二重积分的被积函数中含有根号怎么办？可以尝试改造积分区域实现对根号的去除

一、题目

已知，$D$ $=$ ${(x, y) \mid -1 \leqslant x \leqslant 1,0 \leqslant y \leqslant 2}$, 则 $I=\iint_{D} \sqrt{\left|y-x^{2}\right|} \mathrm{d} x \mathrm{~d} y=?$

难度评级：

二、解析

根据题意我们可以绘制出如图所示的积分区域，并将该积分区域划分为位于 $y=x^{2}$ 下部的 $D_{1}$ 区域和位于 $y=x^{2}$ 上部的 $D_{2}$ 区域：

于是：

$$
I=\iint_{D} \sqrt{\left|y-x^{2}\right|} \mathrm{~d} x \mathrm{~d} y=
$$

$$
I=\iint_{D_{1}} \sqrt{\left|y-x^{2}\right|} \mathrm{~d} x \mathrm{~d} y+\iint_{D_{2}} \sqrt{\left|y-x^{2}\right|} \mathrm{~d} x \mathrm{~d} y
$$

又因为：

$$
D_{1} \Rightarrow yx^{2} \Rightarrow
$$

于是：

$$
I=\iint_{D_{1}} \sqrt{x^{2}-y} \mathrm{~d} x \mathrm{~d} y+\iint_{D_{2}} \sqrt{y-x^{2}} \mathrm{~d} x \mathrm{~d} y \Rightarrow
$$

$$
I=\int_{-1}^{1} \mathrm{~d} x \int_{0}^{x^{2}} \sqrt{x^{2}-y} \mathrm{~d} y+\int_{-1}^{1} \mathrm{~d} x \int_{x^{2}}^{2} \sqrt{y-x^{2}} \mathrm{~d} y \Rightarrow
$$

$$
I=2 \int_{0}^{1} \mathrm{~d} x \int_{0}^{x^{2}} \sqrt{x^{2}-y} \mathrm{~d} y+2 \int_{0}^{1} \mathrm{~d} x \int_{x^{2}}^{2} \sqrt{y-x^{2}} \mathrm{~d} y \Rightarrow
$$

更改为幂指形式：

$$
I=2 \int_{0}^{1}\left[\int_{0}^{x^{2}}\left(x^{2}-y\right)^{\frac{1}{2}} \mathrm{~d} y\right] \mathrm{~d} x+2 \int_{0}^{1}\left[\int_{x^{2}}^{2}\left(y-x^{2}\right)^{\frac{1}{2}} \mathrm{~d} y\right] \mathrm{~d} x \Rightarrow
$$

$y$ 是积分变量，$x$ 看做常数（注意正负号）：

$$
I=\left.2 \int_{0}^{1} \frac{ \textcolor{orangered}{ – } 2 }{3}\left(x^{2}-y\right)^{\frac{3}{2}}\right|_{ \textcolor{springgreen}{ y = 0 } } ^{ \textcolor{springgreen}{ y = x^{2} } } \mathrm{~d} x+\left.2 \int_{0}^{1} \frac{2}{3}\left(y-x^{2}\right)^{\frac{3}{2}}\right|_{ \textcolor{springgreen}{ y = x^{2} } } ^{ \textcolor{springgreen}{ y = 2 }} \mathrm{~d} x \Rightarrow
$$

$$
I=2 \int_{0}^{1} \frac{ \textcolor{orangered}{ – } 2 }{3} (0 \textcolor{orangered}{ – } x^{3}) \mathrm{~d} x + 2 \int_{0}^{1} \frac{2}{3} [ \left(2-x^{2}\right)^{\frac{3}{2}} – 0 ] \mathrm{~d} x \Rightarrow
$$

$$
I=\frac{\textcolor{orangered}{ + } 4}{3} \int_{0}^{1} x^{3} \mathrm{~d} x+\frac{4}{3} \int_{0}^{1}\left(2-x^{2}\right)^{\frac{3}{2}} \mathrm{~d} x
$$

