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Chapter 16: Problem 9

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.) $$ \begin{array}{l}{\text { Parabolic cylinder between planes } \text { planes } \text { The surface cut from the }} \\ {\text { parabolic cylinder } z=4-y^{2} \text { by the planes } x=0, x=2, \text { and }} \\\ {z=0}\end{array} $$

Short Answer

Expert verified
Parametrize surface: \( x(u,v)=u, y(u,v)=v, z(u,v)=4-v^2 \), for \(0\leq u\leq 2\), \(-2\leq v\leq 2\).

Step by step solution

01

Understand the Problem

We need to parametrize the surface of a parabolic cylinder defined by \( z = 4 - y^2 \) between the planes \( x = 0 \), \( x = 2 \), and \( z = 0 \). A parametrization essentially means expressing the coordinates of points on the surface using parameters.
02

Identify the Variables of the Surface

The parabolic cylinder is expressed as \( z = 4 - y^2 \). Our task is to express this in a parametric form. The constraints of \( x \) and \( z \) will aid in setting the bounds for our parameters.
03

Choose Parameters

We choose two parameters that describe the surface. Let \( x \) be parameterized by \( u \), such that \( x = u \) where \( 0 \leq u \leq 2 \). Let \( y \) be parameterized by \( v \), where \( v \) varies along the parabola. Since this is a cylinder about the y-axis, \( v \) can range freely.
04

Express z in terms of Parameter v

The equation of the surface is given by \( z = 4 - y^2 \). Setting \( y = v \), the parametric form of \( z \) becomes \( z = 4 - v^2 \). Also, since the surface is bounded by \( z = 0 \), we have \( 4 - v^2 \geq 0 \).
05

Determine Bounds on Parameters

From the inequality \( 4 - v^2 \geq 0 \), solve for \( v \), resulting in \( -2 \leq v \leq 2 \). Combining this with \( 0 \leq u \leq 2 \), our parameters are \( u \) within [0, 2] and \( v \) within [-2, 2].
06

Write the Parametrization

The parameterization of the surface can thus be defined as: \( x(u, v) = u \), \( y(u, v) = v \), \( z(u, v) = 4 - v^2 \) where \( 0 \leq u \leq 2 \) and \( -2 \leq v \leq 2 \). This fully describes the surface cut from the parabolic cylinder fitting the given conditions.

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Key Concepts

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

Parabolic Surfaces
Parabolic surfaces are fascinating structures in mathematics and physics due to their unique shape. They are generated by a parabola moving parallel to its axis in space. In simple terms, imagine taking the familiar U-shaped curve of a parabola and extending it into three-dimensional space. This extended shape is what we call a parabolic surface. One common type of parabolic surface is a paraboloid, which comes in two main variations:
  • Elliptic Paraboloid: Looks like a regular parabola rotated around an axis, often used to focus light or sound waves.
  • Hyperbolic Paraboloid: A saddle-shaped surface with one axis curving upward and the other curving downward.
In our exercise, we deal with a parabolic cylinder, a special type of parabolic surface that is essentially a parabola extruded parallel to a line, resulting in a tube-like shape that runs infinitely in one direction. This surface is bound by specific planes, creating a finite region we can analyze.
Cylindrical Coordinates
Cylindrical coordinates offer an alternative to Cartesian coordinates, well-suited for problems involving symmetry around a central axis, like cylinders. They extend polar coordinates by adding a height dimension. In this system, a point in space is defined by:
  • ho (rho): the radial distance from the z-axis, much like the radius in polar coordinates.
  • heta (theta): the angle measured from a fixed direction in the xy-plane, also known as the azimuthal angle.
  • z (z): the height above or below the xy-plane.
These coordinates are particularly advantageous when dealing with cylindrical surfaces, as they naturally reduce the complexity of the equations involved. Although we used parameters in Cartesian form for our exercise, understanding cylindrical coordinates provides a beneficial perspective, especially when dealing with rotational bodies or analyzing physical phenomena like electromagnetic fields.
Bounded Surfaces
Bounded surfaces refer to the confines within which a shape or surface exists. In three-dimensional geometry, a bounded surface is one that is enclosed by limits on the x, y, and z dimensions. This means that there are boundaries or constraints that restrict the extent of the surface. In problems like the one in this exercise, recognizing the bounded nature of surfaces is crucial for accurately applying the mathematics involved. For the surface of our parabolic cylinder, the planes provide those bounds:
  • x = 0 and x = 2: These are the vertical planes that limit the surface in the x-direction.
  • z = 0: This horizontal plane acts as a floor, ensuring the surface doesn't extend below this point.
These boundaries are vital for defining the region of interest and helping ensure that the parametrization respects these limits.
Three-Dimensional Geometry
Three-dimensional geometry involves the study of shapes and figures in the three axes of space: x, y, and z. It's what allows us to comprehend the world in its full volume and form. In simpler terms, while two-dimensional geometry deals with flat shapes like squares and circles, 3D geometry introduces cubes, spheres, and other volumetric forms. Key aspects of three-dimensional geometry include:
  • Planes: Flat surfaces that extend infinitely in two directions. They can intersect to form lines or enclose spaces like rooms.
  • Solids: Closed 3D shapes with both volume and surface area, like a cube or cylinder.
  • Coordinate Systems: Using coordinates like (x, y, z) to describe points in space.
Undertanding three-dimensional geometry is essential when engaging in practical applications ranging from architecture to physics. For our exercise, it assists us in visualizing how the parabolic cylinder interacts with the bounding planes, providing a clearer understanding of the surface we are analyzing.

