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Question Number 123163 by aurpeyz last updated on 23/Nov/20

t h e c u r v e y = x^{2} + 4 i s r o t a t e d o n e

r e v o l u t i o n a b o u t t h e x - a x i s b e t w e e n

t h e l i m i t s x = 1 a n d x = 4 . d e t e r m i n e

t h e v o l u m e o f t h e r e v o l u t i o n p r o d u c e d

Answered by ajfour last updated on 23/Nov/20

V = \int_{1}^{4} π y^{2} d x = π \int_{1}^{4} {(x^{2} + 4)}^{2} d x

= π (\frac{x^{5}}{5} + \frac{8 x^{3}}{3} + 16 x) ∣_{1}^{4}

V o l u m e = \frac{2103 π}{5}

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