Problem 59 - Calculus Early Transcendentals

Question

Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the \(y\) -axis. Use geometry to evaluate the integral.

Step-by-Step Solution

Answer

Answer: The volume of the torus is \(V=72\pi^2\).

Step 1: Visualize the torus and the disks

Draw the circle in the x-y plane. This circle can be represented as \((x - 3)^2 + y^2 = 2^2\). Now, imagine revolving the circle around the y-axis to create the torus. The cross-sections of the torus that are perpendicular to the y-axis are all circles (disks). The interval for integration will be from the bottommost part of the torus (y-axis) to the topmost part, which is the diameter of the cross-section.

Step 2: Calculate the radius of each disk

We will find the radius of each disk as a function of y. For any point on the circle, we can write the equation as \((x - 3)^2 + y^2 = 2^2\). Solve for the x-coordinate of the point to find the radius of the disk: \(x = 3 \pm \sqrt{4 - y^2}\). Since we are interested in the distance from x-axis, we only need the positive part: \(x = 3 + \sqrt{4 - y^2}\).

Step 3: Calculate the volume of each disk as a function of y

The volume of a disk is given by the formula \(\pi r^2h\), where r is the radius and h is the height (thickness) of the disk. In this problem, the thickness of each disk is infinitesimally small (dy), so the volume of each disk would be \(dV = \pi r^2 dy\). As we found the radius of the disk in step 2 as a function of y: \(r(y) = 3 + \sqrt{4 - y^2}\), the volume of the disk as a function of y can be written as: \(dV = \pi (3 + \sqrt{4 - y^2})^2 dy\).

Step 4: Set up the integral for the volume of the torus

To find the total volume of the torus, we need to integrate the volume of the individual disks over the y-coordinate from the lower limit of -2 (the bottom of the torus) to the upper limit of 2 (the top of the torus). So, the integral for the total volume becomes: \(V = \int_{-2}^{2} \pi (3 + \sqrt{4 - y^2})^2 dy\).

Step 5: Use geometry to evaluate the integral

We know that the volume of a torus can be found by using the formula \(V = 2\pi^2 Rr^2\), where R is the distance from the center of the torus to the center of the tube, and r is the radius of the tube. In our case, R = 3 (distance from y-axis to the center of the circle) and r = 2 (radius of the circle). So, \(V = 2\pi^2 (3)(2^2) = 72\pi^2\). This is the volume of the torus formed by revolving the circle of radius 2 centered at (3,0) around the y-axis. So, the final answer is \(V=72\pi^2\).