Question Number 26162 by lizan 123 last updated on 21/Dec/17
$${Find}\:\:{the}\:{volume}\:\:{of}\:{the}\:{solid}\:{generated}\:{when}\:{the}\:{region}\:\:{enclosed} \\ $$$${by}\:\:{y}=\:\sqrt{{x}} \\ $$$$\:{y}=\mathrm{0}\:\:{and}\:\:{x}=\mathrm{9}\:{about}\:{the}\:\:{line}\:{x}=\mathrm{9} \\ $$
Answered by ajfour last updated on 21/Dec/17
Commented by ajfour last updated on 22/Dec/17
$${Volume}\:=\int_{\mathrm{0}} ^{\:\:\mathrm{3}} \pi\left(\mathrm{9}−{x}\right)^{\mathrm{2}} {dy} \\ $$$$\:\:\:\:\:\:\:\:=\pi\int_{\mathrm{0}} ^{\:\:\mathrm{3}} \left(\mathrm{9}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} {dy} \\ $$$$\:\:\:\:\:\:\:=\pi\left(\mathrm{81}{y}−\frac{\mathrm{18}{y}^{\mathrm{3}} }{\mathrm{3}}+\frac{{y}^{\mathrm{5}} }{\mathrm{5}}\right)\mid_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\pi\left(\mathrm{9}×\mathrm{27}−\mathrm{6}×\mathrm{27}+\frac{\mathrm{9}}{\mathrm{5}}×\mathrm{27}\right) \\ $$$$\:\:\:\:\:=\pi\left(\frac{\mathrm{24}×\mathrm{27}}{\mathrm{5}}\right)\:=\frac{\mathrm{648}\pi}{\mathrm{5}}\:{sq}.\:{units} \\ $$
Commented by mrW1 last updated on 22/Dec/17
![or: V=∫_0 ^9 2π(9−x)ydx=2π∫_0 ^9 (√x)(9−x)dx =2π[6x^(3/2) −(2/5)x^(5/2) ]_0 ^9 =2π[6×9^(3/2) −(2/5)×9^(5/2) ] =2π[((12)/5)]×27 =((648π)/5)](Q26196.png)
$${or}: \\ $$$${V}=\int_{\mathrm{0}} ^{\mathrm{9}} \mathrm{2}\pi\left(\mathrm{9}−{x}\right){ydx}=\mathrm{2}\pi\int_{\mathrm{0}} ^{\mathrm{9}} \sqrt{{x}}\left(\mathrm{9}−{x}\right){dx} \\ $$$$=\mathrm{2}\pi\left[\mathrm{6}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{2}}{\mathrm{5}}{x}^{\frac{\mathrm{5}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{9}} \\ $$$$=\mathrm{2}\pi\left[\mathrm{6}×\mathrm{9}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{2}}{\mathrm{5}}×\mathrm{9}^{\frac{\mathrm{5}}{\mathrm{2}}} \right] \\ $$$$=\mathrm{2}\pi\left[\frac{\mathrm{12}}{\mathrm{5}}\right]×\mathrm{27} \\ $$$$=\frac{\mathrm{648}\pi}{\mathrm{5}} \\ $$