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Most popular questions from this chapter

A surface \(S\) lies on the paraboloid \(z=\frac{1}{2} x^{2}+\frac{1}{2} y^{2}\) directly above the triangle in the \(x y\) -plane with vertices \((0,0),(2,0),\) and \((2,4) .\) If the density at a point \((x, y, z)\) on \(S\) is given by \(\delta(x, y, z)=9 x y \mathrm{g} / \mathrm{cm}^{2}\) find the total mass of \(S .\)

Find the area of the surface \(2 x^{3 / 2}+2 y^{3 / 2}-3 z=0\) above the square \(R : 0 \leq x \leq 1,0 \leq y \leq 1,\) in the \(x y\) -plane.

Apply Green's Theorem to evaluate the integrals. \(\oint_{C}(6 y+x) d x+(y+2 x) d y\) \(C :\) The circle \((x-2)^{2}+(y-3)^{2}=4\)

In Exercises \(9-20,\) use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) $$\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}$$ $$\begin{array}{l}{\text { a. Cube } D :\quad \quad \quad \quad \text { The cube cut from the first octant by }} \quad\quad\quad\quad\quad\quad\quad\quad {\text { the planes } x=1, y=1, \text { and } z=1}\end{array} $$ $$b. Cube D :\quad\quad\quad\quad \quad The \quad cube \quad bounded \quad \quad \quad by \quad the \quad\quad\quad\quad\quad\quad\quad\quad\quad planes=\pm 1, y=\pm 1, \text { and } z=\pm 1$$ $$\begin{array}{c}{\text { c. Cylindrical can } D : \text { The region cut from the solid cylinder }} \quad\quad\quad\quad\quad\quad\quad\quad {x^{2}+y^{2} \leq 4 \text { by the planes } z=0 \text { and }} \\\ {z=1}\end{array}$$

Use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(\mathbf{F}\) around the simple closed curve \(C\). Perform the following CAS steps. a. Plot \(C\) in the \(x y\)-plane. b. Determine the integrand \((\partial N / \partial x)-(\partial M / \partial y)\) for the tangential form of Green's Theorem. c. Determine the (double integral) limits of integration from your plot in part (a) and evaluate the curl integral for the circulation. \(\mathbf{F}=\left(2 x^{3}-y^{3}\right) \mathbf{i}+\left(x^{3}+y^{3}\right) \mathbf{j}, \quad C :\) The ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)
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