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Chapter 16: Problem 30

Find parametric equations for the surface obtained by rotating the curve \(x=1 / y, y \geqslant 1,\) about the \(y\) -axis and use them to graph the surface.

Short Answer

Expert verified
Use circle coordinates \((\frac{1}{y} \cos \theta, \frac{1}{y} \sin \theta, y)\) with \(\theta \in [0, 2\pi]\) and \(y \ge 1\).

Step by step solution

01

Understanding the Problem

We need to describe a surface generated by rotating the curve \(x = \frac{1}{y}\) for \(y \geq 1\) about the \(y\)-axis. To do this, we'll find parametric equations that represent this surface.
02

Set Up the Curve in Parametric Form

Consider the curve \(x = \frac{1}{y}\). Let's express \(x\) as a function of \(y\), i.e., \(f(y) = \frac{1}{y}\). This will help us set up our parametric representation.
03

Parameterize the Curve for Rotation

To rotate this curve about the \(y\)-axis, we introduce a parameter \(\theta\) for the angle of rotation. The circle at each height \(y\) will have coordinates \((x \cos \theta, x \sin \theta, y)\).
04

Write the Parametric Equations

Using the parameter \(\theta\), we write the parametric equations for the surface as follows:\[\begin{aligned}x &= \frac{1}{y} \cos \theta, \z &= \frac{1}{y} \sin \theta, \y &= y.\end{aligned}\]Here, \(\theta\) ranges from \(0\) to \(2\pi\) and \(y\) ranges from 1 to infinity.
05

Adjust for Graphing

To graph the surface, recognize that \(y\) must start from 1 (since \(y \ge 1\)). The ranges for \(y\) and \(\theta\) will affect your plot. In practice, \(y\) can be sampled up to a practical value, such as 3 or 5, to visualize the surface.

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Key Concepts

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

Surface of Revolution
The concept of a **surface of revolution** is intriguing. It involves creating a 3D shape from a 2D curve. Simplicity emerges through the elegance of rotation. In essence, you take a curve and spin it around a line (called the axis of rotation). When you do this, you create a "surface" that looks like a smooth, continuous shell. The line around which you rotate your curve significantly impacts the resulting shape. In this exercise, the curve defined by the equation \( x = \frac{1}{y} \) gets rotated around the \( y \)-axis. This axis acts as the spine of the generated surface, maintaining its structural alignment and symmetry.This process creates many familiar objects. For example, rotating a straight line around an axis can create a cone. Hence, surfaces of revolution occupy an important spot in our understanding of geometric shapes.
Curve Rotation
**Curve rotation** gives rise to complex surfaces simply and beautifully. This process involves rotating a curve around a given axis which, in our case, is the \( y \)-axis. To obtain a parametric representation of the surface resulting from this rotation, we use polar coordinates. Here, for each point on the original curve defined by \( x = \frac{1}{y} \), we use an angle parameter \( \theta \) that defines a rotation about the axis.
  • First, choose a constant \( y \).
  • Second, introduce a parameter \( \theta \), which typically ranges from \( 0 \) to \( 2\pi \).
  • This rotation gives coordinates \( (x \cos \theta, x \sin \theta, y) \), effectively wrapping the original curve into a surface.
By understanding this process of rotation, you gain insight into how complex 3D objects are built from simple 2D curves.
Graphing Surfaces
The step of **graphing surfaces** derived from parametric equations sparks creativity and visualization. It's a transition from theoretical mathematics to visual representation, allowing us to "see" what we've conceptualized.In our example, the surface emerges from parametric equations:\[ x = \frac{1}{y} \cos \theta, \quad z = \frac{1}{y} \sin \theta, \quad y = y. \]Here, visualization involves both the range of \( y \) and the angle \( \theta \) to ensure a full rotational appearance. Typically, \( \theta \) spans from \(0\) to \(2\pi\), the complete circle, creating a surface around the \( y \)-axis. The key to practical graphing lies in your choice for \( y \) values.
  • Starting at \( y = 1 \), move upwards to a reasonable point (like \( y = 5 \)) to depict a manageable part of the surface.
  • This controlled range allows better detail without computational overload.
By considering such factors, graphing surfaces can turn abstract mathematical equations into intuitive visuals, fostering deeper understanding.

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Most popular questions from this chapter

Determine whether or not the vector field is conservative. If it is conservative, find a function \(f\) such that \(\mathbf{F}=\nabla f\). $$ \mathbf{F}(x, y, z)=z \cos y \mathbf{i}+x z \sin y \mathbf{j}+x \cos y \mathbf{k} $$

The base of a circular fence with radius \(10 \mathrm{m}\) is given by \(x=10 \mathrm{cos} t, y=10 \mathrm{sin} t .\) The height of the fence at position \((x, y)\) is given by the function \(h(x, y)=4+0.01\left(x^{2}-y^{2}\right),\) so the height varies from \(3 \mathrm{m}\) to \(5 \mathrm{m}\). Suppose that \(1 \mathrm{L}\) of paint covers \(100 \mathrm{m}^{2} .\) Sketch the fence and determine how much paint you will need if you paint both sides of the fence.

Let \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(r=|\mathbf{r}|\). Verify each identity. $$\begin{array}{ll}{\text { (a) } \nabla \cdot \mathbf{r}=3} & {\text { (b) } \nabla \cdot(r \mathbf{r})=4 r} \\ {\text { (c) } \nabla^{2} r^{3}=12 r}\end{array}$$

Find the area of the surface with vector equation \(\mathbf{r}(u, v)=\left\langle\cos ^{3} u \cos ^{3} v, \sin ^{3} u \cos ^{3} v, \sin ^{3} v\right\rangle, 0 \leqslant u \leqslant \pi\) \(0 \leqslant v \leqslant 2 \pi .\) State your answer correct to four decimal places.

Determine whether the points \(P\) and \(Q\) lie on the given surface. $$ \begin{array}{l}{\mathbf{r}(u, v)=\langle u+v, u-2 v, 3+u-v\rangle} \\ { P(4,-5,1), Q(0,4,6)}\end{array} $$
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