volume of solids generated by revolving the region bounded by the curve and line the y-axis y=x ,y = x/2 ,x = 2


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Question Number 215280 by universe last updated on 02/Jan/25

$v o l u m e o f s o l i d s g e n e r a t e d b y r e v o l v i n g$ $t h e r e g i o n b o u n d e d b y t h e c u r v e a n d l i n e$ $t h e y - a x i s$ $y = x, y = x / 2, x = 2$
Commented by mr W last updated on 02/Jan/25

$r o t a t e d a b o u t w h i c h a x i s ?$
Commented by universe last updated on 02/Jan/25

$y - a x i s$
	Answered by MrGaster last updated on 02/Jan/25

	$V = π \int_{a}^{b} {[f (x)]}^{2} - {[g (x)]}^{2} d x$ $V = π \int_{0}^{2} {[x]}^{2} - {[\frac{π}{2}]}^{2} d x$ $V = π \int_{0}^{2} x^{2} - \frac{x^{2}}{4} d x$ $V = π \int_{0}^{2} \frac{4 x^{2}}{4} - \frac{x^{2}}{4} d x$ $V = π \int_{0}^{2} \frac{3 x^{2}}{4} d x$ $V = π {[\frac{3 x^{3}}{12}]}_{0}^{2}$ $V = π (\frac{3 \cdot 2^{3}}{12} - \frac{3 \cdot 0^{3}}{12})$ $V = π (\frac{3 \cdot 8}{12})$ $V = π (\frac{24}{12})$ $V = 2 π$
	Answered by mr W last updated on 02/Jan/25

	$i f r o t a t e d a b o u t y - a x i s :$ $V = 2 π \times \frac{2 \times 2}{3} \times \frac{2 \times 1}{2} = \frac{8 π}{3}$ $i f r o t a t e d a b o u t x - a x i s :$ $V = 2 π (\frac{2}{3} \times \frac{2 \times 2}{2} - \frac{1}{3} \times \frac{2 \times 1}{2}) = 2 π$
	Commented by universe last updated on 02/Jan/25

	$w i t h o u t i n t e g r a l s i r$ $c a n u e x p l a i n l i t t l e b i t$
	Commented by mr W last updated on 02/Jan/25

	$V_{a b o u t y - a x i s} = \int_{A} 2 π x d A = 2 π x_{S} A$ $V_{a b o u t x - a x i s} = \int_{A} 2 π y d A = 2 π y_{S} A$ $(x_{S}, y_{S}) = c e n t e r o f a r e a$ $s e e a l s o Q 158237$
	Commented by universe last updated on 02/Jan/25

	$t h a n k s$
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