반원 단면 입체

직선 $\displaystyle x = t \left( \sqrt{\frac{\pi}{6}} \leq t \leq \sqrt{\frac{\pi}{2}} \right)$를 포함하고 $x$축에 수직인 평면으로 자른 단면의 넓이를 $S \left( t \right)$라 하면 $\displaystyle S \left( t \right) = \frac{1}{2} \pi \left( t \sqrt{t \sin t ^{2}} \right) ^{2} = \frac{1}{2} \pi t ^{3} \sin t ^{2}$ 따라서 구하는 부피를 $V$라 하면 $\displaystyle V = \int _{\sqrt{\frac{\pi}{6}}} ^{\sqrt{\frac{\pi}{2}}} {S' \left( t \right) dt} = \int _{\sqrt{\frac{\pi}{6}}} ^{\sqrt{\frac{\pi}{2}}} {\frac{1}{2} \pi t ^{3} \sin t ^{2} dt}$ $t ^{2} = s$라 하면 $\displaystyle \frac{ds}{dt} = 2 t$이고 $\displaystyle t = \sqrt{\frac{\pi}{6}}$일 때 $\displaystyle s = \frac{\pi}{6}$, $\displaystyle t = \sqrt{\frac{\pi}{2}}$일 때 $\displaystyle s = \frac{\pi}{2}$이므로 $\displaystyle V = \int _{\frac{\pi}{6}} ^{\frac{\pi}{2}} {\frac{\pi}{4} \times s \sin s ds} = \frac{\pi}{4} \int _{\frac{\pi}{6}} ^{\frac{\pi}{2}} {s \sin s ds}$ $u \left( s \right) = s$, $v' \left( s \right) = \sin s$라 하면 $u' \left( s \right) = 1$, $v \left( s \right) = - \cos s$이므로 $V$$\displaystyle = \frac{\pi}{4} \times \left\{ \begin{array}{l} \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} - s \cos s \end{array} \right] _{\frac{\pi}{6}} ^{\frac{\pi}{2}} - \int _{\frac{\pi}{6}} ^{\frac{\pi}{2}} {\left( - \cos s \right) ds} \end{array} \right\}$ $\displaystyle = \frac{\pi}{4} \times \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} - s \cos s + \sin s \end{array} \right] _{\frac{\pi}{6}} ^{\frac{\pi}{2}}$ $\displaystyle = \frac{\sqrt{3} \pi ^{2} + 6 \pi}{48}$