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Problem 532

# 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.

Expert verified
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.
See the step by step solution

## 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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