6.2 Volumes/1: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
\pi\int_1^2\left[(2-\frac{1}{2})^{2}\right]dy & = \pi\int_1^2\left[(4-2x+\frac{1}{4}x^2)\right]dy = \pi\int_0^1\left[(2y^2-y^4)\right]dy \\[2ex]
\pi\int_1^2\left[(2-\frac{1}{2})^{2}\right]dx & = \pi\int_1^2\left[(4-2x+\frac{1}{4}x^2)\right]dx = \\[2ex]


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

Revision as of 03:22, 24 November 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} y=2-\frac{1}{2}x, x-axis\\[1ex] y=0\\[1ex] x=1\\[1ex] x=2\\[1ex] \end{align} }


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}})^{2}\right]dx&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dx=\\[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]=\pi \left[{\frac {10}{15}}-{\frac {3}{15}}\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}