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Question Number 47510 by Rio Michael last updated on 11/Nov/18

t h e c u r v e y = 3 x^{2} + 4 (x = 2 a n d x = 3) i s r o t a t e d a b o u t t h e

x . a x i s . f i n d t h e v o l u m e o f t h e s o l i d g e n e r a t e d . L e a v e y o u r

a n s w e r i n t e r m s o f π .

Commented by mr W last updated on 11/Nov/18

V = \int_{x 1}^{x 2} 2 π x \times y \times d x = 2 π \int_{2}^{3} (3 x^{3} + 4 x) d x

= 2 π {[\frac{3 x^{4}}{4} + 2 x^{2}]}_{2}^{3}

= 2 π [\frac{3 (3^{4} - 2^{4})}{4} + 2 (3^{2} - 2^{2})]

= \frac{235 π}{2}

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