$\def\bm#1{{\boldsymbol{#1}}}$

ベクトル解析演習6

今回は重積分の計算練習をしましょう。

要点のまとめ

もし閉領域$D$が$\{ (x,y) | a\leq x \leq b,~\varphi_1(x) \leq y \leq \varphi_2(x) \}$と$\{ (x,y) | c\leq y \leq d,~\psi_1(y) \leq x \leq \psi_2(y) \}$の2通りであらわせるとしたら、
\[
\int_a^b \left\{ \int_{\varphi_1(x)}^{\varphi_2(x)} f(x,y) dy \right\} dx = \int_c^d \left\{ \int_{\psi_1(y)}^{\psi_2(y)} f(x,y) dx \right\} dy
\]
です。これを積分順序の変更と言います。

問題6

次の領域$D$に関する重積分を計算しなさい。

(1)
\[
\iint_D \dfrac{dxdy}{(1+y^2)(3+x)} , ~ D=\{(x,y)|1\leq x \leq 2, 1\leq y \leq \sqrt{3}\}
\]

(2)
\[
\iint_D \sin{(x+y)}dxdy , ~ D=\{(x,y)|0 \leq x \leq \dfrac{\pi}{6}, x\leq y \leq 2x\}
\]

(3)
\[
\iint_D (x^2+2xy)dxdy , ~ D=\{(x,y)|0\leq x \leq 3y, 0\leq y \leq 2\}
\]

(4)
\[
\iint_D y\sqrt{x} dxdy , ~ D=\{(x,y)|x+y\leq1, 0\leq x, 0\leq y\}
\]

(5)
\[
\iint_D (1+xy) dxdy , ~ D=\{(x,y)|x^2 \leq y \leq x\}
\]

(6)
\[
\iint_D \sqrt{x-y}dxdy , ~ D=\{(x,y)|0\leq y\leq x-x^2\}
\]

(7)
\[
\iint_D sqrt{x^2-y}dxdy , ~ D=\{(x,y)|x^2\leq y, 0\leq x, y\leq 8-x^2\}
\]

解答6

(1)
\[
\iint_D \frac{dx}{(y^2+1)(3+x)} dxdy=\int^{\sqrt{3}}_{1}\left\{\int^{2}_{1}\frac{dx}{(y^2+1)(3+x)}\right\}dy=\log{\frac{5}{4}}\int^{\sqrt{3}}_{1}\frac{dy}{(y^2+1)}=\frac{\pi}{12}\log{\frac{5}{4}}
\]

(2)
\[
\iint_D \sin{(x+y)} dxdy=\int^{\frac{\pi}{6}}_{0}\left\{\Bigl[-\cos{(x+y)}\Bigr]^{y=2x}_{y=x}\right\}dx=\left[\frac{1}{2}\sin{2x}-\frac{1}{3}\sin{3x}\right]^{\frac{\pi}{6}}_{0}=\frac{\sqrt{3}}{4}-\frac{1}{3}
\]

(3)
\[
\iint_D (x^2+2xy) dxdy=\int^{2}_{0}\left\{\int^{3y}_{0}(x^2+2xy)dx\right\}dy=\int^{2}_{0}18y^3dy=72
\]

(4)
$D=\{(x,y)|=0\leq x\leq 1, 0\leq y\leq 1-x\}$
従って
\[
\iint_D y\sqrt{x} dxdy=\int^{1}_{0}\left\{\int^{1-x}_{0}y\sqrt{x}dy\right\}dx=\int^{1}_{0}\left\{\frac{1}{2}\sqrt{x}(1-x)^2\right\}dx=\frac{8}{105}
\]

(5)
今、$x-y\geq0 , 0\leq y\leq x-x^2=x(1-x)$より、$D=\{(x,y)|=0\leq x\leq 1, x^2\leq y\leq x\}$
従って
\[
\iint_D (1+xy) dxdy=\int^{1}_{0}\left\{\int^{x}_{x^2}(1+xy)dy\right\}dx=\int^{1}_{0}\left(-\frac{1}{2}x^5+\frac{1}{2}x^3-x^2+x\right)dx=\frac{5}{24}
\]

(6)
今、$x-y\geq0 , 0\leq y\leq x-x^2=x(1-x)$より$0\leq x\leq 1$
従って
\[
\iint_D \sqrt{x-y} dxdy=\int^{1}_{0}\left(\int^{x-x^2}_{0}\sqrt{x-y}dy\right)dx=\int^{1}_{0}\left\{\left[\frac{-2}{3}(x-y)^{\frac{3}{2}}\right]^{y=x-x^2}_{y=0}\right\}dx=\frac{1}{10}
\]

(7)
$D=\{(x,y)|=0\leq x\leq 2,x^2\leq y \leq8-x^2\}$
従って
\[
\iint_D \sqrt{x^2+y} dxdy=\int^{2}_{0}\left\{\int^{8-x^2}_{x^2}(x^2+y)^{\frac{1}{2}}dy\right\}dx=\int^{2}_{0}\left[\frac{2}{3}(x^2+y)^\frac{3}{2}\right]^{y=8-x^2}_{y=x^2}dx
\]
\[
=\int^{2}_{0}\left(\frac{32}{3}\sqrt{2}-\frac{4}{3}\sqrt{2}x^3\right)dx=\left[\frac{32}{3}\sqrt{2}x-\frac{1}{3}\sqrt{2}x^4\right]^{2}_{0}=16\sqrt{2}
\]