The disk $( x - 4 ) ^ { 2 } + y ^ { 2 } \leq 1$ is revolved about the $y$-axis to generate a torus. Find its volume. (Hint: $\int _ { - 1 } ^ { 1 } \sqrt { 1 - y ^ { 2 } } d y = \frac { 1 } { 2 } \pi$, since it is the area of a semicircle of radius 1 .)
A) $8 \pi ^ { 2 }$
B) $2 \pi ^ { 2 }$
C) $4 \pi ^ { 2 }$
D) $4 \pi$

Question

Quizplus · Accepted Answer

The volume of the torus generated by revolving the disk $(x-4)^2 + y^2 \leq 1$ around the y-axis can be found using the formula for the volume of a solid of revolution, specifically the method of disks or washers. The radius of the circular cross-section of the torus is the distance from the y-axis to any point on the circle, which is $x-4$, and since the circle's equation is given in a form that centers it at $(4,0)$ with a radius of 1, the major radius of the torus (distance from the center of the tube to the center of the torus) is 4. The minor radius (radius of the tube) is 1. The volume of the torus is given by $V = 2\pi^2Rr^2$, where $R$ is the major radius and $r$ is the minor radius. Substituting $R = 4$ and $r = 1$ gives $V = 2\pi^2(4)(1)^2 = 8\pi^2$.

The Disk (x−4)2+y2≤1( x - 4 ) ^ { 2 } + y ^ { 2 } \leq 1(x−4)2+y2≤1

The Disk $( x - 4 ) ^ { 2 } + y ^ { 2 } \leq 1$