与えられた重積分を、変数変換を用いて計算する問題です。具体的には以下の5つの積分を解きます。 (1) $\iint_D x^2 dxdy$, $D = \{(x,y); 0 \le x - y \le 1, 0 \le x + y \le 1\}$ (2) $\iint_D (x - y)^2 dxdy$, $D = \{(x, y); -1 \le x + 2y \le 1, -1\le x - y \le 1\}$ (3) $\iint_D (x^2 + y^2)dxdy$, $D = \{(x,y); x^2 + y^2 \le 1\}$ (4) $\iint_D \sqrt{a^2-x^2-y^2} dxdy$, $D= \{(x, y); x^2 + y^2 \le a^2, x \ge 0, y \ge 0\} \quad (a>0)$ (5) $\iint_D xy dxdy$, $D = \{(x,y); \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1\} \quad (a>0,b>0)$

## 問題の回答

1. 問題の内容

与えられた重積分を、変数変換を用いて計算する問題です。具体的には以下の5つの積分を解きます。

(1) $\iint_D x^2 dxdy$, $D = \{(x,y); 0 \le x - y \le 1, 0 \le x + y \le 1\}$

(2) $\iint_D (x - y)^2 dxdy$, $D = \{(x, y); -1 \le x + 2y \le 1, -1\le x - y \le 1\}$

(3) $\iint_D (x^2 + y^2)dxdy$, $D = \{(x,y); x^2 + y^2 \le 1\}$

(4) $\iint_D \sqrt{a^2-x^2-y^2} dxdy$, $D= \{(x, y); x^2 + y^2 \le a^2, x \ge 0, y \ge 0\} \quad (a>0)$

(5) $\iint_D xy dxdy$, $D = \{(x,y); \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1\} \quad (a>0,b>0)$

2. 解き方の手順

それぞれの積分について、変数変換と計算手順を以下に示します。

**(1) $\iint_D x^2 dxdy$, $D = \{(x,y); 0 \le x - y \le 1, 0 \le x + y \le 1\}$**

* 変数変換: $u = x - y$, $v = x + y$ とおく。このとき、$x = \frac{u+v}{2}$, $y = \frac{v-u}{2}$。

* ヤコビアン: $\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{vmatrix} = \frac{1}{4} - (-\frac{1}{4}) = \frac{1}{2}$.

* 積分範囲: $0 \le u \le 1$, $0 \le v \le 1$

* 積分計算:

$\iint_D x^2 dxdy = \int_0^1 \int_0^1 (\frac{u+v}{2})^2 |\frac{1}{2}| dudv = \frac{1}{8} \int_0^1 \int_0^1 (u^2 + 2uv + v^2) dudv = \frac{1}{8} \int_0^1 [\frac{u^3}{3} + u^2v + uv^2]_0^1 dv = \frac{1}{8} \int_0^1 (\frac{1}{3} + v + v^2) dv = \frac{1}{8} [\frac{v}{3} + \frac{v^2}{2} + \frac{v^3}{3}]_0^1 = \frac{1}{8} (\frac{1}{3} + \frac{1}{2} + \frac{1}{3}) = \frac{1}{8} (\frac{2+3+2}{6}) = \frac{7}{48}$.

**(2) $\iint_D (x - y)^2 dxdy$, $D = \{(x, y); -1 \le x + 2y \le 1, -1\le x - y \le 1\}$**

* 変数変換: $u = x - y$, $v = x + 2y$ とおく。このとき、$x = \frac{2u+v}{3}$, $y = \frac{v-u}{3}$。

* ヤコビアン: $\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{vmatrix} = \frac{2}{9} - (-\frac{1}{9}) = \frac{1}{3}$.

* 積分範囲: $-1 \le u \le 1$, $-1 \le v \le 1$

* 積分計算:

$\iint_D (x-y)^2 dxdy = \int_{-1}^1 \int_{-1}^1 u^2 |\frac{1}{3}| dudv = \frac{1}{3} \int_{-1}^1 \int_{-1}^1 u^2 dudv = \frac{1}{3} \int_{-1}^1 [\frac{u^3}{3}]_{-1}^1 dv = \frac{1}{3} \int_{-1}^1 (\frac{1}{3} - (-\frac{1}{3})) dv = \frac{1}{3} \int_{-1}^1 \frac{2}{3} dv = \frac{2}{9} [v]_{-1}^1 = \frac{2}{9} (1 - (-1)) = \frac{4}{9}$.

