R = x 1 2 r = x 3 {\displaystyle {\begin{aligned}R=x^{\frac {1}{2}}r=x^{3}\end{aligned}}}
v = π ∫ 0 1 [ ( x 1 2 ) 2 − ( x 3 ) 2 ] d x = π ∫ 0 1 [ ( x − x 6 ) ] d x = π [ x 2 2 − x 7 7 ] | 0 1 = π [ 1 2 − 1 7 ] = π [ 7 − 2 14 ] = 5 π 14 {\displaystyle {\begin{aligned}v=\pi \int _{0}^{1}\left[(x^{\frac {1}{2}})^{2}-(x^{3})^{2}\right]dx&=\pi \int _{0}^{1}\left[(x-x^{6})\right]dx\\[2ex]&=\pi \left[{\frac {x^{2}}{2}}-{\frac {x^{7}}{7}}\right]{\Bigg |}_{0}^{1}\\[2ex]&=\pi \left[{\frac {1}{2}}-{\frac {1}{7}}\right]=\pi \left[{\frac {7-2}{14}}\right]\\[2ex]&={\frac {5\pi }{14}}\end{aligned}}}
![{\displaystyle {\begin{aligned}v=\pi \int _{0}^{1}\left[(x^{\frac {1}{2}})^{2}-(x^{3})^{2}\right]dx&=\pi \int _{0}^{1}\left[(x-x^{6})\right]dx\\[2ex]&=\pi \left[{\frac {x^{2}}{2}}-{\frac {x^{7}}{7}}\right]{\Bigg |}_{0}^{1}\\[2ex]&=\pi \left[{\frac {1}{2}}-{\frac {1}{7}}\right]=\pi \left[{\frac {7-2}{14}}\right]\\[2ex]&={\frac {5\pi }{14}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b60feb23c871418e2f1f59ecfcd236849c7fdec0)