第8章 8.7

(1)

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

$\int_0^2\int_0^x xy^2\,dy\,dx = \int_0^2\left[\frac{1}{3}xy^3\right]_0^x dx = \int_0^2 \frac{1}{3}x^4\,dx = \left[\frac{1}{15}x^5\right]_0^2 = \frac{32}{15}$

(2)

$x$ $D: x^2+y^2 \leq 4 \quad x+y \geq 2$

$x+y=2$ から $y=2-x$

$x^2+y^2 \leq 4$ に代入して $x^2+4-4x+x^2 \leq 4$

$x^2-2x \leq 0$ $0 \leq x \leq 2$ これから

$\int_0^2\int_{2-x}^{\sqrt{4-x^2}} x\,dy\,dx = \int_0^2 x(\sqrt{4-x^2}-2+x)\,dx = \int_0^2\left(\frac{3}{2}\cdot 2x\sqrt{4-x^2}\cdot\frac{1}{3} - 2x + x^2\right)dx$

$= \left[-\frac{1}{3}\sqrt{(4-x^2)^3} - x^2 + \frac{1}{3}x^3\right]_0^2 = -4+\frac{8}{3}+\frac{8}{3} = \frac{4}{3}$

(3)

$\sqrt{4x^2-y^2}$ $D: 0 \leq y \leq x \leq 1$

$\int_0^1\int_0^x \sqrt{4x^2-y^2}\,dy\,dx = \int_0^1\left[\frac{1}{2}\left(y\sqrt{4x^2-y^2}+4x^2\sin^{-1}\frac{y}{2x}\right)\right]_0^x dy$

$= \int_0^1 \frac{1}{2}\left(x\sqrt{3x^2}+4x^2\sin^{-1}\frac{1}{2}\right)dx = \int_0^1 \frac{1}{2}\left(\sqrt{3}x^2+\frac{2}{3}\pi x^2\right)dx = \left[\frac{1}{2}\left(\frac{\sqrt{3}}{3}x^3+\frac{2}{9}\pi x^3\right)\right]_0^1 = \frac{\sqrt{3}}{6}+\frac{\pi}{9}$

(4)

$x^2$ $D: x^2 \leq y \leq 2-x$

$\int_{-2}^1\int_{x^2}^{2-x} x^2\,dy\,dx = \int_{-2}^1\left[x^2y\right]_{x^2}^{2-x}dx = \int_{-2}^1(2x^2-x^3-x^4)\,dx = \left[\frac{2}{3}x^3-\frac{1}{4}x^4-\frac{1}{5}x^5\right]_{-2}^1 = \frac{63}{20}$