6.2 Volumes/1: Difference between revisions
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(Created page with "<math> R = 1 </math> <math> \begin{align} r + f(y) &= 1\\[1ex] r &= 1-f(y)\\[1ex] r &= 1-y^2 \\ \end{align} </math> <math> \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 = \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]= \pi\left[\frac{10}{15}-\frac{3}{15}\right] \\[2ex] &= \frac{7\pi}{15} \end{align} <...") |
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<math> | <math> | ||
y=2-\frac{1}{2}x, | |||
y = 0, x = 1, x = 2; \text{about the x-axis} | |||
</math> | </math> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
\pi\ | \pi\int_1^2\left[\left(2-\frac{1}{2}x\right)^2\right]dx & = \pi\int_1^2\left[\left(4-2x+\frac{1}{4}x^2\right)\right]dx \\[2ex] | ||
&= \pi\left[\frac{ | &= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex] | ||
&= \pi\left[\frac{ | &= \pi\left[\left(4(2)-(2)^2+\frac{1}{12}(2)^3\right)-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[2ex] | ||
&= \frac{ | &= \pi\left[\left(8-4+\frac{8}{12}\right)-\left(4-1+\frac{1}{12}\right)\right] \\[2ex] | ||
&= \pi\left[4+\frac{8}{12}-3-\frac{1}{12}\right]= \pi\left[1+\frac{7}{12}\right] \\[2ex] | |||
&= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex] | |||
&= \frac{19\pi}{12} | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 04:10, 24 November 2022
![{\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[\left(2-{\frac {1}{2}}x\right)^{2}\right]dx&=\pi \int _{1}^{2}\left[\left(4-2x+{\frac {1}{4}}x^{2}\right)\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[\left(4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}\right)-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[\left(8-4+{\frac {8}{12}}\right)-\left(4-1+{\frac {1}{12}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{12}}\right]=\pi \left[1+{\frac {7}{12}}\right]\\[2ex]&=\pi \left[{\frac {12}{12}}+{\frac {7}{12}}\right]=\pi \left[{\frac {19}{12}}\right]\\[2ex]&={\frac {19\pi }{12}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b626b7619f18339f05d8e4a2e4256b6d9eed3a0a)