Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 90446 by jagoll last updated on 23/Apr/20

find the volume of the solid formed by rotating the area trapped by y = sin x and the x−axis around the line y=3 for 0<x<π

$$\mathrm{find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid} \\ $$ $$\mathrm{formed}\:\mathrm{by}\:\mathrm{rotating}\:\mathrm{the}\:\mathrm{area} \\ $$ $$\mathrm{trapped}\:\mathrm{by}\:\mathrm{y}\:=\:\mathrm{sin}\:\mathrm{x}\:\mathrm{and}\:\mathrm{the} \\ $$ $$\mathrm{x}−\mathrm{axis}\:\mathrm{around}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{3} \\ $$ $$\mathrm{for}\:\mathrm{0}<\mathrm{x}<\pi\: \\ $$

Commented byjohn santu last updated on 23/Apr/20

vol = π ∫_0 ^π V(x)^2 −v(x)^2 dx = π∫_0 ^π 3^2 −(3−sin x)^2 dx = π∫_0 ^π 6sin x−sin^2 x dx = π∫_0 ^π (6sin x−(1/2)+(1/2)cos 2x) dx now easy to solve

$${vol}\:=\:\pi\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:{V}\left({x}\right)^{\mathrm{2}} −{v}\left({x}\right)^{\mathrm{2}} \:{dx}\: \\ $$ $$=\:\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\mathrm{3}^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} \:{dx} \\ $$ $$=\:\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\mathrm{6sin}\:{x}−\mathrm{sin}\:^{\mathrm{2}} {x}\:{dx} \\ $$ $$=\:\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\left(\mathrm{6sin}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right)\:{dx} \\ $$ $${now}\:{easy}\:{to}\:{solve} \\ $$

Commented byjagoll last updated on 23/Apr/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com