6.2 Volumes/1

$y=2-{\frac {1}{2}}x$

y=0, x=1, x=2; about the x-axis

${\begin{aligned}\pi \int _{1}^{2}\left[\left(2-{\frac {1}{2}}x\right)^{2}\right]dx&=\pi \int _{1}^{2}\left[\left(4-2x+{\frac {1}{4}}x^{2}\right)\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[8-4+{\frac {8}{12}}-\left(4-1+{\frac {1}{12}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{12}}\right]=\pi \left[1+{\frac {7}{12}}\right]\\[2ex]&=\pi \left[{\frac {12}{12}}+{\frac {7}{12}}\right]=\pi \left[{\frac {19}{12}}\right]\\[2ex]&={\frac {19\pi }{12}}\end{aligned}}$