y = 2 − 1 2 x , x − a x i s y = 0 x = 1 x = 2 {\displaystyle {\begin{aligned}y=2-{\frac {1}{2}}x,x-axis\\[1ex]y=0\\[1ex]x=1\\[1ex]x=2\\[1ex]\end{aligned}}}
π ∫ 1 2 [ ( 2 − 1 2 x ) 2 ] d y = π ∫ 1 2 [ ( 4 − 2 x + 1 4 x 2 ) ] d x = π [ 4 x − x 2 + 1 12 x 3 ] | 1 2 = π [ 4 ( 2 ) − ( 2 ) 2 + 1 12 ( 2 ) 3 − 4 ( 1 ) − ( 1 ) 2 + 1 12 ( 1 ) 3 ] = π [ 10 15 − 3 15 ] = 7 π 15 {\displaystyle {\begin{aligned}\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[{\frac {10}{15}}-{\frac {3}{15}}\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}
![{\displaystyle {\begin{aligned}y=2-{\frac {1}{2}}x,x-axis\\[1ex]y=0\\[1ex]x=1\\[1ex]x=2\\[1ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ad570f5891fb3315e2cff706966f174d13d9edcd)
![{\displaystyle {\begin{aligned}\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[{\frac {10}{15}}-{\frac {3}{15}}\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/eff449cfa53ba083ace949f4f25b342d351a98fc)