정적분의 값으로 반지름 a 구하기

$x = a \sin \theta$ $\displaystyle \left( - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \right)$로 놓으면 $dx = a \cos \theta d \theta$이고 $x = 0$일 때 $\theta = 0$, $x = a$일 때 $\displaystyle \theta = \frac{\pi}{2}$ $\displaystyle \sqrt{a ^{2} - x ^{2}} = \sqrt{a ^{2} ( 1 - \sin ^{2} \theta )} = a \cos \theta$ ($a > 0$, $\cos \theta \geq 0$)이므로 $\displaystyle \int _{0} ^{a} \sqrt{a ^{2} - x ^{2}} dx = \int _{0} ^{\frac{\pi}{2}} a ^{2} \cos ^{2} \theta d \theta = \frac{a ^{2}}{2} \int _{0} ^{\frac{\pi}{2}} ( 1 + \cos 2 \theta ) d \theta$ $\displaystyle = \frac{a ^{2}}{2} \left[ \theta + \frac{1}{2} \sin 2 \theta \right] _{0} ^{\frac{\pi}{2}} = \frac{\pi a ^{2}}{4}$ $\displaystyle \frac{\pi a ^{2}}{4} = 4 \pi$에서 $a ^{2} = 16$ 따라서 $a = 4$ ($a > 0$)