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Chapter 16: Problem 25

Find a parametric representation for the surface. The part of the sphere \(x^{2}+y^{2}+z^{2}=36\) that lies between the planes \(z=0\) and \(z=3 \sqrt{3}\)

Short Answer

Expert verified
Use \(x = 6\sin\theta\cos\phi\), \(y = 6\sin\theta\sin\phi\), \(z = 6\cos\theta\) with \(0 \leq \theta \leq \frac{\pi}{6}\).

Step by step solution

01

Sphere Equation Reformulation

We start with the given equation of the sphere: \( x^2 + y^2 + z^2 = 36 \). We are asked to find the parametric representation of the surface that lies in the sphere and between the planes \(z=0\) and \(z=3\sqrt{3}\). First, we note that the equation is that of a sphere centered at the origin with radius 6.
02

Express Spherical Coordinates

To find a parametric representation, we use spherical coordinates: \( x = ho \sin\theta \cos\phi \), \( y = ho \sin\theta \sin\phi \), and \( z = ho \cos\theta \). For a sphere of radius 6, set \( \rho = 6 \). The angles \( \theta \) and \( \phi \) are angles in the sphere, where \( \theta \) is the polar angle from the positive z-axis and \( \phi \) is the azimuthal angle from the positive x-axis.
03

Setting z-Coordinate Conditions

Given the range \( z = 0 \) to \( z = 3\sqrt{3} \), we need to determine the values of \( \theta \). Since \( z = 6\cos\theta \), set \( 6 \cos \theta = 3 \sqrt{3} \). Solving gives \( \cos \theta = \frac{\sqrt{3}}{2} \), thus \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{6} \).
04

Determine Parametric Equations

The parametric equations are derived from the expressions of spherical coordinates: \( x = 6 \sin \theta \cos \phi \), \( y = 6 \sin \theta \sin \phi \), \( z = 6 \cos \theta \). With \( \theta \) ranging from \( 0 \) to \( \frac{\pi}{6} \), and \( \phi \) ranging from \( 0 \) to \( 2\pi \) representing the full rotation around the z-axis.

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Key Concepts

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

Spherical Coordinates
In three-dimensional space, spherical coordinates are a system that allows us to locate a point with a combination of radius and angular measurements. These coordinates are specifically useful when dealing with spherical shapes.
  • Imagine a point on a sphere. To find this point, we use three values: the distance from the center of the sphere, and two angles.
  • The first angle, \( \theta \), is the polar angle. It measures the inclination from the positive z-axis.
  • The second angle, \( \phi \), is the azimuthal angle 鈥� think of it like measuring around the equator from the positive x-axis.
  • Lastly, the radius \( \rho \) measures the distance from the origin to the point itself.
The conversion from Cartesian coordinates \((x, y, z)\) to spherical coordinates is crucial for solving problems involving spheres. We use these formulas:
  • \( x = \rho \sin\theta \cos\phi \)
  • \( y = \rho \sin\theta \sin\phi \)
  • \( z = \rho \cos\theta \)
This method allows us to simplify complex geometric problems by using angles and rotations instead of distances along axes.
For spheres, it becomes easier to describe and manage points on the surface.
Surface of a Sphere
The surface of a sphere is a perfectly round, symmetrical object in three dimensions, every point on which is equidistant from a fixed central point. This distance is known as the sphere's radius.
  • The general equation of a sphere in Cartesian coordinates is \( x^2 + y^2 + z^2 = r^2 \), where \( r \) is the radius.
  • Any point on the sphere satisfies this equation, ensuring it's located precisely \( r \) units away from the center.
  • In our exercise, the sphere's equation is \( x^2 + y^2 + z^2 = 36 \), giving it a radius of 6, since \( r^2 = 36 \).
By applying conditions on the z-coordinate, specifically being between two planes, we are focusing on a spherical cap.
This is a portion of the sphere characterized by a truncated top or bottom, depending on the plane constraints. It is crucial to understand these conditions when solving problems involving partial surfaces.
Parametric Equations
Parametric equations provide a powerful way to represent curves and surfaces. In the context of spheres, they express the surface using parameters instead of x, y, and z coordinates explicitly.
  • For a sphere with radius 6, a point can be described using parameters \( \theta \) (polar angle) and \( \phi \) (azimuthal angle).
  • This avoids complex algebra since the relationships are already integrated into the equations: \( x = 6 \sin\theta \cos\phi \), \( y = 6 \sin\theta \sin\phi \), \( z = 6 \cos\theta \).
The parameters \( \theta \) and \( \phi \) guide how the surface unfolds in 3D space:
  • \( \theta \) determines the angle relative to the z-axis, and thus the 'height' on the sphere's surface we are considering.
  • \( \phi \) represents the rotation around this vertical axis.
By specifying limits on \( \theta \) (for example, from \( 0 \) to \( \frac{\pi}{6} \)), you define a section of the sphere, such as a spherical cap.
Over the full \( 0 \) to \( 2\pi \) range for \( \phi \), you cover the full rotation, ensuring a complete loop around the sphere.

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Most popular questions from this chapter

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

Find the exact area of the surface \(z=1+2 x+3 y+4 y^{2}\) \(1 \leqslant x \leqslant 4,0 \leqslant y \leqslant 1\)

Find the area of the part of the sphere \(x^{2}+y^{2}+z^{2}=4 z\) that lies inside the paraboloid \(z=x^{2}+y^{2} .\)

Find the area of the surface. The helicoid (or spiral ramp) with vector equation \(\mathbf{r}(u, v)=u \cos v \mathbf{i}+u \sin v \mathbf{j}+v \mathbf{k}, \quad 0 \leqslant u \leqslant 1\) \(0 \leqslant v \leqslant \pi\)

Let \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(r=|\mathbf{r}|\). If \(\mathbf{F}=\mathbf{r} / r^{p},\) find div \(\mathbf{F} .\) Is there a value of \(p\) for which \(\operatorname{div} \mathbf{F}=0 ?\)
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