6.2 Volumes/25: Difference between revisions

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<math>
<math>
\pi\int_0^1\left[(1)^2-(1-y^2)^2\right]dy = \pi\left[y-\frac{y^5}{5}\right]\Bigg|_0^1 = \pi\left[1-\frac{1}{5}\right]=\frac{4\pi}{5}
\begin{align}
 
\pi\int_0^1\left[(1)^2-(1-y^2)^2\right]dy = \pi\int_0^1\left[(1-(1-2y^2+y^4)\right]dy = \\[2ex]
 
\pi\int_0^1\left[(2y^2-y^4)\right]dy \\
 
\pi\left[\frac{2y^3}{3}-\frac{y^5}{5}\right]\Bigg|_0^1 = \pi\left[\frac{2}{3}-\frac{1}{5}\right]= \\
 
\pi\left[\frac{10}{15}-\frac{3}{15}\right] = \frac{7\pi}{15}
 
\end{align}
</math>
</math>

Revision as of 02:55, 12 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\pi \int _{0}^{1}\left[(1)^{2}-(1-y^{2})^{2}\right]dy=\pi \int _{0}^{1}\left[(1-(1-2y^{2}+y^{4})\right]dy=\\[2ex]\pi \int _{0}^{1}\left[(2y^{2}-y^{4})\right]dy\\\pi \left[{\frac {2y^{3}}{3}}-{\frac {y^{5}}{5}}\right]{\Bigg |}_{0}^{1}=\pi \left[{\frac {2}{3}}-{\frac {1}{5}}\right]=\\\pi \left[{\frac {10}{15}}-{\frac {3}{15}}\right]={\frac {7\pi }{15}}\end{aligned}}}