2022考研数二第15题解析：极坐标系定积分的计算

一、题目

已知曲线 $L$ 的极坐标方程为 $r = \sin 3 \theta \left( 0 \leqslant \theta \leqslant \frac{\pi}{3} \right)$, 则 $L$ 围成有界区域的面积为 $\underline{\qquad\qquad}$.

二、解析

解法 1

本题中的被积函数在极坐标系下也很规整简单，所以，不用转为直角坐标系下的积分进行计算.

接着，根据《利用定积分计算以极坐标系为基准的平面图形面积的公式》和《点火公式》可知：

$$
\begin{aligned}
S & = \int_{0}^{\frac{\pi}{3}} \frac{1}{2} \sin^{2} 3 \theta \mathrm{~d} \theta \\ \\
& = \int_{0}^{\frac{\pi}{3}} \frac{1}{6} \sin^{2} 3 \theta \mathrm{~d} 3 \theta \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{u = 3 \theta} \\ \\
& = \frac{1}{6} \int_{0}^{\pi} \sin^{2} u \mathrm{~d} u \\ \\
& = \frac{1}{6} \cdot 2 \cdot \textcolor{pink}{ \int_{0}^{\frac{\pi}{2}} \sin^{2} u \mathrm{~d} u } \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\text{点火公式}} \\ \\
& = \frac{1}{6} \times 2 \times \textcolor{pink}{ \frac{\pi}{2} \times \frac{1}{2} } \\ \\
& = \textcolor{lightgreen}{ \frac{\pi}{12} }
\end{aligned}
$$

解法 2

本题中的被积函数在极坐标系下也很规整简单，所以，不用转为直角坐标系下的积分进行计算.

接着，根据《利用定积分计算以极坐标系为基准的平面图形面积的公式》可知：

$$
\begin{aligned}
S & = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \sin^{2} 3 \theta \mathrm{~d}\theta \\ \\
& = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \frac{1 – \cos 6 \theta}{2} \mathrm{~d}\theta \\ \\
& = \frac{1}{4} \left(\frac{\pi}{3} – \int_{0}^{\frac{\pi}{3}} \cos 6 \theta \mathrm{~d}\theta\right) \\ \\
& = \frac{\pi}{12} + \left. \frac{1}{24} \sin 6 \theta \right|_{0}^{\frac{\pi}{3}} \\ \\
& = \frac{\pi}{12} + 0 – 0 \\ \\
& = \textcolor{lightgreen}{ \frac{\pi}{12} }
\end{aligned}
$$