x軸で回転 y=0y = 0y=0 のとき 0=1−x0 = 1 - \sqrt{x}0=1−x x=1x = 1x=1 これから V=π∫01(1−x)2 dx=π∫01(1−2x+x) dx=π[x−43x3+12x2]01=π(1−43+12)V = \pi\int_0^1 (1-\sqrt{x})^2\,dx = \pi\int_0^1 (1-2\sqrt{x}+x)\,dx = \pi\left[x - \frac{4}{3}\sqrt{x^3} + \frac{1}{2}x^2\right]_0^1 = \pi\left(1 - \frac{4}{3} + \frac{1}{2}\right)V=π∫01(1−x)2dx=π∫01(1−2x+x)dx=π[x−34x3+21x2]01=π(1−34+21) =π(6−8+36)=π6= \pi\left(\frac{6-8+3}{6}\right) = \frac{\pi}{6}=π(66−8+3)=6π y軸で回転 x=0x = 0x=0 のとき y=1y = 1y=1 これから V=π∫01(1−y)4 dy=π[−15(1−y)5]01=π⋅15=π5V = \pi\int_0^1 (1-y)^4\,dy = \pi\left[-\frac{1}{5}(1-y)^5\right]_0^1 = \pi \cdot \frac{1}{5} = \frac{\pi}{5}V=π∫01(1−y)4dy=π[−51(1−y)5]01=π⋅51=5π