LineIntegral

例題 Greenの定理を用いて次の線積分の値を求めよ．

$\displaystyle \int_{C}(x^2y dx - xy^2dy), C:$ 単位円周

解答 1. Greenの定理 $C$ を単一閉曲線とし， $\Omega$ をその周および内部からなる閉領域とする．関数 $P(x,y)$ , $Q(x,y)$ が $\Omega$ で連続な偏導関数をもつとき，

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

Greenの定理を用いると

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

ここでジャコビアン

を求めると $J = \frac{\partial(x,y)}{\partial(r,\theta)} = \left\vert\begin{array}{cc} \fr... ...y}{\partial r} & \frac{\partial y}{\partial \theta} \end{array}\right\vert = r$ .

これより

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

この文書について...