How do you find the area enclosed by $y = sin x$ $y=sin x$ and the x-axis for $0 \leq x \leq π$ $0≤x≤pi$ 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 = $\frac{π^{2}}{2}$ $pi^2/2$ cubic units.

Area = $\int y d x = \int sin x d x$ $intydx=intsin x dx$ , between the limits $x = 0 and x = π$ $x=0 and x=pi$

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

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

Volume = $π \int y^{2} d x = π \int {sin}^{2} x d x$ $piint y^2dx= pi int sin^2 x dx$ , between the limits $x = 0 and x = π$ $x=0 and x=pi$

$= \frac{π}{2} \int (1 - cos 2 x) d x$ $=pi/2int (1-cos 2x)dx$ , between the limits $x = 0 and x = π$ $x=0 and x=pi$

$= \frac{π}{2} [x - \frac{sin 2 x}{2}]$ $= pi/2[x- (sin 2x)/2 ]$ , between the limits

$= \frac{π}{2} [(π - 0) - (0 - 0)] = \frac{π^{2}}{2}$ $=pi/2 [ (pi-0)- (0-0) ] = pi^2/2$ cubic units/

How do you find the area enclosed by y=sin xy=sinxy=sin x and the x-axis for 0≤x≤pi0≤x≤π0≤x≤pi and the volume of the solid of revolution, when this area is rotated about the x axis?