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
$$
\boxed{\dfrac{3\pi a^{2}}{4}}.
$$
难度：★★☆☆☆