重積分の基礎

■問題

次の2重積分を求めよ．その際領域 $3$\displaystyle{D}$ の図を描くこと．

$3$\displaystyle{ \hspace{1}{\int\int }_{D} {y}^{2}dxdy }$ 　　 $3$\displaystyle{ $ D:{x}^{2}+{y}^{2}\leq 9,x\geq 0 $ }$

■答

$3$\displaystyle{\frac{\hspace{2} 81 \hspace{2}}{\hspace{2}8\hspace{2}}{\pi}}$

■ヒント

極座標表示に変数変換する．

■解説

$3$\displaystyle{ \{\begin{array}{lll}x=r\cos {\theta}& \vspace{6}\\ y=r\sin {\theta}& \vspace{6}\\ \end{array} }$ とおき変数変換する．このときヤコビアン $3$J$ は

$3$\displaystyle{J=\frac{\hspace{2} {\partial} $ x,y $ \hspace{2}}{\hspace{2} {\partial} $ r,{\theta} $ \hspace{2}}= \| \begin{array} \frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}r \hspace{2}} & \frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}{\theta} \hspace{2}} & \vspace{6}\\ \frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}r \hspace{2}} & \frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}{\theta} \hspace{2}} & \vspace{6}\\ \end{array} \| }$ $3$\displaystyle{= \| \begin{array} \cos {\theta} & - r\sin {\theta} & \vspace{6}\\ \sin {\theta} & r\cos {\theta} & \vspace{6}\\ \end{array} \| }$ $3$\displaystyle{=r{\cos }^{2}{\theta}+r{\sin }^{2}{\theta}=r}$

よって，

$3$\displaystyle{ dxdy\rightarrow }$ $3$\displaystyle{ \left|{J}\right|drd{\theta}=\left|{r}\right|drd{\theta}=rdrd{\theta} }$

となる．

積分領域は

$3$\displaystyle{ D:{x}^{2}+{y}^{2}\leq 9,x\geq 0 }$ 　 →　 $3$\displaystyle{{D}^{\prime }:0\leq r\leq 3,0\leq {\theta}\leq {\pi}}$

となる．

$3$\displaystyle{ \hspace{1}{\int\int }_{D} {y}^{2}dxdy }$

$3$\displaystyle{=\displaystyle{\int }_{0}^{3}\left({ \displaystyle{\int }_{ - \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }{ \left({ r\sin {\theta} }\right) }^{2}rd{\theta} }\right)dr}$

$3$\displaystyle{=\displaystyle{ \int _{0}^{3} $ \displaystyle{ \int _{ - \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } {r}^{3}{ \sin }^{2}{\theta}d{\theta} } $ }dr}$

$3$\displaystyle{=\displaystyle{ \int _{0}^{3} $ 2{r}^{3}\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{2}{\theta}d{\theta} } $ }dr}$

$3$\displaystyle{=\displaystyle{ \int _{0}^{3} 2{r}^{3}\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }dr}$

$3$\displaystyle{=\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}\displaystyle{ \int _{0}^{3} {r}^{3}dr }}$

$3$\displaystyle{=\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} \[ \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}{r}^{4} \] _{0}^{3}}$

$3$\displaystyle{=\frac{\hspace{2} 81 \hspace{2}}{\hspace{2}8\hspace{2}}{\pi}}$

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最終更新日： 2023年8月4日