{\left({\left(x-1\right)}^{2}+4{y}^{2}\right)}^{2}}$
$=\frac{{\left(x-1\right)}^{2}-4{y}^{2}}{{\left({\left(x-1\right)}^{2}+4{y}^{2}\right)}^{2}}$Let D be the interior of C, that is, D is a circular disk given by ${x}^{2}+{y}^{2}\le 27$. By Green's Theorem, the given line integral can be obtained as${\oint }_{C}Ldx+Mdy=\int {\int }_{D}\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy$