Consider a circle with radius [tex]r[/tex] centered at some point [tex](R+r,0)[/tex] on the [tex]x[/tex]-axis. This circle has equation
[tex](x-(R+r))^2+y^2=r^2[/tex]
Revolve the region bounded by this circle across the [tex]y[/tex]-axis to get a torus. Using the shell method, the volume of the resulting torus is
[tex]\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx[/tex]
where [tex]2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}[/tex].
So the volume is
[tex]\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx[/tex]
Substitute
[tex]x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt[/tex]
and the integral becomes
[tex]\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt[/tex]
Notice that [tex]\sin t\cos^2t[/tex] is an odd function, so the integral over [tex]\left[-\frac\pi2,\frac\pi2\right][/tex] is 0. This leaves us with
[tex]\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt[/tex]
Write
[tex]\cos^2t=\dfrac{1+\cos(2t)}2[/tex]
so the volume is
[tex]\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}[/tex]
