/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Find a parametrization of the su... [FREE SOLUTION] | 91Ó°ÊÓ

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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.) Spherical cap The upper portion cut from the sphere \(x^{2}+y^{2}+z^{2}=8\) by the plane \(z=-2\)

Short Answer

Expert verified
The parametrization is: \(x(\theta,\phi) = \sqrt{8} \sin \theta \cos \phi\), \(y(\theta,\phi) = \sqrt{8} \sin \theta \sin \phi\), \(z(\theta) = \sqrt{8} \cos \theta\), with \(0 \leq \theta \leq \frac{3\pi}{4}\) and \(0 \leq \phi < 2\pi\).

Step by step solution

01

Understand the Problem

We are given a sphere equation \(x^2 + y^2 + z^2 = 8\) and a plane \(z = -2\). The task is to find a parametrization of the surface which is the upper portion above the plane intersection.
02

Convert to Spherical Coordinates

For a sphere, the corresponding spherical coordinates are given by \(x = \sqrt{8} \sin \theta \cos \phi\), \(y = \sqrt{8} \sin \theta \sin \phi\), and \(z = \sqrt{8} \cos \theta\). Here, \( \theta \) and \( \phi \) are the spherical angles.
03

Determine Limits for Parameters

Since we need the upper portion cut by the plane \(z = -2\), solve \(z = \sqrt{8} \cos \theta = -2\) to find the \(\theta\). We find that \(\cos \theta = -\frac{1}{\sqrt{2}}\), therefore, \(\theta\) ranges from \(0\) to \(\frac{3\pi}{4}\). The parameter \(\phi\) will naturally range from \(0\) to \(2\pi\) because it represents the azimuthal angle going around the z-axis.
04

Write Parametrization

Using the spherical coordinates and parameter limits, the parametrization of the spherical cap is: \(\begin{align*}x(\theta, \phi) &= \sqrt{8} \sin \theta \cos \phi, \y(\theta, \phi) &= \sqrt{8} \sin \theta \sin \phi, \z(\theta) &= \sqrt{8} \cos \theta, \\end{align*}\) where \(0 \leq \theta \leq \frac{3\pi}{4}\) and \(0 \leq \phi < 2\pi\).

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Key Concepts

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

Spherical Coordinates
Spherical coordinates are a way of representing points in three-dimensional space. Instead of using the familiar Cartesian coordinates \(x, y, z\), spherical coordinates use three values: \(r\), \(\theta\), and \(\phi\).
  • \(r\) is the radial distance from the origin to the point.
  • \(\theta\) is the polar angle, measured from the positive z-axis.
  • \(\phi\) is the azimuthal angle, measured from the positive x-axis around the z-axis.
The conversion from spherical to Cartesian coordinates is done as follows: - \(x = r \sin \theta \cos \phi\) - \(y = r \sin \theta \sin \phi\) - \(z = r \cos \theta\) These coordinates are especially useful for problems involving spheres, where symmetry about the origin can greatly simplify calculations. They help in transforming complex three-dimensional problems into more manageable forms, as seen in our problem of parametrizing part of a sphere. Each point on the sphere can be uniquely identified using spherical coordinates, making them a powerful tool for this type of problem.
Sphere Equation
A sphere is a perfectly round three-dimensional geometrical object. The equation of a sphere in Cartesian coordinates is \(x^2 + y^2 + z^2 = r^2\), where \(r\) is the radius of the sphere. In our given exercise, the sphere's equation is \(x^2 + y^2 + z^2 = 8\). From this, we can deduce that the sphere has a radius of \(\sqrt{8}\). This equation represents all points that are exactly \(\sqrt{8}\) units from the center (0, 0, 0) of the sphere. The equation is derived from the distance formula, ensuring every point satisfies this spherical surface condition. In this particular problem, the sphere is intersected by the plane \(z = -2\), highlighting a hemispherical section from the original complete sphere. This intersecting plane effectively "caps" the sphere, showing only the upper part above \(z = -2\), which is what we seek to parametrize using spherical coordinates.
Parametrization of Surfaces
Parametrization is a technique used to express surfaces in terms of parameters. This allows for the transformation of complex surfaces into equations that can be more easily manipulated and understood. For our problem, we're interested in parametrizing a spherical cap, which is the part of the sphere cut off by the plane \(z = -2\). Using spherical coordinates, we can express our point on this surface as a function of two parameters, \(\theta\) and \(\phi\). The parametrization for a surface involves establishing equations for each Cartesian coordinate \((x, y, z)\) in terms of spherical coordinates: - \(x(\theta, \phi) = \sqrt{8} \sin \theta \cos \phi\) - \(y(\theta, \phi) = \sqrt{8} \sin \theta \sin \phi\) - \(z(\theta, \phi) = \sqrt{8} \cos \theta\) These equations account for the shape and size of the surface by relating the spherical angles \(\theta\) and \(\phi\) to the Cartesian coordinates. The limits for \(\theta\) are derived from the fact that \(z\) cannot fall below the plane \(z = -2\), leading to \(0 \leq \theta \leq \frac{3\pi}{4}\). Meanwhile, \(\phi\) spans from 0 to \(2\pi\), covering the full circle around the z-axis. Through parametrization, complex geometric forms can be described in a way that highlights their dimensions and shape straightforwardly.

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Most popular questions from this chapter

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Cylinder } \mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \text { outward through the portion of }} \\ {\text { the cylinder } x^{2}+y^{2}=1 \text { cut by the planes } z=0 \text { and } z=a}\end{array} \)

Find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. \(\begin{array}{l}{\mathbf{F}(x, y, z)=y x^{2} \mathbf{i}-2 \mathbf{j}+x z \mathbf{k}} \\ {\text {S: rectangular surface } y=0, \quad-1 \leq x \leq 2, \quad 2 \leq z \leq 7}, \\ {\text { direction }-\mathbf{j}}\end{array}\)

Work and area Suppose that \(f(t)\) is differentiable and positive for \(a \leq t \leq b .\) Let \(C\) be the path \(\mathbf{r}(t)=t \mathbf{i}+f(t) \mathbf{j}, a \leq t \leq b\) and \(\mathbf{F}=y \mathbf{i} .\) Is there any relation between the value of the work integral $$ \int_{C} \mathbf{F} \cdot d \mathbf{r} $$ and the area of the region bounded by the \(t\) -axis, the graph of \(f\) and the lines \(t=a\) and \(t=b ?\) Give reasons for your answer.

Definite integral as a line integral Suppose that a nonnegative function \(y=f(x)\) has a continuous first derivative on \([a, b] .\) Let \(C\) be the boundary of the region in the \(x y\) -plane that is bounded below by the \(x\) -axis, above by the graph of \(f,\) and on the sides by the lines \(x=a\) and \(x=b .\) Show that $$\int_{a}^{b} f(x) d x=-\oint_{C} y d x$$

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, \boldsymbol{v})=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0}\) . Find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. Parabolic cylinder \(\quad\) The parabolic cylinder surface \(\mathbf{r}(x, y)=\) \(x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty

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