问题 #
$$ V_y = 2 \pi a^3 \int_{0}^{2 \pi} (t - \sin t)(1 - \cos t)(1 - \cos t) \mathop{}\mathrm{d} t $$想法 #
显然，展开后求定积分即可。群友们也是这样做的。
但思来想去，总感觉应该有更快的方法，然后发现了这么个做法：
$$ \begin{aligned} V_y &= 2 \pi a^3 \int_{0}^{2 \pi} (t - \sin t)(1 - \cos t)(1 - \cos t) \mathop{}\mathrm{d} t \\ &= 8 \pi a^3 \int_{0}^{2 \pi } (t - \pi - \sin t + \pi) \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \\ &= 8 \pi a^3 \left[ \int_{0}^{2 \pi } (t - \pi) \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t - \int_{0}^{2 \pi } \sin t \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t + \int_{0}^{2 \pi } \pi \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \right] \end{aligned} $$$$ \begin{aligned} V_y &= 8 \pi^2 a^3 \int_{0}^{2 \pi } \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \end{aligned} $$$$ \begin{aligned} V_y &= 16 \pi^2 a^3 \int_{0}^{ \pi } \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \\ &= 32 \pi^2 a^3 \int_{0}^{ \frac{\pi}{2} } \cos^4 u \mathop{}\mathrm{d} u \\ &= 32 \pi^2 a^3 \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \\ &= 6 \pi^3a^3 \end{aligned} $$启发 #
$$ \int_{0}^{\pi} xf(\sin x) \mathop{}\mathrm{d} x = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathop{}\mathrm{d} x \qquad f(x)\text{ 连续} $$从直观上来证明的话，就是因为 $\sin x$ 是关于 $\displaystyle \frac{\pi}{2}$ 对称的，而且 $f(x)$ 是连续的，所以 $f(\sin x)$ 也是关于 $\displaystyle \frac{\pi}{2}$ 对称的。而 $x - \displaystyle \frac{\pi}{2}$ 则关于 $\left(\displaystyle \frac{\pi}{2} , 0\right)$ 对称
$$ g(x) = (x - \frac{\pi}{2}) f(\sin x) $$$$ g(\frac{\pi}{2} + x) = x f(\cos x) = -[-xf(\cos x)] = -g(\frac{\pi}{2} - x) $$$$ \int_{0}^{\pi} xf(\sin x) \mathop{}\mathrm{d} x = \int_{0}^{\pi} (x - \frac{\pi}{2} + \frac{\pi}{2})f(\sin x) \mathop{}\mathrm{d} x = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathop{}\mathrm{d} x $$