体積積分のブービートラップ – 2022年東大 数学 第5問

V = \int_1^2 2\pi (2-t)\sqrt{ 4- t^2 } dt

\begin {aligned}
 V &= \int_\frac{1}2^1 2\pi (2-2s)\sqrt{ 4- 4s^2 }  \cdot 2ds \\
& = 16 \pi \int_\frac{1}2^1  (1-s)\sqrt{ 1- s^2 } ds
\end {aligned}

\begin{aligned}
&\int _{\frac{1}2 }^1 \sqrt{ 1 -s^2} ds \\
= & \int _{\frac{ \pi}6}^{\frac{\pi}2} \cos^2 \theta d \theta \\
= & \int _{\frac{ \pi}6}^{\frac{\pi}2} \left (\frac{\cos2 \theta +1} 2 \right ) d \theta \\
= & \left [ \frac{\sin 2 \theta}4 \right ]_{\frac{ \pi}6}^{\frac{\pi}2} + \frac{1}2 \left (\frac{\pi}2 - \frac{ \pi}6 \right ) \\
= & -\frac{\sqrt{3}}8 + \frac{ \pi}6
\end{aligned}

\begin{aligned}
&\int _{\frac{1}2 }^1 s\sqrt{ 1 -s^2} ds \\
= & \int _{\frac{ \pi}6}^{\frac{\pi}2} \sin \theta\cos^2 \theta d \theta \\
=& -\int_{\frac{ \sqrt 3 } 2}^0 x^2 dx \\
= & \frac{\sqrt 3}8

\end{aligned}

\begin{aligned}
V  &= 16 \pi( -\frac{\sqrt{3}}8 + \frac{ \pi}6 -\frac{\sqrt{3}}8) \\
& = -4 \sqrt{3} \pi+\frac{8 \pi^2}3
\end{aligned}