问题及解答

求 $\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^R \sqrt{R^2-(r\cos\theta)^2}r\mathrm{d}r$.

解.

\[
\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^R \sqrt{R^2-(r\cos\theta)^2}r\mathrm{d}r=\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^R \sqrt{R^2-r^2\cos^2\theta}\cdot\frac{1}{2}\mathrm{d}r,
\]

令 $t=r^2\cos^2\theta$, 则积分等于

\[
\begin{split}
&\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^{R^2\cos^2\theta}\sqrt{R^2-t}\cdot\frac{1}{2}\frac{1}{\cos^2\theta}\mathrm{d}t\\
=&\frac{1}{2}\int_0^{\frac{\pi}{2}}\sec^2\theta\mathrm{d}\theta\cdot\int_0^{R^2\cos^2\theta}\sqrt{R^2-t}\mathrm{d}t\\
=&\frac{1}{2}\int_0^{\frac{\pi}{2}}\sec^2\theta\mathrm{d}\theta\cdot\Bigl[-\frac{2}{3}(R^2-t)^{\frac{3}{2}}\Bigr]\biggr|_{R^2\cos^2\theta}^{0}\\
=&\frac{1}{2}\int_0^{\frac{\pi}{2}}\sec^2\theta\mathrm{d}\theta\cdot\frac{2}{3}\cdot\Bigl(R^3-(R^2-R^2\cos^2\theta)^{\frac{3}{2}}\Bigr)\\
=&\frac{1}{3}\int_0^{\frac{\pi}{2}}\sec^2\theta\mathrm{d}\theta\cdot(R^3-R^3\sin^3\theta)\\
=&\frac{1}{3}R^3\cdot\int_0^{\frac{\pi}{2}}\sec^2\theta\cdot(1-\sin^3\theta)\mathrm{d}\theta\\
=&\frac{1}{3}R^3\cdot\int_0^{\frac{\pi}{2}}\frac{1-\sin^3\theta}{\cos^2\theta}\mathrm{d}\theta,
\end{split}
\]

其中 

\[
\begin{split}
\int_0^{\frac{\pi}{2}}\frac{1-\sin^3\theta}{\cos^2\theta}\mathrm{d}\theta&=\int_0^{\frac{\pi}{2}}\frac{(1-\sin\theta)(1+\sin\theta+\sin^2\theta)}{1-\sin^2\theta}\mathrm{d}\theta\\
&=\int_0^{\frac{\pi}{2}}\frac{1+\sin\theta+\sin^2\theta}{1+\sin\theta}\mathrm{d}\theta\\
&=\frac{\pi}{2}+\int_0^{\frac{\pi}{2}}\frac{\sin^2\theta}{1+\sin\theta}\mathrm{d}\theta.
\end{split}
\]

下面计算积分 $\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\sin^2\theta}{1+\sin\theta}\mathrm{d}\theta$.

\[
\begin{split}
\int_{0}^{\frac{\pi}{2}}\frac{\sin^2\theta}{1+\sin\theta}\mathrm{d}\theta&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2\theta-1+1}{1+\sin\theta}\mathrm{d}\theta\\
&=\int_{0}^{\frac{\pi}{2}}\Bigl(\frac{1}{1+\sin\theta}-(1-\sin\theta)\Bigr)\mathrm{d}\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin\theta}\mathrm{d}\theta-(\theta+\cos\theta)\biggr|_{0}^{\frac{\pi}{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin\theta}\mathrm{d}\theta+1-\frac{\pi}{2}.
\end{split}
\]

对于积分 $\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin\theta}\mathrm{d}\theta$, 我们可以采用万能变换, 令 $u=\tan\frac{\theta}{2}$, 则 $\sin\theta=\dfrac{2u}{1+u^2}$, $\mathrm{d}\theta=\dfrac{2\mathrm{d}u}{1+u^2}$, 于是

\[
\begin{split}
\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin\theta}\mathrm{d}\theta&=\int_0^1 \frac{1}{1+\frac{2u}{1+u^2}}\cdot\frac{2\mathrm{d}u}{1+u^2}\\
&=\int_0^1 \frac{2}{(1+u)^2}\mathrm{d}u\\
&\xlongequal{t=1+u}2\int_1^2 \frac{1}{t^2}\mathrm{d}t=1.
\end{split}
\]

因此

\[
\int_{0}^{\frac{\pi}{2}}\frac{\sin^2\theta}{1+\sin\theta}\mathrm{d}\theta=1+1-\frac{\pi}{2}=2-\frac{\pi}{2}.
\]

因此, 原积分为

\[
\frac{R^3}{3}\cdot\Bigl(\frac{\pi}{2}+2-\frac{\pi}{2}\Bigr)=\frac{2}{3}R^3.
\]

一个更简单的计算方法是将这个累次积分转换为二重积分. 注意到积分区域为第一象限中的四分之一圆盘, 记为 $D$.

 使用极坐标 $x=r\cos\theta$, $y=r\sin\theta$, 则该累次积分可写为

\[
\begin{split}
\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^R \sqrt{R^2-(r\cos\theta)^2}r\mathrm{d}r&=\iint_{D}\sqrt{R^2-x^2}\mathrm{d}x\mathrm{d}y\\
&=\int_0^R\mathrm{d}x\int_{0}^{\sqrt{R^2-x^2}}\sqrt{R^2-x^2}\mathrm{d}y\\
&=\int_0^R (R^2-x^2)\mathrm{d}x\\
&=R^3-\frac{1}{3}x^3\biggr|_{0}^{R}\\
&=\frac{2}{3}R^3.
\end{split}
\]