Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 60

Volume of a shape bounded by curve y=f(x) and x-axis between a leq x leq b revolving about y-axis is given by
V_y=2pi int_a^b xy dx
In order to use the above formula we first need to write y as a function of x.
(x-h)^2+y^2=r^2
y=+-sqrt(r^2-(x-h)^2)
The positive part describes upper half of the circle (blue) while the negative part (red) describes the lower semicircle.

In the graph above r=2 and h=5.
Since the both halves have equal ares the resulting volumes will also be equal for each half. Therefore, we can calculate volume of whole torus as two times the semi-torus (solid obtained by revolving a semicircle).
Bounds of integration will be points where the semicircle touches the x-axis.
4pi int_(h-r)^(h+r)x sqrt(r^2-(x-h)^2)dx=
Substitute x-h=r sin t => x=r sin t+h => dx=r cos t dt, t_l=-pi/2, t_u=pi/2
t_l and t_u denote new lower and upper bounds of integration.
4pi r int_(-pi/2)^(pi/2)(r sin t+h)sqrt(r^2-r^2 sin^2 t)cos t dt=
4pi r int_(-pi/2)^(pi/2)(r sin t+h)r sqrt(1-sin^2 t)cos t dt=
Use the fact that sqrt(1-sin^2 t)=cos t.
4pi r^2 int_(-pi/2)^(pi/2)(r sin t+h)cos^2 t dt=
4pi r^3 int_(-pi/2)^(pi/2) cos^2 t sin t dt+4pi r^2h int_(-pi/2)^(pi/2)cos^2 t dt=
Let us calculate each integral separately
I_1=4pi r^3 int_(-pi/2)^(pi/2) cos^2 t sin t dt=
Substitute u=cos t => du=sin t dt, u_l=0, u_u=0.
4pir^3 int_0^0 u^3/3 du=0
I_2=4pi r^2h int_(-pi/2)^(pi/2)cos^2 t dt=
Rewrite the integral using the following formula cos^2 theta=(1+cos2theta)/2.
2pi r^2h int_(-pi/2)^(pi/2)(1+cos 2t)dt=2pi r^2h(t+1/2sin2t)|_(-pi/2)^(pi/2)=
2pi r^2 h(pi/2+0+pi/2-0)=2pi^2r^2 h
The volume of the torus generated by revolving the given region about y-axis is 2pi^2r^2h.
The image below shows the torus generated by revolving region bounded by circle (x-5)^2+y^2=2^2 i.e. r=2, h=5 about y-axis. The part generated by revolving y=sqrt(r^2-(x-h)^2) is colored blue while the negative part is colored red.