How do you find the area enclosed by #y=sin x# and the x-axis for #0≤x≤pi# and the volume of the solid of revolution, when this area is rotated about the x axis?

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Answer 1 · 2016-05-04T11:38:23

Area = 2 areal units.
Volume of the solid of revolution = #pi^2/2# cubic units.

Area = #intydx=intsin x dx#, between the limits #x=0 and x=pi#

= #[- cos x ]#, between the limits

#= [ - cos pi + cos 0 ] = [ 1 + 1 ] = 2# areal units.

Volume = #piint y^2dx= pi int sin^2 x dx#, between the limits #x=0 and x=pi#

#=pi/2int (1-cos 2x)dx#, between the limits #x=0 and x=pi#

#= pi/2[x- (sin 2x)/2 ]#, between the limits

#=pi/2 [ (pi-0)- (0-0) ] = pi^2/2# cubic units/