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}- | &= \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] | ||
&= \frac{7\pi}{15} | &= \frac{7\pi}{15} | ||
\end{align} | \end{align} | ||
</math> | </math> |
Revision as of 03:29, 24 November 2022