6.2 Volumes/29: Difference between revisions
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(Created page with "<math>y^2+R=1</math> <math>\int_{0}^{1}\pi(1-y^2)^2-\pi(1-\sqrt[31]{y})^2dy=\pi\sqrt{\int_{0}^{1}}1-2y^2+y^4dy</math> <math>\int_{0}^{1}1-2-\sqrt[3]{y}</math>") |
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<math>\int_{0}^{1}\pi(1-y^2)^2-\pi(1-\sqrt[31]{y})^2dy=\pi\sqrt{\int_{0}^{1}}1-2y^2+y^4dy</math> | <math>\int_{0}^{1}\pi(1-y^2)^2-\pi(1-\sqrt[31]{y})^2dy=\pi\sqrt{\int_{0}^{1}}1-2y^2+y^4dy</math> | ||
<math>\int_{0}^{1}1-2-\sqrt[3]{y}</math> | <math>\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}</math> | ||
<math>=\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})</math> | |||
<math>=\pi(\frac{18}{15}x\frac{2}{10})x3=\pi(\frac{16}{30}=\pi(\frac{13}{30})=</math> | |||
<math>=(\frac{13\pi}{30})</math> |
Latest revision as of 02:04, 29 November 2022