人邮高数 第7章 第7-2-11题
📝 题目
11．求闭曲线 $\Gamma:\left(x^{2}+y^{2}\right)^{3}=a^{2}\left(x^{4}+y^{4}\right)(a\gt 0)$ 所围区域的面积．
💡 答案解析
[AI解答]
[AI解答] 我们要求闭曲线 $$ \Gamma:\left(x^{2}+y^{2}\right)^{3}=a^{2}\left(x^{4}+y^{4}\right),\quad a>0 $$ 所围区域的面积。采用极坐标变换 $$ x = r\cos\theta,\quad y = r\sin\theta, $$ 则 $$ x^{2}+y^{2}=r^{2},\quad x^{4}+y^{4}=r^{4}(\cos^{4}\theta+\sin^{4}\theta). $$ 代入曲线方程得 $$ (r^{2})^{3}=a^{2}r^{4}(\cos^{4}\theta+\sin^{4}\theta) \implies r^{6}=a^{2}r^{4}(\cos^{4}\theta+\sin^{4}\theta). $$ 当 $r\neq 0$ 时，两边除以 $r^{4}$ 得 $$ r^{2}=a^{2}(\cos^{4}\theta+\sin^{4}\theta). $$ 因此极坐标方程为 $$ r = a\sqrt{\cos^{4}\theta+\sin^{4}\theta}. $$ 由于曲线对称于极轴和 $\pi/2$ 轴，我们只需要考虑 $\theta\in[0,2\pi)$ 的整个周期，面积公式为 $$ S = \frac{1}{2}\int_{0}^{2\pi} r^{2}\,d\theta. $$ 代入 $r^{2}$ 表达式： $$ S = \frac{1}{2}\int_{0}^{2\pi} a^{2}(\cos^{4}\theta+\sin^{4}\theta)\,d\theta = \frac{a^{2}}{2}\int_{0}^{2\pi}(\cos^{4}\theta+\sin^{4}\theta)\,d\theta. $$ 利用三角恒等式 $$ \cos^{4}\theta+\sin^{4}\theta = (\cos^{2}\theta+\sin^{2}\theta)^{2} - 2\sin^{2}\theta\cos^{2}\theta = 1 - \frac{1}{2}\sin^{2}2\theta. $$ 又因为 $\sin^{2}2\theta = \frac{1-\cos4\theta}{2}$，所以 $$ \cos^{4}\theta+\sin^{4}\theta = 1 - \frac{1}{2}\cdot\frac{1-\cos4\theta}{2} = 1 - \frac{1}{4} + \frac{1}{4}\cos4\theta = \frac{3}{4} + \frac{1}{4}\cos4\theta. $$ 于是积分 $$ \int_{0}^{2\pi}(\cos^{4}\theta+\sin^{4}\theta)\,d\theta = \int_{0}^{2\pi}\left(\frac{3}{4}+\frac{1}{4}\cos4\theta\right)d\theta. $$ 由于 $\int_{0}^{2\pi}\cos4\theta\,d\theta=0$，因此 $$ \int_{0}^{2\pi}(\cos^{4}\theta+\sin^{4}\theta)\,d\theta = \frac{3}{4}\cdot 2\pi = \frac{3\pi}{2}. $$ 代入面积公式得 $$ S = \frac{a^{2}}{2}\cdot\frac{3\pi}{2} = \frac{3\pi a^{2}}{4}. $$
因此，曲线所围区域的面积为 $$ \boxed{\dfrac{3\pi a^{2}}{4}}. $$
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