What is the volume of the solid generated by rotating about the y-axis the region enclosed by y=sinx and the x-axis, from x=0 to x=π?
Thank you for your help!!
Volume = π[integral] (sin x)^2 dx from 0 to π
The tricky part is to integrate sin^2 x
start with
cos 2x = 1 - 2sin^2 x
sin^2 x = 1/2 - (1/2)cos 2x
= (1/2) (1 - cos 2x)
the integral of that would be (1/2) (x - (1/2)sin 2x)
I am sure you can take it from there
( I got π/2)
To find the volume of the solid generated by rotating the region enclosed by the curve y = sin(x) and the x-axis about the y-axis, you can use the disk method.
The disk method involves integrating the cross-sections of the solid, which are essentially infinite cylinders of infinitesimal thickness. Each cylinder has a radius equal to the distance from the y-axis to the curve y = sin(x), and the height of each cylinder is dx, which represents the thickness of the slice being rotated.
To find the radius of each cylinder, we need to express the curve y = sin(x) in terms of x. In this case, it is already in terms of x, so the radius of each disk is simply sin(x).
Now, we can set up the integral to find the volume:
V = ∫[a,b] π*(sin(x))^2 dx
Where [a, b] represents the interval of x-values over which the region is being rotated. In this case, the interval is from x = 0 to x = π, so the integral becomes:
V = ∫[0, π] π*(sin(x))^2 dx
Now, you can evaluate this definite integral to find the volume of the solid generated by the rotation.