eLineIntegral

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,

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

この文書について...