6.2 Volumes/25: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
Line 4: Line 4:
\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 = \pi\int_0^1\left[(2y^2-y^4)\right]dy \\[2ex]
\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 = \pi\int_0^1\left[(2y^2-y^4)\right]dy \\[2ex]


&= \pi\left[\frac{2y^3}{3}-\frac{y^5}{5}\right]\Bigg|_0^1 = \pi\left[\frac{2}{3}-\frac{1}{5}\right]= \\[2ex]
&= \pi\left[\frac{2y^3}{3}-\frac{y^5}{5}\right]\Bigg|_0^1 \\[2ex]
&= \pi\left[\frac{2}{3}-\frac{1}{5}\right]= \\[2ex]


\pi\left[\frac{10}{15}-\frac{3}{15}\right] = \frac{7\pi}{15}
\pi\left[\frac{10}{15}-\frac{3}{15}\right] = \frac{7\pi}{15}

Revision as of 02:56, 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=\pi \int _{0}^{1}\left[(2y^{2}-y^{4})\right]dy\\[2ex]&=\pi \left[{\frac {2y^{3}}{3}}-{\frac {y^{5}}{5}}\right]{\Bigg |}_{0}^{1}\\[2ex]&=\pi \left[{\frac {2}{3}}-{\frac {1}{5}}\right]=\\[2ex]\pi \left[{\frac {10}{15}}-{\frac {3}{15}}\right]={\frac {7\pi }{15}}\end{aligned}}}