変数変換

■問題

適当な変数変換を行って次の重積分を計算せよ．

${\displaystyle {\displaystyle {\iint }_D\left(x^2+y^2\right)e^{-x-y}}dxdy}$　　${\displaystyle (D:-2<x+y<2,-2<x-y<2)}$

■答

${\displaystyle \frac{2}{3}\left(5e^2-\frac{17}{e^2}\right)}$

■ヒント

${\displaystyle x+y=u}$ ，${\displaystyle x-y=v}$ とおく変数変換により，積分の計算を簡単化する。ヤコビアンの計算式も参考にする．

■解き方

領域${\displaystyle D}$　 領域${\displaystyle D^{\prime }}$

${\displaystyle x+y=u}$，${\displaystyle x-y=v}$ とおくと

${\displaystyle x=\frac{1}{2}\left(u+v\right)}$，${\displaystyle y=\frac{1}{2}\left(u-v\right)}$

となる．よって，この変数変換によるヤコビアンは

${\displaystyle J\left(u,v\right)=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right|=\left|\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right|=-\frac{1}{2}}$

となる．変数変換によって${\displaystyle \left(x,y\right)}$ の領域$D$ は${\displaystyle \left(u,v\right)}$ の領域${\displaystyle D^{\prime }}$

${\displaystyle D^{\prime }:-2\leqq u\leqq 2,-2\leqq v\leqq 2}$ (長方形)

に変換される．よって

(与式)${\displaystyle ={\displaystyle {\iint }_{D^{\prime }}\frac{1}{2}\left(u^2+v^2\right)}{e}^{-u}·\frac{1}{2}dudv}$

${\displaystyle =\frac{1}{4}{\displaystyle {\iint }_{D^{\prime }}\left(u^2+v^2\right)}{e}^{-u}dudv}$

${\displaystyle =\frac{1}{4}{\displaystyle {\int }_{-2}^2\left({{\displaystyle \int }}_{-2}^2{e}^{-u}\left(u^2+v^2\right)dv\right)du}}$

${\displaystyle =\frac{1}{4}{\displaystyle {\int }_{-2}^2{e}^{-u}{\left[u^{2}v+\frac{1}{3}{v}^3\right]}_{-2}^{2}du}}$

${\displaystyle =\frac{1}{4}{\displaystyle {\int }_{-2}^2{e}^{-u}}\left(4u^2+\frac{16}{3}\right)du}$

${\displaystyle ={\displaystyle {\int }_{-2}^2{e}^{-u}}\left(u^2+\frac{4}{3}\right)du}$

部分積分法を用いて計算をする

${\displaystyle ={{\displaystyle \int }}_{-2}^2{\left(-e^{-u}\right)}^{\prime }\left(u^2+\frac{4}{3}\right)du}$

${\displaystyle ={\left[-e^{-u}\left(u^2+\frac{4}{3}\right)\right]}_{-2}^2}$${\displaystyle -{{\displaystyle \int }}_{-2}^2\left(-e^{-u}\right)·2udu}$

${\displaystyle =\left\{-e^{-2}\left(4+\frac{4}{3}\right)+e^2\left(4+\frac{4}{3}\right)\right\}}$${\displaystyle +{\displaystyle {\int }_{-2}^{2}2u{e}^{-u}du}}$

最後の項の積分は部分積分法を用いて計算をする

${\displaystyle =\frac{16}{3}{e}^2-\frac{16}{3}{e}^{-2}-{\displaystyle {\int }_{-2}^{2}2u{\left(e^{-u}\right)}^{\prime }du}}$

${\displaystyle =\frac{16}{3}{e}^2-\frac{16}{3}{e}^{-2}-{\left[2u{e}^{-u}\right]}_{-2}^2}$${\displaystyle +{\displaystyle {\int }_{-2}^{2}2e^{-u}}du}$

${\displaystyle =\frac{16}{3}{e}^2-\frac{16}{3}{e}^{-2}-}$${\displaystyle \left\{4e^{-2}-\left(-4e^2\right)\right\}}$${\displaystyle +{\displaystyle {\int }_{-2}^{2}2e^{-u}}du}$

${\displaystyle =\frac{16}{3}{e}^2-\frac{16}{3}{e}^{-2}-4e^{-2}-4e^2}$${\displaystyle +{\displaystyle {\int }_{-2}^{2}2e^{-u}}du}$

${\displaystyle =\frac{4}{3}{e}^2-\frac{28}{3}{e}^{-2}+2{\displaystyle {\int }_{-2}^2{e}^{-u}}du}$

${\displaystyle =\frac{4}{3}{e}^2-\frac{28}{3}{e}^{-2}-2{\left[e^{-u}\right]}_{-2}^2}$

${\displaystyle =\frac{4}{3}{e}^2-\frac{28}{3}{e}^{-2}-2\left(e^{-2}-e^2\right)}$

${\displaystyle =\frac{10}{3}{e}^2-\frac{34}{3}{e}^{-2}}$

${\displaystyle =\frac{2}{3}\left(5e^2-\frac{17}{e^2}\right)}$

　

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最終更新日： 2023年10月19日