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Chapter 16: Problem 16

In Exercises \(1-16,\) find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.) Circular cylinder band The portion of the cylinder \(y^{2}+\) \((z-5)^{2}=25\) between the planes \(x=0\) and \(x=10\)

Short Answer

Expert verified
The parametrization of the cylinder band is \( \mathbf{r}(x, \theta) = (x, 5\sin(\theta), 5 + 5\cos(\theta)) \) with \( 0 \leq x \leq 10 \) and \( 0 \leq \theta < 2\pi \).

Step by step solution

01

Identify the surface equation

The surface described in the problem is a circular cylinder given by the equation \( y^2 + (z - 5)^2 = 25 \). This equation represents a cylinder centered at \( z = 5 \) with a radius of 5, extending along the \( x \)-axis.
02

Define the parametric variables

To parametrize the surface, we'll use \( x \) as a free parameter to cover the range along the \( x \)-axis, and an angle \( \theta \) to cover the circular cross-section of the cylinder. Let \( \theta \) range from \( 0 \) to \( 2\pi \) to complete the circle.
03

Parameterize the y and z coordinates

For the circular cross-section at a fixed \( x \), we can write: \( y = 5 \sin(\theta) \) and \( z = 5 + 5 \cos(\theta) \). These equations utilize trigonometric functions to represent points on the circle centered at \( z = 5 \).
04

Combine everything into a single parametrization

The parametrization of the cylinder surface is given by the vector-valued function: \[ \mathbf{r}(x, \theta) = (x, 5\sin(\theta), 5 + 5\cos(\theta)) \] \(x\) ranges from 0 to 10, and \(\theta\) ranges from 0 to \(2\pi\).
05

Specify the range of parameters

Ensure that the parameters \( x \) and \( \theta \) follow the specified intervals: \( 0 \leq x \leq 10 \) and \( 0 \leq \theta < 2\pi \), which defines the entire band segment of the cylinder between the planes.

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Key Concepts

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

Circular Cylinder
A circular cylinder is a 3-dimensional shape characterized by having two parallel circular bases joined by a curved surface. In mathematics, it is often represented using equations because of its simple geometry.
In the problem we are considering, the circular cylinder is given by the equation \( y^2 + (z-5)^2 = 25 \), which describes a cylinder centered along the \( x \)-axis with a constant radius of 5. If you visualize this, think of a soup can laying on its side with the center running along the \( x \)-axis.
The portion of the cylinder we are interested in is between the planes \( x=0 \) and \( x=10 \). This condition limits the cylinder to a segment, forming a band or strip rather than the entire infinite cylinder.
  • A cylinder can be visualized as the rotation of a rectangle around an axis.
  • The cylinder extends infinitely unless restricted by additional planes or boundaries.
Parametric Equations
Parametric equations are a powerful mathematical tool for describing curves and surfaces by using parameters to express the coordinates of the points. For a cylinder, these equations help us capture the essence of the shape.
In the case of a circular cylinder, two parameters are typically used:
  • \( x \): To account for movement along the length of the cylinder, which is parallel to the \( x \)-axis.
  • \( \theta \): To represent angles that form the circle's perimeter within any given plane of \( x \).
The beauty of parametric equations is that they allow you to express complex curves in terms of simpler, more manageable functions. For the cylinder, we crafted the parametric form using
\( y = 5\sin(\theta) \) and \( z = 5 + 5\cos(\theta) \)
This expression traces out a circle in the planes perpendicular to the \( x \)-axis, centered at \( z = 5 \), as \( \theta \) ranges from \( 0 \) to \( 2\pi \). As \( x \) changes, different cross sections of the cylinder are described.
Trigonometric Functions
Trigonometric functions, like sine and cosine, are key tools in creating parametric equations for cylindrical surfaces. These functions emerge naturally when dealing with circular shapes because they relate angles to circle coordinates.
In our scenario, the cylinder's cross section requires us to use:
  • \( y = 5 \sin(\theta) \): This equation provides the \( y \)-coordinate of a point along the circular cross section.
  • \( z = 5 + 5 \cos(\theta) \): The \( z \)-coordinate centers around \( z = 5 \) and varies as the cosine function dictates.
These functions help accurately map every point along the cylinder's face. By varying \( \theta \) from \( 0 \) to \( 2\pi \), each point on the circle is described. Sine and cosine effectively draw the complete circle because:
  • \( \sin(\theta) \) reaches every vertical position from -5 to 5.
  • \( \cos(\theta) \) covers the entire horizontal span symmetrically around the center.
These trigonometric functions simplify describing points around a circle and are essential for handling circular geometry in three dimensions.

