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問題5.1.2a

積分を計算せよ。

（１）${\displaystyle {\iint }_D\sin (2x+y)dxdy}$　$D:0\leqq x\leqq {\displaystyle \frac{\pi }{2}}$、$0\leqq y\leqq {\displaystyle \frac{\pi }{2}}$

（２）${\displaystyle {\iint }_D(x^{2}y+y^2)dxdy}$　$D:1\leqq x\leqq 2$、$2\leqq y\leqq 3$

（３）${\displaystyle {\iint }_{D}xdxdy}$　$D:x^2+y^2\leqq 1$、$x\geqq 0$

（４）${\displaystyle {\iint }_D\sqrt{a^2-y^2}dxdy}$　$D:x^2+y^2\leqq a^2$

 

《ポイント》

累次積分の計算練習です。

 

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《解答例》

（１）${\displaystyle {\iint }_D\sin (2x+y)dxdy}$　$D:0\leqq x\leqq {\displaystyle \frac{\pi }{2}}$、$0\leqq y\leqq {\displaystyle \frac{\pi }{2}}$

$\begin{array}{rl}& {\displaystyle {\iint }_D\sin (2x+y)dxdy}\\ & ={\displaystyle {\int }_0^{\frac{\pi }{2}}[-\cos (2x+y){]}_0^{\frac{\pi }{2}}dx}\\ & ={\displaystyle {\int }_0^{\frac{\pi }{2}}\{-\cos (2x+{\displaystyle \frac{\pi }{2}})+\cos 2x\}dx}\\ & ={\displaystyle {\int }_0^{\frac{\pi }{2}}(\sin 2x+\cos 2x)dx}\\ & ={[-{\displaystyle \frac{1}{2}}\cos 2x+{\displaystyle \frac{1}{2}}\sin 2x]}_0^{\frac{\pi }{2}}\\ & =-{\displaystyle \frac{1}{2}}(-1-1)+{\displaystyle \frac{1}{2}}(0-0)\\ & =1\cdots \cdots \text{(答)}\end{array}$

 

（２）${\displaystyle {\iint }_D(x^{2}y+y^2)dxdy}$　$D:1\leqq x\leqq 2$、$2\leqq y\leqq 3$

$\begin{array}{rl}& {\displaystyle {\iint }_D(x^{2}y+y^2)dxdy}\\ & ={\displaystyle {\int }_2^3{[\frac{1}{3}{x}^{3}y+x{y}^2]}_1^{2}dy}\\ & ={\displaystyle {\int }_2^3(\frac{7}{3}y+y^2)dy}\\ & ={[\frac{7}{6}{y}^2+\frac{1}{3}{y}^3]}_2^3\\ & ={\displaystyle \frac{7}{6}}(63-28)+{\displaystyle \frac{1}{3}}(27-8)\\ & ={\displaystyle \frac{73}{6}}\cdots \cdots \text{(答)}\end{array}$

《別解：先に$y$で積分した場合》

$\begin{array}{rl}& {\displaystyle {\iint }_D(x^{2}y+y^2)dxdy}\\ & ={\displaystyle {\int }_1^2{[\frac{1}{2}{x}^2{y}^2+\frac{1}{3}{y}^3]}_2^{3}dx}\\ & ={\displaystyle {\int }_1^2(\frac{5}{2}{x}^2+\frac{19}{3})dx}\\ & ={[\frac{5}{6}{x}^3+\frac{19}{3}x]}_1^2\\ & ={\displaystyle \frac{5}{6}}(8-1)+{\displaystyle \frac{19}{3}}(2-1)\\ & ={\displaystyle \frac{73}{6}}\cdots \cdots \text{(答)}\end{array}$

 

（３）${\displaystyle {\iint }_{D}xdxdy}$　$D:x^2+y^2\leqq 1$、$x\geqq 0$

$x$の範囲は $0\leqq x\leqq \sqrt{1-y^2}$ であるから、

$\begin{array}{rl}& {\displaystyle {\iint }_{D}xdxdy}\\ & ={\displaystyle {\int }_{-1}^{1}dy{\int }_0^{\sqrt{1-y^2}}xdx}\\ & ={\displaystyle {\int }_{-1}^1{\left[{\displaystyle \frac{1}{2}}{x}^2\right]}_0^{\sqrt{1-y^2}}dy}\\ & ={\displaystyle {\displaystyle \frac{1}{2}}{\int }_{-1}^1(1-y^2)dy}\\ & ={\displaystyle \frac{1}{2}}{[y-\frac{1}{3}{y}^3]}_{-1}^1\\ & ={[y-\frac{1}{3}{y}^3]}_0^1\\ & ={\displaystyle \frac{2}{3}}\cdots \cdots \text{(答)}\end{array}$

 

（４）${\displaystyle {\iint }_D\sqrt{a^2-y^2}dxdy}$　$D:x^2+y^2\leqq a^2$

$-\sqrt{a^2-y^2}\leqq x\leqq \sqrt{a^2-y^2}$、$-a\leqq y\leqq a$ より、

$\begin{array}{rl}& {\displaystyle {\iint }_D\sqrt{a^2-y^2}dxdy}\\ & ={\displaystyle {\int }_{-a}^a{\left[x\sqrt{a^2-y^2}\right]}_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}}dy}\\ & ={\displaystyle {\int }_{-a}^{a}2(a^2-y^2)dy}\\ & =4{\displaystyle {\int }_0^a(a^2-y^2)dy}\\ & =4{[a^{2}y-{\displaystyle \frac{1}{3}}{y}^3]}_0^a\\ & ={\displaystyle \frac{8}{3}}{a}^3\cdots \cdots \text{(答)}\end{array}$

※積分計算の途中で偶関数の性質を利用しています。

 

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復習例題は設定していません。

 

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