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Chapter 6: Problem 38

The circle \(x=a \cos t, y=a \sin t, 0 \leq t \leq 2 \pi\), is revolved about the line \(x=b, 0

Short Answer

Expert verified
The surface area is \(4\pi^2 ab\).

Step by step solution

01

Identify the parameters

The circle is defined parametrically by the equations \(x = a \cos t\) and \(y = a \sin t\), where \(0 \leq t \leq 2\pi\). It is revolved around the line \(x = b\) to form a torus. Here, \(a\) is the radius of the circle, and \(b\) is the distance from the center of the circle to the line of revolution. Given \(0 < a < b\), the circle lies strictly to the left of the line \(x = b\).
02

Formula for the Surface Area of a Torus

The surface area \(A\) of a torus formed by revolving a circle of radius \(a\) around a line \(b\) units from the center of the circle, is given by: \[ A = 4\pi^2 ab \] This formula is derived from considering the circle tracing a circular path of length \(2\pi b\) through the revolution.
03

Substitute the Parameters

Substitute \(a\) and \(b\) into the formula:\[ A = 4\pi^2 ab \] This represents substituting into the known formula as all circles with specified parameters will generate a torus with this area.

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Key Concepts

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

Parametric Equations
Parametric equations are a powerful tool in mathematics, often used to define geometric figures using parameters. In the given exercise, a circle is described by the parametric equations:
  • \(x = a \cos t\)
  • \(y = a \sin t\)
The parameter \(t\) varies from 0 to \(2\pi\) and helps plot every point on the circle as \(t\) changes.
This representation separates complex shapes into simple components, encoding geometry through equations rather than relying solely on Cartesian coordinates. Parametric equations give us flexibility, allowing us to represent curves that are difficult to express with standard equation forms.
Circle Revolution
Revolving a geometric figure around a line in 3D space is fascinating, and it helps in understanding real-world structures. In this exercise, the circle is revolved around the line \(x = b\). This process is known as circle revolution.
When revolved, every point on the circle sweeps a circular path around this line, resulting in a toroid shape. It's akin to holding a hoop and spinning it about a central line.
Key points to note:
  • The line of revolution (\(x = b\)) does not intersect the circle, as \(a < b\).
  • The entire circle traces a circular path in space, forming what we call a torus.
Understanding circle revolution helps visualize and compute areas and volumes of 3D shapes.
Torus Geometry
A torus, often seen in doughnut shapes, is a surface of revolution. Its geometry can be challenging but intriguing. Here, the key is to maintain a constant equidistant path from the axis of revolution.
Characteristics of a torus like the one in this exercise:
  • Two radii: the radius of the original circle \(a\), and the distance \(b\) from the circle's center to the line of revolution.
  • Its surface can be imagined as the continuous circular sweep of the initial circle in a plane, tracking around an axis at a fixed range.
These properties help in visualizing and performing calculations on toroidal structures, revealing the aesthetic and mathematical unity in its shape.
Calculus Applications
Calculus plays a crucial role in calculating areas of shapes like the torus. Here, we're focusing on finding the surface area, which is derived using integral calculus concepts.
The provided formula for the torus's surface area, \(A = 4\pi^2 ab\), is obtained through a process related to wrapping the circular path along its own axis.
Why calculus?
  • Integrals help calculate total areas and volumes where simple arithmetic fails due to curvature or non-linear paths.
  • The use of \(2\pi\) in the formula reflects the circular path's involvement鈥攖he circumference of the path that the circle follows during revolution.
Through calculus, we can navigate complex surfaces and derive meaningful results, vital for both theoretical and applied sciences.

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Most popular questions from this chapter

The expected value of a function \(g(X)\) of a continuous random variable \(X\) having PDF \(f(x)\) is defined to be \(E[g(X)]=\) \(\int_{A}^{B} g(x) f(x) d x\). If the PDF of \(X\) is \(f(x)=\frac{15}{512} x^{2}(4-x)^{2}\), \(0 \leq x \leq 4\), find \(E(X)\) and \(E\left(X^{2}\right)\).

The time in minutes that it takes a worker to complete a task is a random variable with PDF \(f(x)=k(2-|x-2|)\), \(0 \leq x \leq 4\). (a) Find the value of \(k\) that makes this a valid PDF. (b) What is the probability that it takes more than 3 minutes to complete the task? (c) Find the expected value of the time to complete the task. (d) Find the CDF \(F(x)\). (e) Let \(Y\) denote the time in seconds required to complete the task. What is the CDF of \(Y\) ? Hint: \(P(Y \leq y)=\) \(P(60 X \leq y)\)

Find the area of the region trapped between \(y=x e^{-x^{2}}\) and \(y=x / 4\). Hint: There are two separate regions.

In Problems 1-12, find the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(x=\sqrt{2 y}+1, y=2, x=0, y=0 ;\) about the line \(y=3\)

In Problems 11-30, sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer: \(y=e^{2 x}, y=0\), between \(x=0\) and \(x=\ln 2\)
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