Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 33

The region bounded by y=sin(x) , y =0 , x=0 ,and x=pi revolved about the x-axis is shown on the attached image. We may apply Disk Method wherein we use a rectangular strip representation such that it is perpendicular to the axis of rotation.
The vertical orientation of the rectangular strip shows the thickness of strip =dx.
That will be the basis to use the formula of the Disc method in a form of:
V = int_a^b A(x) dx where A(x) = pir^2 and r =y_(above)-y_(below) .
The r is radius of the disc which is the same as the length of the rectangular strip.
Then, r = sin(x)=0 = sin(x) with boundary values of x from x=0 to x=pi.
The integral will be:
V = int_0^pi (sin(x))^2 dx
V = x/2-(sin(x)cos(x))/2|_0^pi
Using the definite integral formula: int_a^b f(x) dx = F(b) - F(a) , we get:
V =[pi/2-(sin(pi)cos(pi))/2] -[0/2-(sin(0)cos(0))/2]
V = [pi/2-(0*(-1))/2] -[0-(0*1)/2]
V = [pi/2-0]-[0-0]
V = pi/2