8.设D_(1)=(x,y)mid 0le x,yle 1,D_(2)=(x,y)mid 0le xle 1,0le yle sqrt(x),D_(3)=(x,y)mid 0le xle 1,x^2le yle 1,I_(i)=iintlimits_(D_{i)}(x-y)dxdy,i=1,2,3.则（）A. I_(1) > I_(2) > I_(3)B. I_(2) > I_(1) > I_(3)C. I_(1) > I_(3) > I_(2)D. I_(3) > I_(1) > I_(2)

本题考查二重积分的计算以及比较大小，解题思路是分别计算出$I_1$、$I_2$、$I_3$的值，然后再比较它们的大小。

计算$I_1$

已知$D_{1}=\{(x,y)\mid 0\le x,y\le 1\}$，则$I_{1}=\iint\limits_{D_{1}}(x - y)dxdy$可化为：
$\begin{align*}I_{1}&=\int_{0}^{1}dx\int_{0}^{1}(x - y)dy\\&=\int_{0}^{1}\left[xy - \frac{1}{2}y^2\right]_{0}^{1}dx\\&=\int_{0}^{1}\left(x - \frac{1}{2}\right)dx\\&=\left[\frac{1}{2}x^2 - \frac{1}{2}x\right]_{0}^{1}\\&=\frac{1}{2} - \frac{1}{2}\\&= 0\end{align*}$

计算$I_2$

已知$D_{2}=\{(x,y)\mid 0\le x\le 1,0\le y\le \sqrt{x}\}$，则$I_{2}=\iint\limits_{D_{2}}(x - y)dxdy$可化为：
$\begin{align*}I_{2}&=\int_{0}^{1}dx\int_{0}^{\sqrt{x}}(x - y)dy\\&=\int_{0}^{1}\left[xy - \frac{1}{2}y^2\right]_{0}^{\sqrt{x}}dx\\&=\int_{0}^{1}\left(x\sqrt{x} - \frac{1}{2}x\right)dx\\&=\int_{0}^{1}\left(x^{\frac{3}{2}} - \frac{1}{2}x\right)dx\\&=\left[\frac{2}{5}x^{\frac{5}{2}} - \frac{1}{4}x^2\right]_{0}^{1}\\&=\frac{2}{5} - \frac{1}{4}\\&=\frac{8}{20} - \frac{5}{20}\\&=\frac{3}{20}\end{align*}$

计算$I_3$

已知$D_{3}=\{(x,y)\mid 0\le x\le 1,x^{2}\le y\le 1\}$，则$I_{3}=\iint\limits_{D_{3}}(x - y)dxdy$可化为：
$\begin{align*}I_{3}&=\int_{0}^{1}dx\int_{x^2}^{1}(x - y)dy\\&=\int_{0}^{1}\left[xy - \frac{1}{2}y^2\right]_{x^2}^{1}dx\\&=\int_{0}^{1}\left(x - \frac{1}{2} - x^3 + \frac{1}{2}x^4\right)dx\\&=\left[\frac{1}{2}x^2 - \frac{1}{2}x - \frac{1}{4}x^4 + \frac{1}{10}x^5\right]_{0}^{1}\\&=\frac{1}{2} - \frac{1}{2} - \frac{1}{4} + \frac{1}{10}\\&=-\frac{1}{4} + \frac{1}{10}\\&=-\frac{5}{20} + \frac{2}{20}\\&=-\frac{3}{20}\end{align*}$

比较$I_1$、$I_2$、$I_3$的大小

因为$\frac{3}{20} > 0 > -\frac{3}{20}$，即$I_{2} > I_{1} > I_{3}$。