The x-axis is the axis of rotation so y is the radius of the cylindrical shells, while some function of x is the height of the cylinder. The volume of an infinitesimal shell is 2πyhdy, where h is to be evaluated for each shell. x=sin-1(y/5) and x=y-1. The line and curve meet when 5sin(x)=x+1, and the curve intersects the x-axis at x=0 and x=π.
If the area enclosed by the line and the curve is the region to be rotated then there's a problem, because the solution of 5sin(x)=x+1 (which identifies the x coordinates of the intersections) is not straightforward. The question doesn't make it clear which region is to be rotated or provide any limits.
Assuming that the enclosed region is to be rotated we first solve 5sin(x)=x+1. Newton's iterative method can be used. Let f(x)=x+1-5sin(x), then f'(x)=1-5cos(x).
xn+1=xn-f(xn)/f'(xn) starting with x0=0.
x1=-1/(-4)=¼; x2=0.2533..., x3=0.253378......., x∞=0.253378094206 (approx). We take this as a solution.
Now start with x0=2 and we get eventually x=2.39515169335. This is the other solution.
To find y, we simply add 1 so the intersection points are (0.253378,1.253378) and (2.395152,3.395152).
The limits of integration are y=1.253378 and 3.395152.
Volume V=2π∫y(y-1-sin-1(y/5))dy between the limits above.
More to follow in due course...