6.2 Volumes/1: Difference between revisions
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&= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex] | &= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex] | ||
&= \pi\left[4(2)-(2)^2+\frac{1}{12}(2)^3-4(1)-(1)^2+\frac{1}{12}(1)^3\right]= \pi\left[8-4+\frac{8}{12}-(4-1+\frac{1}{2})\right] \\[2ex] | &= \pi\left[4(2)-(2)^2+\frac{1}{12}(2)^3-(4(1)-(1)^2+\frac{1}{12}(1)^3)\right]= \pi\left[8-4+\frac{8}{12}-(4-1+\frac{1}{2})\right] \\[2ex] | ||
&= \frac{7\pi}{15} | &= \frac{7\pi}{15} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 03:29, 24 November 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[(2-{\frac {1}{2}}x)^{2}\right]dy&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}-(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3})\right]=\pi \left[8-4+{\frac {8}{12}}-(4-1+{\frac {1}{2}})\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}