Find the volume of the solid if the cross sections are squares perpendicular to the x-axis

The region is bounded by y= sin(x) and y= 0. Set up in the integral, and then compute the volume.

I'm assuming that you only want the first bump of the sin(x) function, because otherwise the volume would be infinity. Thus we know that the limits on x are extending from 0 to π.

V = ∫ A(x) dx from 0 to π

Now, the cross sections are squares, so we just need to find the length of one side in the xy plane and then square it to get the area. The length of a side in the xy plane is just sin(x) though, so:

A(x) = sin^2(x)

V = ∫ sin^2(x) dx from 0 to π

Now apply the identity sin^2(x) = (1/2)(1 - cos(2x))

V = (1/2) ∫ (1 - cos(2x)) dx
V = (1/2)[x - sin(2x)]

Evaluate the bounds and note that the lower limit of 0 will be 0:

V = (1/2)[{π} - sin(2{π})]
V = π / 2
V ≈ 1.5707

Done!

Not too sure