y 2 + R = 1 {\displaystyle y^{2}+R=1}
∫ 0 1 π ( 1 − y 2 ) 2 − π ( 1 − y 31 ) 2 d y = π ∫ 0 1 1 − 2 y 2 + y 4 d y {\displaystyle \int _{0}^{1}\pi (1-y^{2})^{2}-\pi (1-{\sqrt[{31}]{y}})^{2}dy=\pi {\sqrt {\int _{0}^{1}}}1-2y^{2}+y^{4}dy}
∫ 0 1 1 − 2 − y 3 + y 2 3 d y = [ y − 2 y 3 3 + y 5 5 ] | 0 1 − [ y − 6 y 4 + 3 y 5 ] | 0 1 {\displaystyle \int _{0}^{1}1-2-{\sqrt[{3}]{y}}+y{\frac {2}{3}}dy=[y-{\frac {2y^{3}}{3}}+{\frac {y^{5}}{5}}]{\bigg |}_{0}^{1}-[y-{\frac {6y}{4}}+{\frac {3y}{5}}]{\bigg |}_{0}^{1}}
= π [ ( 1 − 2 3 + 1 5 ) − ( 1 − 6 4 + 3 5 ) ] = π [ 15 15 − 10 15 + 13 15 ) − ( 20 20 − 30 20 + 12 20 ) {\displaystyle =\pi [(1-{\frac {2}{3}}+{\frac {1}{5}})-(1-{\frac {6}{4}}+{\frac {3}{5}})]=\pi [{\frac {15}{15}}-{\frac {10}{15}}+{\frac {13}{15}})-({\frac {20}{20}}-{\frac {30}{20}}+{\frac {12}{20}})}
= π ( 18 15 x 2 10 ) x 3 = π ( 16 30 = π ( 13 30 ) = {\displaystyle =\pi ({\frac {18}{15}}x{\frac {2}{10}})x3=\pi ({\frac {16}{30}}=\pi ({\frac {13}{30}})=}
= ( 13 π 30 ) {\displaystyle =({\frac {13\pi }{30}})}
![{\displaystyle \int _{0}^{1}\pi (1-y^{2})^{2}-\pi (1-{\sqrt[{31}]{y}})^{2}dy=\pi {\sqrt {\int _{0}^{1}}}1-2y^{2}+y^{4}dy}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/381123d9afc2500e0788626cfc78e1a6eac13bb7)
![{\displaystyle \int _{0}^{1}1-2-{\sqrt[{3}]{y}}+y{\frac {2}{3}}dy=[y-{\frac {2y^{3}}{3}}+{\frac {y^{5}}{5}}]{\bigg |}_{0}^{1}-[y-{\frac {6y}{4}}+{\frac {3y}{5}}]{\bigg |}_{0}^{1}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cd06a519086bdae5c11829171f6508934bb5747c)
![{\displaystyle =\pi [(1-{\frac {2}{3}}+{\frac {1}{5}})-(1-{\frac {6}{4}}+{\frac {3}{5}})]=\pi [{\frac {15}{15}}-{\frac {10}{15}}+{\frac {13}{15}})-({\frac {20}{20}}-{\frac {30}{20}}+{\frac {12}{20}})}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e0dbc4e962223471312003a898493515e2048bd7)
