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A domain \(D\) in space is given. Parametrize each of the bounding surfaces of \(D\). \(D\) is the domain bounded by the paraboloid \(z=4-x^{2}-4 y^{2}\) and the plane \(z=0\).

Short Answer

Expert verified
Parametrize the paraboloid with \((x, y, z) = (2\cos(\theta), \sin(\theta), 4-4\cos^2(\theta)-\sin^2(\theta))\) and the plane as \((x, y, z) = (2\cos(\theta), \sin(\theta), 0)\).

Step by step solution

01

Understand the Surfaces Defining the Domain

The given domain \(D\) is bounded by a paraboloid and a plane. The paraboloid is described by the equation \( z = 4 - x^2 - 4y^2 \), and the plane is given by \( z = 0 \). This means that \(D\) is the region between these two surfaces.
02

Determine the Intersection of the Surfaces

Find where the paraboloid \( z = 4 - x^2 - 4y^2 \) intersects with the plane \( z = 0 \). Set \( 4 - x^2 - 4y^2 = 0 \) to find the boundary in the \(xy\)-plane. This simplifies to \( x^2 + 4y^2 = 4 \), representing an ellipse in the \(xy\)-plane.
03

Parametrize the Bounding Paraboloid

To parametrize the portion of the paraboloid, we express \(x\), \(y\), and \(z\) in terms of parameters. Using elliptical cylindrical coordinates: \( x = 2\cos(\theta) \), \( y = \sin(\theta) \), with \( \theta \) ranging from \(0\) to \(2\pi\) and \(-2 \leq \sqrt{4 - x^2 - 4y^2} \leq 2\). Thus, \( z = 4-x^2 - 4y^2 \).
04

Parametrize the Bounding Plane

Since the plane \( z = 0 \) intersects the paraboloid along the ellipse \( x^2 + 4y^2 = 4 \), we can use the parametric equations of this ellipse for the plane as well: \( x = 2\cos(\theta) \), \( y = \sin(\theta) \), \( z = 0 \) with \( \theta \) ranging from \(0\) to \(2\pi\).

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Key Concepts

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

Paraboloid
A paraboloid is a three-dimensional geometric shape that is an extension of a parabola into three dimensions. In the context of this exercise, the paraboloid is given by the equation \( z = 4 - x^2 - 4y^2 \). This is a classic example of an elliptic paraboloid, characterized by:
  • Its bowl-like shape, opening downwards in this case because of the negative coefficients in front of \(x^2\) and \(4y^2\).
  • The highest point being at \((x, y, z) = (0, 0, 4)\), known as the vertex of the paraboloid.
To understand a paraboloid, think about how a quadratic curve in two dimensions (a parabola) rotates around an axis to create this 3D surface.
When we set the paraboloid to intersect with the plane \(z = 0\), we essentially find the level curve of the paraboloid. This level curve is an ellipse, formed by setting \(4 - x^2 - 4y^2 = 0\) which simplifies to \(x^2 + 4y^2 = 4\).
This shows how a 3D surface can project interesting shapes in lower dimensions, helping us understand the parameters that control curves like ellipses within the paraboloid.
Ellipse
An ellipse is a type of conic section that looks like a stretched circle. It has two axes: a major axis and a minor axis, corresponding to its longer and shorter diameters. In the solution, the ellipse appears as the intersection of the paraboloid and the plane \(z = 0\).
This ellipse has the equation \(x^2 + 4y^2 = 4\), where:
  • \(x^2\) term dominates along the major axis, and \(4y^2\) term is along the minor axis, because the coefficients dictate how the ellipse stretches in each direction.
  • The boundary points occur where this equation holds true, forming a closed curve on the \(xy\)-plane.
Parametrizing this ellipse involves using trigonometric functions since it can be thought of as a rotated circle. We can set \(x = 2\cos(\theta)\) and \( y = \sin(\theta)\) with \(\theta\) ranging from \(0\) to \(2\pi\), giving us a full path around the ellipse.
This parametrization maps out the ellipse entirely as \(\theta\) cycles through its range, outlining the boundary shared by the paraboloid and the plane.
Cylindrical Coordinates
Cylindrical coordinates extend two-dimensional polar coordinates to three dimensions. They are particularly useful for dealing with problems involving symmetry around an axis, like the situation of the paraboloid in this exercise.
In elliptical cylindrical coordinates, used here, variables are defined to model the ellipse precisely:
  • \(x = a\cos(\theta)\): Representing the horizontal distance from the origin, weighted by some scaling factor \(a\), here \(a=2\).
  • \(y = b\sin(\theta)\): Representing the vertical distance, weighted by a different factor \(b\), here implicitly 1.
  • \(z = 4 - x^2 - 4y^2\): Provides the height value, showcasing how the vertical dimension of the paraboloid behaves.
These coordinates simplify the parameters needed to define the surfaces in this exercise, by working with angles and radii, inherent to the ellipse’s symmetry.
The parametric equations derived from these coordinates allow us to effectively model both the boundary of the paraboloid and the intersecting plane in a unified, simplified manner.

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Most popular questions from this chapter

Compute the flux of \(\vec{F}\) across \(\mathcal{S}\). (If \(\mathcal{S}\) is not a closed surface, choose \(\vec{n}\) so that it has a positive z-component, unless otherwise indicated.) \(\mathcal{S}\) is the closed surface composed of \(\mathcal{S}_{1},\) whose boundary is the ellipse in the \(x-y\) plane described by \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) and \(\mathcal{S}_{2},\) part of the elliptical paraboloid \(f(x, y)=1-\frac{x^{2}}{25}-\frac{y^{2}}{9}\) (see graph); \(\vec{F}=\langle 5,2,3\rangle .\)

In Exercises \(13-16,\) a vector field \(\vec{F}\) and a closed curve \(C,\) enclosing a region \(R,\) are given. Verify Green's Theorem by evaluating \(\oint_{C} \vec{F} \cdot d \vec{r}\) and \(\iint_{R}\) curl \(\vec{F} d A,\) showing they are equal. \(\vec{F}=\langle x-y, x+y\rangle ; C\) is the closed curve composed of the parabola \(y=x^{2}\) on \(0 \leq x \leq 2\) followed by the line segment from (2,4) to (0,0).

In Exercises \(9-12,\) a closed curve \(C\) that is the boundary of a surface \(S\) is given along with a vector field \(\vec{F}\). Verify Stokes' Theorem on \(C ;\) that is, show \(\oint_{c} \vec{F} \cdot d \vec{r}=\iint_{S}(\operatorname{curl} \vec{F}) \cdot \vec{n} d S\). $$ \begin{aligned} &C \text { is the curve parametrized by } \vec{r}(t)=\left\langle\cos t, \sin t, e^{-1}\right\rangle\\\ &\text { and } \mathcal{S} \text { is the portion of } z=e^{-x^{2}-y^{2}} \text { enclosed by } c ; \vec{F}=\\\ &\langle-y, x, 1\rangle . \end{aligned} $$

In Exercises \(13-16,\) find the work performed by the force field \(\vec{F}\) moving a particle along the path \(C\). \(\vec{F}=\left\langle y, x^{2}\right\rangle \mathrm{N} ; C\) is the portion of \(y=\sqrt{x}\) from (0,0) to \((1,1),\) where distances are measured in meters.

In Exercises \(15-18,\) a parametrized curve \(C\) in space is given. Find the area above the \(x\) -y plane that is under \(C\). $$ C: \vec{r}(t)=\langle 3 t, 4 t, t\rangle \text { for } 0 \leq t \leq 1 $$

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