Solve the problem.
-The disk $( x - 6 ) ^ { 2 } + y ^ { 2 } \leq 1$ 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) $3 \pi ^ { 2 }$
B) $6 \pi$
C) $12 \pi ^ { 2 }$
D) $6 \pi ^ { 2 }$

Question

Quizplus · Accepted Answer

When the disk $(x-6)^2+y^2\leq 1$ is revolved about the $y$-axis, the result is a torus with radii 1 and 6. We can think of the torus as being made up of many thin disks. Each disk has radius $\sqrt{1-y^2}$ and thickness $dx$, where $x$ is the distance from the center of the torus.

The volume of a single disk is given by the formula for the volume of a cylinder: $$\pi(\text{radius})^2(\text{thickness}).$$ In this case, the radius of the disk is $\sqrt{1-y^2}$, and the thickness is $dx$, so the volume of the disk is $$\pi(\sqrt{1-y^2})^2\cdot dx = \pi(1-y^2)\cdot dx.$$

To find the total volume of the torus, we must add up the volumes of all the disks. Since we are revolving around the $y$-axis, we need to integrate with respect to $y$, not $x$. The limits of integration for $y$ are from $-1$ to 1, since the original disk has its center at $(6,0)$ and radius 1.

Therefore, the total volume of the torus is $$\int_{-1}^1 \pi(1-y^2)\cdot dy = \pi\left[y-\frac{1}{3}y^3\right]_{-1}^1 = \pi\left[(1-\frac{1}{3}) - (-1 + \frac{1}{3})\right]=12\pi.$$

Therefore, the answer is choice C, $12\pi^2$.

Solve the Problem (x−6)2+y2≤1( x - 6 ) ^ { 2 } + y ^ { 2 } \leq 1(x−6)2+y2≤1

Solve the Problem $( x - 6 ) ^ { 2 } + y ^ { 2 } \leq 1$