题目

$D = \{(x,y)||x| + |y| \le \dfrac{\pi}{2}\}$,比较积分大小:
$$
I_1=\iint\limits_D\sqrt{x^2+y^2}d\sigma,I_2=\iint\limits_D\sin\sqrt{x^2+y^2}d\sigma,I_3=\iint\limits_D(1-\cos\sqrt{x^2+y^2})d\sigma
$$

解答

由于积分区域相同,故只需比较被积函数大小

令 $x^2+y^2 = u^2$

$$
I_1 = \iint\limits_D u d\sigma,I_2 = \iint\limits_D \sin u d\sigma,
I_3 = \iint\limits_D (1-\cos u) d\sigma,
$$

令 $f(x) = x -1+ \cos x$,则 $f(0) = 0, f’(x) = 1 - \sin x > 0$

故 $x > 1-\cos x$

令 $f(x) = \sin x -1+ \cos x$,则 $f(0) = 0, f’(x) = \cos x - \sin x$

易得:$f(x)$ 在 $[0, \dfrac{\pi}{4}]$ 单增,在 $[\dfrac{\pi}{4}, \dfrac{\pi}{2}]$ 单减,且 $f(0) = 0, f(\dfrac{\pi}{2}) = 0$

故 $f(x) > 0 \Rightarrow \sin x > 1-\cos x$

综上:$x > \sin x > 1- \cos x$

故:$I_1 > I_2 > I_3$