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

Question

19778 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-18T22:02:00

#4pi^2#

#(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=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#