第8章 8.17

(1)

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

$\int_0^1\int_0^y x^2 y\,dxdy=\int_0^1\left[\frac13 x^3 y\right]_0^y dy=\int_0^1 \frac13 y^4 dy=\left[\frac{1}{15}y^5\right]_0^1=\frac{1}{15}$

(2)

$1+x+y$ $D:x\geq 0\quad y\geq 0\quad x+y\leq 1$

$\int_0^1\int_0^{1-x} 1+x+y\,dydx=\int_0^1\left[y+xy+\frac12 y^2\right]_0^{1-x}dx=\int_0^1\left(1-x+x-x^2+\frac12-x+\frac12 x^2\right)dx$

$=\int_0^1\left(-\frac12 x^2-x+\frac32\right)dx=\left[-\frac16 x^3-\frac12 x^2+\frac32 x\right]_0^1=\frac56$

(3)

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

$\int_0^1\int_{y^2}^y x^2+y^2\,dxdy=\int_0^1\left[\frac13 x^3+y^2 x\right]_{y^2}^y dy=\int_0^1 -\frac13 y^6-y^4+\frac43 y^3\,dy$

$=\left[-\frac{1}{21}y^7-\frac15 y^5+\frac13 y^4\right]_0^1=\frac{3}{35}$

(4)

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

$\int_0^1\int_{x^2}^x \frac{1}{\sqrt{1-x}}\,dydx=\int_0^1\left[\frac{1}{\sqrt{1+x}}y\right]_{x^2}^x dx=\int_0^1 \frac{x}{\sqrt{1+x}}-\frac{x^2}{\sqrt{1+x}}\,dx$

$\sqrt{1+x}=t$ とおくと $dx=2tdt$ $x:0\to 1$ $t:1\to\sqrt2$ これから

$=\int_1^{\sqrt2}\frac{(t^2-1)(1-t^2+1)}{t}2tdt=\int_1^{\sqrt2}2(-t^4+3t^2-2)dt=2\left[-\frac15 t^5+t^3-2t\right]_1^{\sqrt2}$

$=2\cdot\left(-\frac{4\sqrt2}{5}+2\sqrt2-2\sqrt2+\frac15-1+2\right)=\frac{12-8\sqrt2}{5}$