其中：

$$
\int_{0}^{1} x^{3} \mathrm{~d} x=\left.\frac{1}{4} x^{4}\right|_{0} ^{1}=\frac{1}{4}
$$

又：

$$
\int_{0}^{1}\left(2-x^{2}\right)^{\frac{3}{2}} \mathrm{~d} x \Rightarrow x=\sqrt{2} \sin t \Rightarrow x \in(0,1) \Rightarrow
$$

$$
\sin t \in\left(0, \frac{\sqrt{2}}{2}\right) \Rightarrow t \in\left(0, \frac{\pi}{4}\right) \Rightarrow
$$

$$
\int_{0}^{1}\left(2-x^{2}\right)^{\frac{3}{2}} \mathrm{~d} x=\int_{0}^{\frac{\pi}{4}} \sqrt{\left(2-2 \sin ^{2} t\right)^{3}} \mathrm{~d} (\sqrt{2} \sin t)
$$

$$
\int_{0}^{\frac{\pi}{4}} \sqrt{\left(2 \cos ^{2} t\right)^{3}} \mathrm{~d} (\sqrt{2} \sin t)=
$$

$$
2 \sqrt{2} \cdot \sqrt{2} \int_{0}^{\frac{\pi}{4}} \cos ^{3} t \cdot \cos t \mathrm{~d} t=
$$

$$
4 \int_{0}^{\frac{\pi}{4}} \cos ^{4} t \mathrm{~d} t=
$$

上面的式子没办法用点火公式，也没办法直接积分，因此只能用倍角公式降幂：

$$
4 \int_{0}^{\frac{\pi}{4}}\left(\cos ^{2} t\right)^{2} \mathrm{~d} t \Rightarrow
$$

$$
\textcolor{yellow}{
\cos 2 \alpha=2 \cos ^{2} \alpha-1 \Rightarrow \cos ^{2} \alpha=\frac{1}{2}(1+\cos 2 \alpha) } \Rightarrow
$$

$$
4 \int_{0}^{\frac{\pi}{4}}\left[\frac{1}{2}\left(1+\cos ^{2} 2 t\right)\right]^{2} \mathrm{~d} t=
$$

$$
4 \int_{0}^{\frac{\pi}{4}} \frac{1}{4}\left(1+\cos ^{2} 2 t+2 \cos 2 t\right) \mathrm{~d} t=
$$

$$
\frac{\pi}{4}+\int_{0}^{\frac{\pi}{4}} \cos ^{2} 2 t \mathrm{~d} t + 2 \int_{0}^{\frac{\pi}{4}} \cos 2 t \mathrm{~d} t=
$$

再次使用倍角公式降幂：

$$
\frac{\pi}{4}+\int_{0}^{\frac{\pi}{4}}\left[\frac{1}{2}(1+\cos 4 t)\right] \mathrm{~d} t+2 \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 t \mathrm{~d} (2 t) =
$$

$$
\frac{\pi}{4}+\frac{1}{2} \cdot \frac{\pi}{4}+\left.\frac{1}{4} \sin 4 t\right|_{0} ^{\frac{\pi}{4}}+\left.\frac{2}{2} \sin 2 t\right|_{0} ^{\frac{\pi}{4}}=
$$

$$
\frac{\pi}{4}+\frac{\pi}{8}+\frac{1}{4}(0-0)+\frac{2}{2}(1-0)=
$$

$$
\frac{3 \pi}{8}+1
$$

综上可知：

$$
I=\frac{4}{3} \times \frac{1}{4}+\frac{4}{3}\left(\frac{3 \pi}{8}+1\right) \Rightarrow
$$

$$
I=\frac{1}{3}+\frac{\pi}{2}+\frac{4}{3}=\frac{5}{3}+\frac{\pi}{2}
$$

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