6.2 Volumes/1: Difference between revisions

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&= \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]
&= \frac{7\pi}{15}
&= \frac{7\pi}{15}


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

Revision as of 03:31, 24 November 2022


Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \pi\int_1^2\left[(2-\frac{1}{2}x)^2\right]dy & = \pi\int_1^2\left[(4-2x+\frac{1}{4}x^2)\right]dx \\[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+\frac{8}{12}-3-\frac{1}{3}\right} = \pi\left[1+\frac{7}{12}\right] &= \frac{7\pi}{15} \end{align} }