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?

1 Answer
May 4, 2016

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

Explanation:

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/