y = 2 − 1 2 x , x − a x i s y = 0 x = 1 x = 2 {\displaystyle y=2-{\frac {1}{2}}x,x-axisy=0x=1x=2}
π ∫ 1 2 [ ( 2 − 1 2 ) 2 ] d y = π ∫ 1 2 [ ( 4 − 2 x + 1 4 x 2 ) ] d y = π ∫ 0 1 [ ( 2 y 2 − y 4 ) ] d y = π [ 2 y 3 3 − y 5 5 ] | 0 1 = π [ 2 3 − 1 5 ] = π [ 10 15 − 3 15 ] = 7 π 15 {\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[(2-{\frac {1}{2}})^{2}\right]dy&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dy=\pi \int _{0}^{1}\left[(2y^{2}-y^{4})\right]dy\\[2ex]&=\pi \left[{\frac {2y^{3}}{3}}-{\frac {y^{5}}{5}}\right]{\Bigg |}_{0}^{1}\\[2ex]&=\pi \left[{\frac {2}{3}}-{\frac {1}{5}}\right]=\pi \left[{\frac {10}{15}}-{\frac {3}{15}}\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[(2-{\frac {1}{2}})^{2}\right]dy&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dy=\pi \int _{0}^{1}\left[(2y^{2}-y^{4})\right]dy\\[2ex]&=\pi \left[{\frac {2y^{3}}{3}}-{\frac {y^{5}}{5}}\right]{\Bigg |}_{0}^{1}\\[2ex]&=\pi \left[{\frac {2}{3}}-{\frac {1}{5}}\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/d00b1e6c08ed6eda26c5f27ab782fd1ac58a9b9a)