题目

设 $D$ 是由 $0\le x \le 1, 0 \le y \le 1$ 所确定的平面区域，求 $\iint\limits_D\sqrt{x^2+y^2}dxdy$

解答

被积函数 里有 $x^2+y^2$，考虑转换成 极坐标

$$
\begin{aligned}
\text{原式} \quad
= \quad &
\int_0^{\frac{\pi}{4}}d\theta\int_0^{\frac{1}{\cos\theta}} r^2dr +
\int_{\frac{\pi}{4}}^\frac{\pi}{2}d\theta\int_0^{\frac{1}{\sin\theta}} r^2dr
\\
= \quad &
\frac{1}{3}\bigg(\int_0^{\frac{\pi}{4}} \sec^3\theta d\theta + \int_{\frac{\pi}{4}}^\frac{\pi}{2} \csc^3\theta d\theta\bigg)
\\
= \quad &
\frac{2}{3}\bigg(\int_0^{\frac{\pi}{4}} \sec^3\theta d\theta\bigg)
\\
\end{aligned}
$$

$$
\begin{aligned}
\int \sec^3x dx = \tan x\sec x - \int \tan^2 x \sec x dx = \tan x\sec x - \int (\sec^3 - \sec x) dx \\
=\tan x\sec x - I + \int\sec xdx \Rightarrow I = \frac{1}{2}\tan x \sec x + \frac{1}{2} \ln (\tan x + \sec x) + C
\end{aligned}
$$

$$
\text{原式} =
\frac{1}{3}\tan x \sec x + \frac{1}{3} \ln (\tan x + \sec x) \bigg|_0^{\frac{\pi}{4}} = \frac{\sqrt{2}}{3}+\frac{1}{3}\ln (1+\sqrt{2})
$$