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

Question

Emily Ammerman · Accepted Answer

#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=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#

Charles Arocha · Answer

undefined To find the volume of the torus formed by revolving the curve $(x-2)^2 + y^2 = 1$ about the y-axis, you can use the formula for the volume of a torus, which is $\pi^2 R r^2$, where $R$ is the distance from the center of the torus to the center of the tube (the distance from the y-axis to the center of the circle), and $r$ is the radius of the tube (the radius of the circle).

First, you need to identify the radii of the torus. The distance from the center of the torus to the center of the circle is $2$ units, so $R = 2$. The radius of the circle forming the torus is $1$ unit (from the equation $(x-2)^2 + y^2 = 1$), so $r = 1$.

Substitute these values into the formula: $\pi^2 \times 2 \times 1^2$.

This simplifies to $2\pi^2$ cubic units. Therefore, the volume of the torus is $2\pi^2$ cubic units.