**(3) $\iint_D (x^2 + y^2)dxdy$, $D = \{(x,y); x^2 + y^2 \le 1\}$**

* 変数変換: 極座標変換 $x = r\cos\theta$, $y = r\sin\theta$。このとき、$x^2 + y^2 = r^2$。

* ヤコビアン: $\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r$.

* 積分範囲: $0 \le r \le 1$, $0 \le \theta \le 2\pi$

* 積分計算:

$\iint_D (x^2 + y^2) dxdy = \int_0^{2\pi} \int_0^1 r^2 |r| dr d\theta = \int_0^{2\pi} \int_0^1 r^3 dr d\theta = \int_0^{2\pi} [\frac{r^4}{4}]_0^1 d\theta = \int_0^{2\pi} \frac{1}{4} d\theta = \frac{1}{4} [\theta]_0^{2\pi} = \frac{1}{4} (2\pi) = \frac{\pi}{2}$.

**(4) $\iint_D \sqrt{a^2-x^2-y^2} dxdy$, $D= \{(x, y); x^2 + y^2 \le a^2, x \ge 0, y \ge 0\} \quad (a>0)$**

* 変数変換: 極座標変換 $x = r\cos\theta$, $y = r\sin\theta$。このとき、$x^2 + y^2 = r^2$。

* ヤコビアン: $r$

* 積分範囲: $0 \le r \le a$, $0 \le \theta \le \frac{\pi}{2}$

* 積分計算:

$\iint_D \sqrt{a^2 - x^2 - y^2} dxdy = \int_0^{\pi/2} \int_0^a \sqrt{a^2 - r^2} |r| dr d\theta = \int_0^{\pi/2} d\theta \int_0^a r \sqrt{a^2 - r^2} dr$

$\int_0^{\pi/2} d\theta = \frac{\pi}{2}$

$\int_0^a r \sqrt{a^2 - r^2} dr$ に対して $t = a^2 - r^2$ と置換すると、$dt = -2r dr$, $rdr = -\frac{1}{2} dt$。$r=0$ のとき $t=a^2$, $r=a$ のとき $t=0$。

$\int_0^a r \sqrt{a^2 - r^2} dr = \int_{a^2}^0 \sqrt{t} (-\frac{1}{2}) dt = \frac{1}{2} \int_0^{a^2} \sqrt{t} dt = \frac{1}{2} [\frac{2}{3} t^{3/2}]_0^{a^2} = \frac{1}{3} (a^2)^{3/2} = \frac{1}{3} a^3$

よって、$\iint_D \sqrt{a^2 - x^2 - y^2} dxdy = \frac{\pi}{2} \cdot \frac{a^3}{3} = \frac{\pi a^3}{6}$.

**(5) $\iint_D xy dxdy$, $D = \{(x,y); \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1\} \quad (a>0,b>0)$**

* 変数変換: $x = ar\cos\theta$, $y = br\sin\theta$。このとき、$\frac{x^2}{a^2} + \frac{y^2}{b^2} = r^2$。

* ヤコビアン: $\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} a\cos\theta & -ar\sin\theta \\ b\sin\theta & br\cos\theta \end{vmatrix} = abr\cos^2\theta + abr\sin^2\theta = abr$.

* 積分範囲: $0 \le r \le 1$, $0 \le \theta \le 2\pi$

* 積分計算:

$\iint_D xy dxdy = \int_0^{2\pi} \int_0^1 (ar\cos\theta)(br\sin\theta) |abr| dr d\theta = a^2b^2 \int_0^{2\pi} \int_0^1 r^2 \cos\theta \sin\theta dr d\theta = a^2b^2 \int_0^{2\pi} \cos\theta \sin\theta d\theta \int_0^1 r^3 dr = a^2b^2 [\frac{\sin^2\theta}{2}]_0^{2\pi} [\frac{r^4}{4}]_0^1 = a^2b^2 (0)(\frac{1}{4}) = 0$.

3. 最終的な答え

(1) $\frac{7}{48}$

(2) $\frac{4}{9}$

(3) $\frac{\pi}{2}$

(4) $\frac{\pi a^3}{6}$

(5) $0$