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 r*ectangular strip representation such that it is perpendicular to the axis of rotation.*

The vertical orientation of...

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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 r*ectangular 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`