Question Number 192396 by moh777 last updated on 16/May/23
Answered by mehdee42 last updated on 16/May/23
![4x−x^2 =3⇒x=1,3 v_1 =π∫_0 ^1 (4x−x^2 )^2 dx=π∫_0 ^1 (16x^2 −8x^3 +x^4 )dx=π[((16)/3)x^3 −2x^4 +(x^5 /5)]_0 ^1 v_2 =18π v_3 =π∫_3 ^4 (4x−x^2 )^2 dx=π[((16)/3)x^3 −2x^4 +(x^5 /5)]_3 ^4 v=v_1 +v_2 +v_3](https://www.tinkutara.com/question/Q192400.png)
$$\mathrm{4}{x}−{x}^{\mathrm{2}} =\mathrm{3}\Rightarrow{x}=\mathrm{1},\mathrm{3} \\ $$$${v}_{\mathrm{1}} =\pi\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{4}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} {dx}=\pi\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{16}{x}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{3}} +{x}^{\mathrm{4}} \right){dx}=\pi\left[\frac{\mathrm{16}}{\mathrm{3}}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{4}} +\frac{{x}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${v}_{\mathrm{2}} =\mathrm{18}\pi \\ $$$${v}_{\mathrm{3}} =\pi\int_{\mathrm{3}} ^{\mathrm{4}} \left(\mathrm{4}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} {dx}=\pi\left[\frac{\mathrm{16}}{\mathrm{3}}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{4}} +\frac{{x}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{3}} ^{\mathrm{4}} \\ $$$${v}={v}_{\mathrm{1}} +{v}_{\mathrm{2}} +{v}_{\mathrm{3}} \\ $$
Commented by mehdee42 last updated on 17/May/23
