微積5.1.2b

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問題5.1.2b

積分を計算せよ。

（５）${\displaystyle {\iint }_{D}x{y}^{2}dxdy}$　$D:0\leqq y\leqq x\leqq 1$

（６）${\displaystyle {\iint }_D(2x-y)dxdy}$　$D:x\leqq y\leqq 2x$、$x+y\leqq 3$

（７）${\displaystyle {\iiint }_{D}zdxdydz}$　$D:0\leqq x\leqq 1$、$0\leqq y\leqq 1-x$、$0\leqq z\leqq 1-x-y$

（８）${\displaystyle {\iiint }_{D}ydxdydz}$　$D:x\geqq 0$、$y\geqq 0$、$z\geqq 0$、$x+2y+3z\leqq 6$

 

《ポイント》

累次積分の計算練習です。

 

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《解答例》

（５）${\displaystyle {\iint }_{D}x{y}^{2}dxdy}$　$D:0\leqq y\leqq x\leqq 1$

$\begin{array}{rl}& {\displaystyle {\iint }_{D}x{y}^{2}dxdy}\\ & ={\displaystyle {\int }_0^{1}dy{\int }_y^{1}x{y}^{2}dx}\\ & ={\displaystyle {\int }_0^1{\left[{\displaystyle \frac{1}{2}}{x}^2{y}^2\right]}_y^{1}dy}\\ & ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1(y^2-y^4)dy}\\ & ={\displaystyle \frac{1}{2}}{[{\displaystyle \frac{1}{3}}{y}^3-{\displaystyle \frac{1}{5}}{y}^5]}_0^1\\ & ={\displaystyle \frac{1}{15}}\cdots \cdots \text{(答)}\end{array}$

 

（６）${\displaystyle {\iint }_D(2x-y)dxdy}$　$D:x\leqq y\leqq 2x$、$x+y\leqq 3$

領域$D$を図示すると以下のようになる。

$\begin{array}{rl}& {\displaystyle {\iint }_D(2x-y)dxdy}\\ & ={\displaystyle {\int }_0^{1}dx{\int }_x^{2x}(2x-y)dy+{\int }_1^{\frac{3}{2}}dx{\int }_x^{-x+3}(2x-y)dy}\\ & ={\displaystyle {\int }_0^1{[2xy-{\displaystyle \frac{1}{2}}{y}^2]}_x^{2x}dx+{\int }_1^{\frac{3}{2}}{[2xy-{\displaystyle \frac{1}{2}}{y}^2]}_x^{-x+3}dx}\\ & ={\displaystyle {\int }_0^1{\displaystyle \frac{1}{2}}{x}^{2}dx+{\int }_1^{\frac{3}{2}}(-4x^2+9x-{\displaystyle \frac{9}{2}})dx}\\ & ={\displaystyle \frac{1}{2}}{\left[{\displaystyle \frac{1}{3}}{x}^3\right]}_0^1+{[-{\displaystyle \frac{4}{3}}{x}^3+{\displaystyle \frac{9}{2}}{x}^2-{\displaystyle \frac{9}{2}}x]}_1^{\frac{3}{2}}\\ & ={\displaystyle \frac{1}{6}}+(-{\displaystyle \frac{9}{8}}+{\displaystyle \frac{4}{3}})\\ & ={\displaystyle \frac{3}{8}}\cdots \cdots \text{(答)}\end{array}$

※上記の解答例では$y$から積分を始めていますが、先に$x$で積分することも可能です。

 

（７）${\displaystyle {\iiint }_{D}zdxdydz}$　$D:0\leqq x\leqq 1$、$0\leqq y\leqq 1-x$、$0\leqq z\leqq 1-x-y$

$\begin{array}{rl}& {\displaystyle {\iiint }_{D}zdxdydz}\\ & ={\displaystyle {\int }_0^{1}dx{\int }_0^{1-x}dy{\int }_0^{1-x-y}zdz}\\ & ={\displaystyle {\int }_0^{1}dx{\int }_0^{1-x}dy{\left[{\displaystyle \frac{1}{2}}{z}^2\right]}_0^{1-x-y}}\\ & ={\displaystyle {\int }_0^{1}dx{\int }_0^{1-x}{\displaystyle \frac{1}{2}}(1-x-y{)}^{2}dy}\\ & ={\displaystyle {\int }_0^{1}dx{\int }_0^{1-x}{\displaystyle \frac{1}{2}}(x+y-1{)}^{2}dy}\\ & ={\displaystyle \frac{1}{2}}{\int }_0^1{[{\displaystyle \frac{1}{3}}(x+y-1{)}^3]}_0^{1-x}dx\\ & ={\displaystyle \frac{1}{6}}{\int }_0^1\{0-(x-1{)}^3\}dx\\ & =-{\displaystyle \frac{1}{6}}{\int }_0^1(x-1{)}^{3}dx\\ & =-{\displaystyle \frac{1}{6}}{[{\displaystyle \frac{1}{4}}(x-1{)}^4]}_0^1\\ & =-{\displaystyle \frac{1}{24}}(0-1)\\ & ={\displaystyle \frac{1}{24}}\cdots \cdots \text{(答)}\end{array}$

 

（８）${\displaystyle {\iiint }_{D}ydxdydz}$　$D:x\geqq 0$、$y\geqq 0$、$z\geqq 0$、$x+2y+3z\leqq 6$

領域$D$は連立不等式$$\{\begin{array}{l}0\leqq x\leqq 6-3z-2y\\ 0\leqq y\leqq {\displaystyle \frac{3}{2}}(2-z)\\ 0\leqq z\leqq 2\end{array}$$によって表せるから、
$\begin{array}{rl}& {\displaystyle {\iiint }_{D}ydxdydz}\\ & ={\displaystyle {\int }_0^{2}dz{\int }_0^{\frac{3}{2}(2-z)}dy{\int }_0^{6-3z-2y}ydx}\\ & ={\displaystyle {\int }_0^{2}dz{\int }_0^{\frac{3}{2}(2-z)}[xy{]}_0^{6-3z-2y}dy}\\ & ={\displaystyle {\int }_0^{2}dz{\int }_0^{\frac{3}{2}(2-z)}\{3(2-z)y-2y^2\}dy}\\ & ={\int }_0^2{[{\displaystyle \frac{3}{2}}(2-z)y^2-{\displaystyle \frac{2}{3}}{y}^3]}_0^{\frac{3}{2}(2-z)}dz\\ & ={\int }_0^2\{{\displaystyle \frac{27}{8}}(2-z{)}^3-{\displaystyle \frac{9}{4}}(2-z{)}^3\}dz\\ & ={\displaystyle \frac{9}{8}}{\int }_0^2(2-z{)}^{3}dz\\ & ={\displaystyle \frac{9}{8}}{[-{\displaystyle \frac{1}{4}}(2-z{)}^4]}_0^2\\ & =-{\displaystyle \frac{9}{32}}(0-16)\\ & ={\displaystyle \frac{9}{2}}\cdots \cdots \text{(答)}\end{array}$

 

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復習例題は設定していません。

 

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