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[\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(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[8-4+\frac{8}{12}-\left(4-1+\frac{1}{12}\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[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] | &= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex] |
Latest revision as of 04:10, 24 November 2022