Question Number 91611 by ar247 last updated on 01/May/20
$$\:{find}\:{the}\:{volume}\:{of}\:{the}\:{region}\: \\ $$$${between}\:{curves}\:\left({xy}=\mathrm{4}\:{and}\:{x}+{y}=\mathrm{5}\right) \\ $$$${revolvex}\:{around}\:{the}\:{X}\:{axis} \\ $$
Commented by john santu last updated on 02/May/20
![vol = π ∫ _1 ^4 (5−x)^2 −(4x^(−1) )^2 dx = π [ −(1/3)(5−x)^3 +((16)/x) ]_1 ^4 = π [ (4−(1/3))−(16−((64)/3)) ] = π [ ((11)/3)+((16)/3) ] = 9π](https://www.tinkutara.com/question/Q91623.png)
$${vol}\:=\:\pi\:\int\underset{\mathrm{1}} {\overset{\mathrm{4}} {\:}}\left(\mathrm{5}−{x}\right)^{\mathrm{2}} −\left(\mathrm{4}{x}^{−\mathrm{1}} \right)^{\mathrm{2}} \:{dx}\: \\ $$$$=\:\pi\:\left[\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{5}−{x}\right)^{\mathrm{3}} +\frac{\mathrm{16}}{{x}}\:\right]_{\mathrm{1}} ^{\mathrm{4}} \\ $$$$=\:\pi\:\left[\:\left(\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\left(\mathrm{16}−\frac{\mathrm{64}}{\mathrm{3}}\right)\:\right] \\ $$$$=\:\pi\:\left[\:\frac{\mathrm{11}}{\mathrm{3}}+\frac{\mathrm{16}}{\mathrm{3}}\:\right]\:=\:\mathrm{9}\pi \\ $$