重積分の計算問題

■問題

次の重積分の値を求めよ．

$3$\displaystyle{{\int\int }\limits_{D} $ x+y $ dxdy }$ 　　 $3$\displaystyle{ $ \text{\hspace{1} }D\text{\hspace{1} }:\text{\hspace{1} }\frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+y\leq 2\text{\hspace{1} },\text{\hspace{1} }x\geq 0\text{\hspace{1} },\text{\hspace{1} }y\geq 0\text{\hspace{1} } $ }$

■答

$3$\displaystyle{16}$

■ヒント

領域 $3$\displaystyle{D}$ より変数 $3$\displaystyle{x}$ と変数 $3$\displaystyle{y}$ の積分範囲を決定する．

次に， $3$\displaystyle{x\text{\hspace{1} }}$ を定数とみなして $3$\displaystyle{y\text{\hspace{1} }}$ について積分し，その結果を更に $3$\displaystyle{x\text{\hspace{1} }}$ で積分する．

■解説

領域 $3$\displaystyle{D}$ より

$3$\displaystyle{\frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+y\text{\hspace{1} }}$ $3$\displaystyle{\leq 2}$

$3$\displaystyle{y\text{\hspace{1} }}$ $3$\displaystyle{\leq - \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+2}$

$3$\displaystyle{ y\geq 0\text{\hspace{1} } }$ を考慮すると

$3$\displaystyle{0\leq y\leq - \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+2}$

これを満たす $3$\displaystyle{x}$ の条件は

− x 3 +2 $3$\displaystyle{\geq 0}$

$3$\displaystyle{- \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}\text{\hspace{1} }}$ $3$\displaystyle{\geq - 2}$

$3$\displaystyle{x\text{\hspace{1} }}$ $3$\displaystyle{\leq 6}$

$3$\displaystyle{ x\geq 0\text{\hspace{1} } }$ を考慮すると

$3$\displaystyle{0\leq x\leq 6}$

以上から領域 $3$\displaystyle{D}$ は

$3$\displaystyle{D\text{\hspace{1} }:\text{\hspace{1} }0\leq x\leq 6\text{\hspace{1} },\text{\hspace{1} }0\leq y\leq - \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+2}$

となるので

$3$\displaystyle{{\int\int }\limits_{D} $ x+y $ dxdy }$

$3$\displaystyle{=\int _{0}^{6} $ \int _{0}^{ - \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+2 } \( x+y $ dy \) dx }$

$3$\displaystyle{x}$ を定数とみなして $3$\displaystyle{y}$ について積分する．

$3$\displaystyle{=\int _{0}^{6} \[ xy+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{y}^{2} \] _{ \text{\hspace{1} }0 }^{ \text{\hspace{1} }- \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+2 }dx }$

$3$\displaystyle{=\int _{0}^{6} \{ x\text{\hspace{1} }\cdot \text{\hspace{1} } $ - \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+2 $ +\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{ $ - \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+2 $ }^{2}- $ x\text{\hspace{1} }\cdot \text{\hspace{1} }0+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }0 $ \} dx \text{\hspace{1} }}$

$3$\displaystyle{=\int _{0}^{6} \{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{2}+2x+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}{x}^{2}- \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}x+4 $ - 0 \} dx }$

$3$\displaystyle{=\int _{0}^{6} \{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{2}+2x+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 18 \hspace{2}}{x}^{2}- \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}x+2 \} dx }$

$3$\displaystyle{=\int _{0}^{6} \{ $ \frac{\hspace{2} - 6+1 \hspace{2}}{\hspace{2} 18 \hspace{2}} $ {x}^{2}+ $ \frac{\hspace{2} 6- 2 \hspace{2}}{\hspace{2}3\hspace{2}} $ x+2 \} dx }$

$3$\displaystyle{=\int _{0}^{6} \{ - \frac{\hspace{2}5\hspace{2}}{\hspace{2} 18 \hspace{2}}{x}^{2}+\frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}x+2 \} dx }$

更に $3$\displaystyle{x}$ で積分する．

$3$\displaystyle{= \[ - \frac{\hspace{2}5\hspace{2}}{\hspace{2} 54 \hspace{2}}{x}^{3}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{2}+2x \] _{ \text{\hspace{1} }0 }^{ \text{\hspace{1} }6 }}$

$3$\displaystyle{=- \frac{\hspace{2}5\hspace{2}}{\hspace{2} 54 \hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }{6}^{3}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }{6}^{2}+2\text{\hspace{1} }\cdot \text{\hspace{1} }6- $ \frac{\hspace{2}5\hspace{2}}{\hspace{2} 54 \hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }{0}^{3}- \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }{0}^{2}+2\text{\hspace{1} }\cdot \text{\hspace{1} }0 $ }$

$3$\displaystyle{=- 5\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} 216 \hspace{2}}{\hspace{2} 54 \hspace{2}}+2\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} 36 \hspace{2}}{\hspace{2}3\hspace{2}}+12- 0}$

$3$\displaystyle{=- 5\text{\hspace{1} }\cdot \text{\hspace{1} }4+2\text{\hspace{1} }\cdot \text{\hspace{1} }12+12}$

$3$\displaystyle{=- 20+24+12}$

$3$\displaystyle{=16}$

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最終更新日： 2023年8月3日