Solids of revolution question I am trying to find the volume of the solid below about line C

$$y=\sin x$$
$$0\le x\le\pi$$
$$y=c$$
$$0\le c\le1$$
I divided the regions into $3$ areas. Top, left and right.
Finding volume of top
$$\text{Top's Volume}=\pi\int_0^{\pi} (\sin x-c)^2 dx$$
Since left and right region are the same, i will just find one of them and multiply by two later.
$$\text{Left's Volume}=2\pi\int_0^c (y)(\sin^{-1}y) dy$$
Are my steps correct so far?

We take advantage of the symmetry, finding the volume from $0$ to $\frac{\pi}{2}$, and then doubling. 
The geometry looks different from $x=0$ to $x=\arcsin(c)$ than from $\arcsin(c)$ to $\frac{\pi}{2}$. We use slicing.  
From $0$ to $\arcsin(c)$, the radius of cross-section is $c-\sin x$, so the volume generated is 
$$\int_0^{\arcsin(c)}\pi (c-\sin x)^2\,dx.$$
From $\arcsin(c)$ to $\frac{\pi}{2}$, the radius of cross-section is $\sin x-c$, so the volume generated is 
$$\int_{\arcsin(c)}^{\\pi/2}\pi (\sin x-c)^2\,dx.$$
But note that the integrands for both parts are the same, because of the squaring. We might as well double at this stage. Our full volume is 
$$2\int_0^{\pi/2}\pi (\sin x-c)^2\,dx.$$
The rest is routine. Expand $(\sin x-c)^2$ and integrate. 
