Question Number 211020 by peter frank last updated on 26/Aug/24
Answered by mahdipoor last updated on 26/Aug/24
![dV=(πy^2 )(dx)=π(((x−a)^2 (x−b)^2 )/c^2 )dx V=∫dV=(π/c^2 )∫_a ^( b) (x−a)^2 (x−b)^2 dx =(π/c^2 )[(((x−a)^3 (x−b)^2 )/3)−(((x−a)^4 (x−b))/6)−(((x−b)^5 )/(30))]_a ^b = ((π(a−b)^5 )/(30))](https://www.tinkutara.com/question/Q211025.png)
$${dV}=\left(\pi{y}^{\mathrm{2}} \right)\left({dx}\right)=\pi\frac{\left({x}−{a}\right)^{\mathrm{2}} \left({x}−{b}\right)^{\mathrm{2}} }{{c}^{\mathrm{2}} }{dx} \\ $$$${V}=\int{dV}=\frac{\pi}{{c}^{\mathrm{2}} }\int_{{a}} ^{\:{b}} \left({x}−{a}\right)^{\mathrm{2}} \left({x}−{b}\right)^{\mathrm{2}} {dx} \\ $$$$=\frac{\pi}{{c}^{\mathrm{2}} }\left[\frac{\left({x}−{a}\right)^{\mathrm{3}} \left({x}−{b}\right)^{\mathrm{2}} }{\mathrm{3}}−\frac{\left({x}−{a}\right)^{\mathrm{4}} \left({x}−{b}\right)}{\mathrm{6}}−\frac{\left({x}−{b}\right)^{\mathrm{5}} }{\mathrm{30}}\right]_{{a}} ^{{b}} = \\ $$$$\frac{\pi\left({a}−{b}\right)^{\mathrm{5}} }{\mathrm{30}}\:\:\: \\ $$
Commented by peter frank last updated on 27/Aug/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mahdipoor} \\ $$