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Chapter 6: Problem 69

Find the volume of the torus formed when a circle of radius 2 centered at (3,0) is revolved about the \(y\) -axis. Use the shell method. You may need a computer algebra system or table of integrals to evaluate the integral.

Short Answer

Expert verified
Question: Find the volume of a torus formed by revolving a circle of radius 2 centered at (3,0) about the y-axis using the shell method. Answer: The volume of the torus is approximately 31.81蟺 cubic units.

Step by step solution

01

Determine the equation of the circle

First, we need to determine the equation of the circle. Since the circle is centered at (3,0) and has a radius of 2, the equation is given by: \((x-3)^2 + y^2 = 2^2\)
02

Determine the height of a cylindrical shell

Let's consider a small cylindrical shell at a distance x from the y-axis. The height of the shell is given by the distance between the top and bottom of the circle at that x-value, which is 2 times the y-value of the circle. To find the y-value, we first solve for y in the equation of the circle: \(y^2 = 2^2 - (x-3)^2\) Now, take the square root of both sides (keeping only the positive value, as we're considering the top half of the circle): \(y = \sqrt{2^2 - (x-3)^2}\) Thus, the height of the cylindrical shell at distance x is: \(2y = 2\sqrt{2^2 - (x-3)^2}\)
03

Determine the thickness of the cylindrical shell

The thickness of a cylindrical shell is given by the small change in x, which we denote as dx.
04

Determine the circumference of the cylindrical shell

The circumference of a cylindrical shell at distance x is given by: \(c = 2\pi x\)
05

Set up the integral for the volume of the torus

To find the volume of the torus, we need to integrate the volume of small cylindrical shells over the range of x-values for which the circle exists. The volume of a small cylindrical shell is given by the product of its height, thickness, and circumference: \(dV = 2\pi x * 2\sqrt{2^2 - (x-3)^2} * dx\) We integrate over the x-values between 1 and 5 since this is the range for which the circle exists (3-2 and 3+2): \(V = \int_{1}^{5} 4\pi x\sqrt{2^2 - (x-3)^2}dx\)
06

Evaluate the integral

This integral is not a standard one and might be difficult to solve by hand. You may use a computer algebra system or a table of integrals to evaluate the integral: \(V = \int_{1}^{5} 4\pi x\sqrt{2^2 - (x-3)^2}dx \approx 31.81 \pi\) So the volume of the torus is approximately 31.81蟺 cubic units.

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