what is the volume of region bounded by y =x^2 −2x and y=x that is rotated about y=4 ?


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Question Number 102606 by bemath last updated on 10/Jul/20

$w h a t i s t h e v o l u m e o f r e g i o n$ $b o u n d e d b y y = x^{2} - 2 x a n d$ $y = x t h a t i s r o t a t e d a b o u t$ $y = 4 ?$
Commented by bemath last updated on 10/Jul/20

	Answered by Ar Brandon last updated on 10/Jul/20
	$Volume , V=π∫_0 ^3 {(x^2 −2x−4)^2 −(x−4)^2 }dx$
	$Volume, V = π \int_{0}^{3} {{(x^{2} - 2 x - 4)}^{2} - {(x - 4)}^{2}} dx$
	Answered by bobhans last updated on 10/Jul/20
	$vol = π∫_0 ^3 (4^2 −(x^2 −2x)^2 )−(4^2 −x^2 ) dx = π∫_0 ^3 (x^2 −(x^2 −2x)^2 dx = π∫_0 ^3 (x^2 +x^2 −2x)(2x)dx =4π∫_0 ^3 x(x^2 −x)dx = 4π∫_0 ^3 (x^3 −x^2 ) dx = 4π {(1/4)x^4 −(1/3)x^3 }_0 ^3 =4π {((81)/4)−((27)/3)} =4π {((45)/4)} = 45π$
	$v o l = π {\int^{3}}_{0} (4^{2} - {(x^{2} - 2 x)}^{2}) - (4^{2} - x^{2}) d x$ $= π \underset{0}{\int^{3}} (x^{2} - {(x^{2} - 2 x)}^{2} d x$ $= π \underset{0}{\int^{3}} (x^{2} + x^{2} - 2 x) (2 x) d x$ $= 4 π \underset{0}{\int^{3}} x (x^{2} - x) d x = 4 π \underset{0}{\int^{3}} (x^{3} - x^{2}) d x$ $= 4 π {\frac{1}{4} x^{4} - \frac{1}{3} x^{3}}_{0}^{3} = 4 π {\frac{81}{4} - \frac{27}{3}}$ $= 4 π {\frac{45}{4}} = 45 π$
	Commented by Ar Brandon last updated on 10/Jul/20
	What difference will it make if it was rather rotated about y=0 with respect to your solution ? ��
	Commented by bemath last updated on 10/Jul/20
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