第8章 8.3

(1)

$(x+y)^2$ $D : 0 \leq x \leq 1$ $0 \leq y \leq 1$

$\int_0^1\int_0^1 (x+y)^2\,dydx = \int_0^1\int_0^1 (x^2+2xy+y^2)\,dydx = \int_0^1 \left[x^2y+xy^2+\frac{1}{3}y^3\right]_0^1 dx = \int_0^1 x^2+x+\frac{1}{3}\,dx$

$= \left[\frac{1}{3}x^3+\frac{1}{2}x^2+\frac{1}{3}x\right]_0^1 = \frac{7}{6}$

(2)

$xy$ $D : 0 \leq x \leq 1$ $0 \leq y \leq x^2$

$\int_0^1\int_0^{x^2} xy\,dydx = \int_0^1 \left[\frac{1}{2}xy^2\right]_0^{x^2} dx = \int_0^1 \frac{1}{2}x^5\,dx = \left[\frac{1}{12}x^6\right]_0^1 = \frac{1}{12}$

(3)

$x^2+y^2$ $D : x \geq 0$ $y \geq 0$ $x+y \leq 1$

$\int_0^1\int_0^{1-x} (x^2+y^2)\,dydx = \int_0^1 \left[x^2y+\frac{1}{3}y^3\right]_0^{1-x} dx = \int_0^1 \left\{x^2-x^3+\frac{1}{3}(1-x)^3\right\}dx$

$= \left[\frac{1}{3}x^3-\frac{1}{4}x^4-\frac{1}{12}(1-x)^4\right]_0^1 = \frac{1}{6}$

(4)

$3y^2-xy$ $D : y \leq 2x$ $y \geq 2x^2$

$\int_0^1\int_{2x^2}^{2x} 3y^2-xy\,dydx = \int_0^1 \left[y^3-\frac{1}{2}xy^2\right]_{2x^2}^{2x} dx = \int_0^1 8x^3-2x^3-8x^6+2x^5\,dx = \int_0^1 6x^3-8x^6+2x^5\,dx$

$= \left[\frac{3}{2}x^4-\frac{8}{7}x^7+\frac{1}{3}x^6\right]_0^1 = \frac{29}{42}$

(5)

$e^{x+y}$ $D : 0 \leq x \leq 1$ $0 \leq y \leq 1$

$\int_0^1\int_0^1 e^{x+y}\,dydx = \int_0^1 \left[e^{x+y}\right]_0^1 dx = \int_0^1 e^{x+1}-e^x\,dx = \left[e^{x+1}-e^x\right]_0^1 = e^2-2e+1 = (e-1)^2$

(6)

$x\sqrt{1-y^3} \qquad D: 0 \leq x \leq 1 \quad x \leq y \leq 1$

$\int_0^1 \int_x^1 x\sqrt{1-y^3}\, dy\, dx$

順序を入れ替えて

$= \int_0^1 \int_0^y x\sqrt{1-y^3}\, dx\, dy = \int_0^1 \frac{1}{2} y^2 \sqrt{1-y^3}\, dy = \left[-\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{2}{3}\sqrt{(1-y^3)^3}\right]_0^1 = \frac{1}{9}$