find the volume of the region between curves (xy=4 and x+y=5) revolvex around the X axis


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Question Number 91611 by ar247 last updated on 01/May/20

$f i n d t h e v o l u m e o f t h e r e g i o n$ $b e t w e e n c u r v e s (x y = 4 a n d x + y = 5)$ $r e v o l v e x a r o u n d t h e X a x i s$
Commented by john santu last updated on 02/May/20

$v o l = π \int \underset{1}{\overset{4}{}} {(5 - x)}^{2} - {(4 x^{- 1})}^{2} d x$ $= π {[- \frac{1}{3} {(5 - x)}^{3} + \frac{16}{x}]}_{1}^{4}$ $= π [(4 - \frac{1}{3}) - (16 - \frac{64}{3})]$ $= π [\frac{11}{3} + \frac{16}{3}] = 9 π$
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