📝 题目
例 2 求双纽线 ${r}^{2} = {a}^{2}\cos {2\theta }\left( {a > 0}\right)$ 所围的面积与绕极轴旋转的侧面积.
💡 答案与解析
解 双纽线的图形如图 3.7 所示, 由图形的对称性得图形的面积为
$$ A = 4 \cdot \frac{1}{2}{\int }_{0}^{\frac{\pi }{4}}{a}^{2}\cos {2\theta }\mathrm{d}\theta = {a}^{2}. $$
\begin{center} \includegraphics[max width=0.2\textwidth]{images/028.jpg} \end{center} \hspace*{3em}
图 3.7
图形绕极轴旋转的侧面积为
$$ P = 2 \cdot {2\pi }{\int }_{0}^{\frac{\pi }{4}}r\left( \theta \right) \sin \theta \sqrt{{r}^{2}\left( \theta \right) + {r}^{\prime 2}\left( \theta \right) }\mathrm{d}\theta $$
$$ = 2 \cdot {2\pi }{\int }_{0}^{\frac{\pi }{4}}a\sqrt{\cos {2\theta }} \cdot \sin \theta \sqrt{{a}^{2}\cos {2\theta } + \frac{{a}^{2}{\sin }^{2}{2\theta }}{\cos {2\theta }}}\mathrm{d}\theta $$
$$ = {4\pi }{a}^{2}{\int }_{0}^{\frac{\pi }{4}}\sin \theta \mathrm{d}\theta = {2\pi }{a}^{2}\left( {2 - \sqrt{2}}\right) . $$