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Question Number 123163 by aurpeyz last updated on 23/Nov/20
the curve y=x^2 +4 is rotated one   revolution about the x−axis between  the limits x=1 and x=4. determine  the volume of the revolution produced
$${the}\:{curve}\:{y}={x}^{\mathrm{2}} +\mathrm{4}\:{is}\:{rotated}\:{one}\: \\ $$$${revolution}\:{about}\:{the}\:{x}−{axis}\:{between} \\ $$$${the}\:{limits}\:{x}=\mathrm{1}\:{and}\:{x}=\mathrm{4}.\:{determine} \\ $$$${the}\:{volume}\:{of}\:{the}\:{revolution}\:{produced} \\ $$
Answered by ajfour last updated on 23/Nov/20
V=∫_1 ^(  4) πy^2 dx =π∫_1 ^( 4) (x^2 +4)^2 dx   = π((x^5 /5)+((8x^3 )/3)+16x)∣_1 ^4     Volume = ((2103π)/5)
$${V}=\int_{\mathrm{1}} ^{\:\:\mathrm{4}} \pi{y}^{\mathrm{2}} {dx}\:=\pi\int_{\mathrm{1}} ^{\:\mathrm{4}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} {dx} \\ $$$$\:=\:\pi\left(\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{16}{x}\right)\mid_{\mathrm{1}} ^{\mathrm{4}} \\ $$$$\:\:{Volume}\:=\:\frac{\mathrm{2103}\pi}{\mathrm{5}} \\ $$

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