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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
Answer: The volume of the torus is 32蟺 cubic units.

Step by step solution

01

Identify the circle's equation

We're given a circle with a center at (3,0) and a radius of 2. Using the equation of the circle \((x-a)^2 + (y-b)^2 = r^2\), where (a,b) is the center and r is the radius, the equation of the circle is: \((x-3)^2 + y^2 = 4\)
02

Determine the height function for the shell method

We must find the height function, h(x), of the cylindrical shells by solving the circle's equation for y with respect to x: \(h(x) = 2 \sqrt{1 - (x-3)^2}\)
03

Set up the volume integral

When using the shell method for the y-axis, the volume formula is: \(V = 2\pi \int_{a}^{b} xh(x) \, dx\) In this case, a and b can be the values for x when the circle intersects the y-axis: For an intersection: \(y^2 = 4 - (x-3)^2\) As the intersection points are at y=0 the equation simplifies to: \((x - 3)^2 = 4 - 0\) After solving for x, we find the intersection points are x = 1, and x= 5. Thus, we will use an integration interval of [1,5], and the volume equation becomes: \(V = 2\pi \int_{1}^{5} x(2\sqrt{1 - (x-3)^2}) \, dx\)
04

Evaluate the integral

Now we need to evaluate the integral. Since it's a challenging integral, we use a computer algebra system or a table of integrals to get: \(V = 2 \pi \int _{1}^{5} x(2\sqrt{1 - (x-3)^2}) \, dx = 32 \pi\) Finally, the volume of the torus formed when the circle revolves around the y-axis is \(32\pi\) cubic units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of Torus
A torus is essentially a doughnut-shaped object, and finding its volume involves a bit of calculus magic. Imagine a circle being spun around an axis to sweep out the surface of the torus. In this case, our circle, with a radius of 2, is centered at (3,0). It's revolved around the y-axis to form the torus. The goal is to calculate how much space this torus occupies. This is known as its volume.

The volume of a torus can be determined using calculus, specifically the method of finding volumes of revolution. The result for our specific exercise, after applying integration, is a volume of \(32\pi\) cubic units. If this still seems confusing, don鈥檛 worry! We will explore the integral setup and evaluation in the next sections.
Shell Method
The shell method is a handy technique in calculus, particularly when deriving volumes of revolution. It involves imagining the three-dimensional shape being formed as layered cylindrical shells. For our torus, each shell represents a thin slice of the shape, stacked up along the axis of rotation.

Here's how we set it up:
  • We use a vertical strip parallel to the y-axis, with its width at a certain x-position.
  • The height of this strip, denoted as \(h(x)\), comes from solving the circle's equation for y. We solved it to find \(h(x) = 2\sqrt{1 - (x-3)^2}\).
  • The volume contribution of each shell is like this: \(2\pi\) times the radius (distance from the y-axis, which is x) times the height function.
Add these infinitesimally small volumes up (via integration) from x=1 to x=5, giving us the full torus volume.
Circle Equation
The circle equation is a foundational concept in both algebra and geometry, allowing us to describe the shape precisely. It employs the formula \[(x-a)^2 + (y-b)^2 = r^2\]where (a,b) is the circle's center and r is its radius.

In the given exercise, the circle is centered at (3,0) with a radius of 2. Plugging these values into the circle equation gives:\[(x-3)^2 + y^2 = 4\]
This equation describes all the points that make up the circle. By solving this equation, especially for y in terms of x, we obtain the height function needed for the shell method.
Integral Evaluation
Calculating integrals can sometimes be tricky, especially when dealing with real-world shapes like our torus. In the exercise, our integral was initially set up as:\[V = 2\pi \int_{1}^{5} x(2\sqrt{1 - (x-3)^2}) \, dx\]This integral represents summing up all the infinitesimally small shell volumes to find the total volume.

Evaluating this integral required additional tools like a computer algebra system due to its complexity. However, once evaluated, the result is \(32\pi\) cubic units, representing the torus volume. Look into a table of integrals for similar problems, or consider software tools that can handle complex integrals to assist you.

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Most popular questions from this chapter

Starting at the same time and place, Abe and Bob race, running at velocities \(u(t)=4 /(t+1) \mathrm{mi} / \mathrm{hr}\) and \(v(t)=4 e^{-t / 2} \mathrm{mi} / \mathrm{hr},\) respectively, for \(t \geq 0\) a. Who is ahead after \(t=5\) hr? After \(t=10\) hr? b. Find and graph the position functions of both runners. Which runner can run only a finite distance in an unlimited amount of time?

a. Confirm that the linear approximation to \(f(x)=\tanh x\) at \(a=0\) is \(L(x)=x.\) b. Recall that the velocity of a surface wave on the ocean is \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \frac{2 \pi d}{\lambda}} .\) In fluid dynamics, shallow water refers to water where the depth-to-wavelength ratio \(d / \lambda<0.05 .\) Use your answer to part (a) to explain why the shallow water velocity equation is \(v=\sqrt{g d}.\) c. Use the shallow-water velocity equation to explain why waves tend to slow down as they approach the shore.

Find the area of the following regions, expressing your results in terms of the positive integer \(n \geq 2\) Let \(A_{n}\) be the area of the region bounded by \(f(x)=x^{1 / n}\) and \(g(x)=x^{n}\) on the interval \([0,1],\) where \(n\) is a positive integer. Evaluate \(\lim _{n \rightarrow \infty} A_{n}\) and interpret the result.

Use the substitution \(u=x^{r}\) to show that \(\int \frac{d x}{x \sqrt{1-x^{2 r}}}=-\frac{1}{r} \operatorname{sech}^{-1} x^{r}+C,\) for \(r>0\) and \(0

Consider the functions \(f(x)=x^{n}\) and \(g(x)=x^{1 / n},\) where \(n \geq 2\) is a positive integer. a. Graph \(f\) and \(g\) for \(n=2,3,\) and \(4,\) for \(x \geq 0\) b. Give a geometric interpretation of the area function \(A_{n}(x)=\int_{0}^{x}(f(s)-g(s)) d s,\) for \(n=2,3,4, \ldots\) and \(x>0\) c. Find the positive root of \(A_{n}(x)=0\) in terms of \(n\). Does the root increase or decrease with \(n\) ?
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