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Chapter 18: Chapter 18

Problem 532

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a) Define a surface patch surrounding the point \((0,0,1)\) on the sphere \(x^{2}+y^{2}+z^{2}=1\) b) Show that the paraboloid \(z=x^{2}+y^{2}\) is a surface patch.

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a) A surface patch surrounding the point \((0,0,1)\) on the sphere \(x^2+y^2+z^2=1\) is given by: $$ \begin{cases} x = \sin\theta \cos\phi \\ y = \sin\theta \sin\phi \\ z = \cos\theta \end{cases} , \quad 0 \leq \theta \leq \epsilon, \quad 0 \leq \phi \leq 2\pi, $$ where \(\epsilon\) is a small positive number. b) The paraboloid \(z=x^2+y^2\) is a surface patch with parameterization: $$ \begin{cases} x = u \\ y = v \\ z = u^2 + v^2 \end{cases} , \quad a \leq u \leq b, \quad c \leq v \leq d. $$ where \(a, b, c, d\) are constants.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Understand the concept of a surface patch

A surface patch is a parameterization of a part of a surface in terms of two parameters, usually written as \(u\) and \(v\). In other words, we want to express the coordinates \((x, y, z)\) in terms of \((u, v)\).

Step 2: Define a surface patch for the sphere

We can use spherical coordinates to parameterize the surface of a sphere. The spherical coordinates are given by \((r, \theta, \phi)\), where \(r\) is the radius, \(\theta\) is the polar angle, and \(\phi\) is the azimuthal angle. For a sphere of radius 1, the parameterization can be written as: $$ x = \sin\theta \cos\phi \\ y = \sin\theta \sin\phi \\ z = \cos\theta $$ Since we want to surround the point \((0,0,1)\), we know that \(\theta = 0\), so we can focus on a small neighborhood around this point.

Step 3: Find a suitable range for the parameters

As \(\theta\) is close to \(0\) and \(\phi\) ranges from \(0\) to \(2\pi\), we can choose a small range around \(\theta = 0\) to define the surface patch. For example, let's choose the range: $$ 0 \leq \theta \leq \epsilon \\ 0 \leq \phi \leq 2\pi, $$ where \(\epsilon\) is a small positive number. Now, we have defined a surface patch surrounding the point \((0,0,1)\) on the sphere as: $$ \begin{cases} x = \sin\theta \cos\phi \\ y = \sin\theta \sin\phi \\ z = \cos\theta \end{cases} , \quad 0 \leq \theta \leq \epsilon, \quad 0 \leq \phi \leq 2\pi. $$ #Part b#

Step 1: Choose the parameterization

In this case, we want to show that the paraboloid given by \(z=x^2+y^2\) is a surface patch. We can simply denote \(x=u\) and \(y=v\) and rewrite the equation as: $$ z = u^2 + v^2. $$

Step 2: Define the range for our parameters

In order for the paraboloid to be considered a surface patch, we need to choose a suitable range for the parameters \((u, v)\). Let’s choose an arbitrary range: $$ a \leq u \leq b \\ c \leq v \leq d, $$ where \(a, b, c, d\) are constants.

Step 3: Present the final parameterization

Now, we have parameterized the paraboloid as: $$ \begin{cases} x = u \\ y = v \\ z = u^2 + v^2 \end{cases} , \quad a \leq u \leq b, \quad c \leq v \leq d. $$ This parameterization shows that the paraboloid \(z = x^2 + y^2\) is indeed a surface patch.

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Most popular questions from this chapter

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Compute the coefficients of the first fundamental form in each of the following cases. (1) the plane \(z=0\) (ii) the cylinder \(\mathrm{x}^{2}+\mathrm{y}^{2}=1\) (iii) the sphere \(\mathrm{x}^{2}+\mathrm{y}^{2}+z^{2}=1\).
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Let \(\mathrm{d} s^{2}=\mathrm{E} \mathrm{du}^{2}+2 \mathrm{~F}\) dudv \(+\mathrm{G} \mathrm{dv}^{2}\) be the first fundamental form of a surface patch. Show that the family of curves orthogonal to the family defined by \(\mathrm{M} \mathrm{du}+\mathrm{N} \mathrm{dv}=0\) is determined by $$ (E N-F M) d u+(F N-G M) d v=0 $$
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