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Most popular questions from this chapter

Find the area of the portion of the paraboloid \(x=4-y^{2}-z^{2}\) that lies above the ring \(1 \leq y^{2}+z^{2} \leq 4\) in the \(y z\) -plane.

If \(S\) is the surface defined by a function \(z=f(x, y)\) that has continuous first partial derivatives throughout a region \(R_{x y}\) in the \(x y\) -plane (Figure \(16.49 ),\) then \(S\) is also the level surface \(F(x, y, z)=0\) of the function \(F(x, y, z)=f(x, y)-z\) . Taking the unit normal to \(R_{x y}\) to be \(\mathbf{p}=\mathbf{k}\) then gives $$ \begin{aligned}|\nabla F|=\left|f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right| &=\sqrt{f_{x}^{2}+f_{y}^{2}+1} \\\|\nabla F \cdot \mathbf{p}| &=\left|\left(f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right) \cdot \mathbf{k}\right|=|-1|=1 \end{aligned} $$ and $$ \iint_{R_{\mathrm{xy}}} \frac{|\nabla F|}{|\nabla F \cdot \mathbf{p}|} d A=\iint_{R_{\mathrm{xy}}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y $$ Similarly, the area of a smooth surface \(x=f(y, z)\) over a region \(R_{y z}\) in the \(y z\) -plane is $$ A=\iint_{R_{y x}} \sqrt{f_{y}^{2}+f_{z}^{2}+1} d y d z $$ and the area of a smooth \(y=f(x, z)\) over a region \(R_{x z}\) in the \(x z\) -plane is $$ A=\iint_{R_{\mathrm{xz}}} \sqrt{f_{x}^{2}+f_{z}^{2}+1} d x d z $$ Use Equations \((11)-(13)\) to find the area of the surfaces in Exercises \(39-44 .\) The surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=3\)

In Exercises \(13-18\) , use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(\mathbf{F}\) across the surface \(S\) in the direction of the outward unit normal \(\mathbf{n} .\) $$ \begin{array}{l}{\mathbf{F}=x^{2} y \mathbf{i}+2 y^{3} z \mathbf{j}+3 z \mathbf{k}} \\ {S : \quad \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+r \mathbf{k}} \\ {0 \leq r \leq 1, \quad 0 \leq \theta \leq 2 \pi}\end{array} $$

Zero curl, yet field not conservative Show that the curl of $$ \mathbf{F}=\frac{-y}{x^{2}+y^{2}} \mathbf{i}+\frac{x}{x^{2}+y^{2}} \mathbf{j}+z \mathbf{k} $$ is zero but that $$ \begin{array}{l}{\oint_{C} \mathbf{F} \cdot d \mathbf{r}} \\ {c}\end{array} $$ is not zero if \(C\) is the circle \(x^{2}+y^{2}=1\) in the \(x y\) -plane. (Theorem 6 does not apply here because the domain of \(\mathbf{F}\) is not simply connected. The field \(\mathbf{F}\) is not defined along the \(z\) -axis so there is no way to contract \(C\) to a point without leaving the domain of \(\mathbf{F} . )\)

In Exercises \(21-26,\) find the flux of the field \(\mathbf{F}\) across the portion of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) in the first octant in the direction away from the origin. $$ \mathbf{F}(x, y, z)=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} $$
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