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Example Evaluate the following line integral using Green's theorem．

$\displaystyle \int_{C}(x^2y dx - xy^2dy), C:$   unit circle

Answer 1. Green's theorem Let $C$ be a simply closed curve and $\Omega$ be a closed region with the boundary of $C$. Suppose that $P(x,y)$ and $Q(x,y)$ have the continuous partial derivatives in $\Omega$. Then we have

$\displaystyle \int_{C}Pdx + Qdy = \iint_{\Omega}(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dx dy$

Using Green's thereom, we have

$\displaystyle \int_{c}(xy^2 dx - xy^2 dy)$ $\displaystyle =$ $\displaystyle \int\int_{R}(\frac{\partial(-xy^2)}{\partial{x}} - \frac{\partial(x^2 y)}{\partial{y}})dx dy$  
$\displaystyle \int\int_{R}(-y^2 - x^2)dxdy = -\int\int_{R}(x^2 + y^2)dx dy$      
  $\displaystyle =$ $\displaystyle - \int_{0}^{2\pi}\int_{r =0}^{1} r^2 \vert J\vert dr d\theta$  

Here we find the Jacobian $J$. Then $J = \frac{\partial(x,y)}{\partial(r,\theta)} = \left\vert\begin{array}{cc} \fr... ...y}{\partial r} & \frac{\partial y}{\partial \theta} \end{array}\right\vert = r$.

Thus,

$\displaystyle \int_{c}(xy^2 dx - xy^2 dy)$ $\displaystyle =$ $\displaystyle - \int_{0}^{2\pi}\int_{r =0}^{1} r^2 r dr d\theta$  
  $\displaystyle =$ $\displaystyle - \int_{0}^{2\pi} d\theta \int_{r =0}^{1} r^3 dr = - 2\pi \frac{1}{4} = -\frac{\pi}{2}$