6.2 Volumes/1: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
\pi\int_1^2\left[\left(2-\frac{1}{2}x\right)^2\right]dy & = \pi\int_1^2\left[\left(4-2x+\frac{1}{4}x^2\right)\right]dx \\[2ex]
\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[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex]

Revision as of 03:44, 24 November 2022