How do you find the volume of the torus formed by revolving (x-2)^2 +y^2=1 about the y-axis?

1 Answer
Oct 18, 2015

4pi^2

Explanation:

(x-2)^2+y^2=1 is a circle which center is C(2,0) and radius r=1.

V=pi int_(y_1)^(y_2) (R^2-r^2)dy

Obvious, y_1=-1, y_2=1.

(x-2)^2+y^2=1 => x=2+sqrt(1-y^2) vv x=2-sqrt(1-y^2)

R=2+sqrt(1-y^2)
r=2-sqrt(1-y^2)

V=pi int_(-1)^1 ((2+sqrt(1-y^2))^2-(2-sqrt(1-y^2))^2)dy

V=pi int_(-1)^1 ((4+4sqrt(1-y^2)+1-y^2)-(4-4sqrt(1-y^2)+1-y^2)dy

V=pi int_(-1)^1 (4+4sqrt(1-y^2)+1-y^2-4+4sqrt(1-y^2)-1+y^2)dy

V=pi int_(-1)^1 (4+4sqrt(1-y^2)+1-y^2-4+4sqrt(1-y^2)-1+y^2)dy

V=8pi int_(-1)^1 sqrt(1-y^2)dy

y=sint => dy=costdt
t_1 = arcsin(-1)=-pi/2
t_2 = arcsin1=pi/2
sqrt(1-y^2)=sqrt(1-sin^2t)=sqrt(cos^2t)=cost

V=8pi int_(-pi/2)^(pi/2) cos^2tdt =4pi int_(-pi/2)^(pi/2) (1+cos2t)dt

V=4pi (t+1/2sin2t)|_(-pi/2)^(pi/2)= 4pi * pi = 4pi^2