6.2 Volumes/1: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
y=2-\frac{1}{2}x, x-axis\\[1ex]
y=2-\frac{1}{2}x, \x-axis\\[1ex]
y=0\\[1ex]
y=0\\[1ex]
x=1\\[1ex]
x=1\\[1ex]
Line 16: Line 16:
&= \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]
&= \pi\left[4+\frac{8}{12}-3-\frac{1}{3}\right] = \pi\left[1+\frac{7}{12}\right] \\[2ex]
&= \pi\left[4+\frac{8}{12}-3-\frac{1}{3}\right]= \pi\left[1+\frac{7}{12}\right] \\[2ex]
&= \pi\left[\frac{12}{12}+\frac{7}{12}\right] = \pi\left[\frac{19}{12} \\[2ex]
&= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex]
&= \frac{19\pi}{12}
&= \frac{19\pi}{12}


\end{align}
\end{align}
</math>
</math>

Revision as of 03:34, 24 November 2022

Failed to parse (unknown function "\x"): {\displaystyle \begin{align} y=2-\frac{1}{2}x, \x-axis\\[1ex] y=0\\[1ex] x=1\\[1ex] x=2\\[1ex] \end{align} }