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Chapter 13: Problem 19

\(15-22=\) Find a parametric representation for the surface. The part of the sphere \(x^{2}+y^{2}+z^{2}=4\) that lies above the cone \(z=\sqrt{x^{2}+y^{2}}\)

Short Answer

Expert verified
Parametric equations: \(x = \sqrt{2} \sin \theta \cos \phi\), \(y = \sqrt{2} \sin \theta \sin \phi\), \(z = \sqrt{2} \cos \theta\), with \(0 \leq \theta \leq \pi/4\), \(0 \leq \phi < 2\pi\).

Step by step solution

01

Understand the given surfaces

We are given a sphere with the equation \(x^2 + y^2 + z^2 = 4\) and a cone with the equation \(z = \sqrt{x^2 + y^2}\). We need to find the part of the sphere that lies above the cone.
02

Analyze the cone and sphere intersection

To find where the sphere lies above the cone, we set \(z = \sqrt{x^2 + y^2}\) and substitute it into the sphere's equation: \(x^2 + y^2 + (\sqrt{x^2+y^2})^2 = 4\). Simplifying, this gives \(2(x^2 + y^2) = 4\), or \(x^2 + y^2 = 2\). This is the circle of intersection between the sphere and the cone.
03

Set parametric equations for the spherical cap

To parameterize the surface above the cone (spherical cap), we use spherical coordinates. Let \(x = \sqrt{2} \sin \theta \cos \phi\), \(y = \sqrt{2} \sin \theta \sin \phi\), and \(z = \sqrt{2} \cos \theta\). \(\theta\) ranges from 0 to \(\pi/4\) because the cone equation limits \(z\) to be at least \(\sqrt{x^2 + y^2}\). \(\phi\) ranges from 0 to \(2\pi\).
04

Range determination for parameters

The range for \(\theta\) is determined by the intersection of the cone and sphere: \(\cos \theta = \frac{\sqrt{2}}{2}\), hence \(0 \leq \theta \leq \frac{\pi}{4}\). The full circle in the \(xy\)-plane means \(\phi\) ranges from 0 to \(2\pi\).
05

Write the final parametric representation

The parametric equations for the spherical cap are: \(x = \sqrt{2} \sin \theta \cos \phi\), \(y = \sqrt{2} \sin \theta \sin \phi\), \(z = \sqrt{2} \cos \theta\) for \(0 \leq \theta \leq \pi/4\) 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.

Sphere equations
A sphere is a perfectly round three-dimensional shape, much like a ball. The equation of a sphere in three-dimensional space can be represented as \(x^2 + y^2 + z^2 = r^2\), where \(r\) is the radius of the sphere.
- The center of the sphere is the point \((0,0,0)\), assuming the equation is not altered with linear terms.- All points \((x, y, z)\) that satisfy the sphere equation are equally distant from the center, with the distance being the radius \(r\).
In our original problem, we have a sphere with the equation \(x^2 + y^2 + z^2 = 4\). Here, the sphere is centered at the origin, and the radius \(r\) is 2, since \(r^2 = 4\). Identifying the radius helps in understanding the dimensions of the surface it encloses.
Cone equations
A cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex. The equation for a cone that opens upward or downward can generally be written as \(z = k\sqrt{x^2 + y^2}\), where \(k\) denotes the steepness of the cone.
- In our case, the given cone equation is \(z = \sqrt{x^2 + y^2}\). Here, \(k\) equals 1, meaning the cone is symmetrical and opens upwards.
- The formula expresses how the height \(z\) of the cone varies with the radial distance from the center of its base.
The apex of the cone in this equation is at the origin \((0,0,0)\), and as \(z\) increases, \(x^2 + y^2\) also increases quadratically. This sets the stage for examining the surfaces' intersection.
Spherical coordinates
Spherical coordinates are a system for expressing the positions of points in 3D space, specifically using angles and distances. This is especially useful when working with problems involving spheres.
  • The positions are represented via three coordinates: \(\rho\) (the radial distance), \(\theta\) (the polar angle), and \(\phi\) (the azimuthal angle).
  • For our exercise, however, we modify this system? by setting \(\rho\) constant and focusing on \(\theta\) and \(\phi\).

    • In the parameterization of the portion of the sphere that lies above the cone, we use:
    • \(x = \sqrt{2} \sin \theta \cos \phi\)
    • \(y = \sqrt{2} \sin \theta \sin \phi\)
    • \(z = \sqrt{2} \cos \theta\)
    We note that \(\theta\) ranges from 0 to \(\pi/4\) due to the constraints set by the cone's height. Meanwhile, \(\phi\) ranges from 0 to \(2\pi\), allowing for a full revolution around the axis.
Intersection of surfaces
The intersection of surfaces refers to the points where two or more surfaces meet. Analyzing these intersections can solve problems about shapes and volumes formed by their overlaps.
In the context of this exercise:
  • The sphere and cone intersect along a circle. Since both surfaces are symmetrical about the z-axis.
  • To find the intersection, equating the cone and sphere equations, we get \(z = \sqrt{x^2 + y^2}\) and substitute into \(x^2 + y^2 + z^2 = 4\).
  • Simplification provides \(x^2 + y^2 = 2\), which is the circle of intersection on the plane where these surfaces share common points.
    • The parametric settings must respect this circle, ensuring the representation accurately describes the region where the sphere is above the cone.

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Most popular questions from this chapter

\(25-30=\) Prove each identity, assuming that \(S\) and \(E\) satisfy the conditions of the Divergence Theorem and the scalar functions and components of the vector fields have continuous second-order partial derivatives. $$\iint_{S} D_{\mathrm{n}} f d S=\iiint_{E} \nabla^{2} f d V$$

\(29-32=\) Find an equation of the tangent plane to the given parametric surface at the specified point. If you have software that graphs parametric surfaces, use a computer to graph the surface and the tangent plane. $$x=u+v, \quad y=3 u^{2}, \quad z=u-v ; \quad(2,3,0)$$

Use the Divergence Theorem to evaluate \(\iint_{S} \mathbf{F} \cdot d \mathbf{S},\) where \(\mathbf{F}(x, y, z)=z^{2} x \mathbf{i}+\left(\frac{1}{3} y^{3}+\tan z\right) \mathbf{j}+\left(x^{2} z+y^{2}\right) \mathbf{k}\) and \(S\) is the top half of the sphere \(x^{2}+y^{2}+z^{2}=1\) [Hint: Note that \(S\) is not a closed surface. First compute integrals over \(S_{1}\) and \(S_{2},\) where \(S_{1}\) is the disk \(x^{2}+y^{2} \leqslant 1\) oriented downward, and \(S_{2}=S \cup S_{1} . ]\)

Maxwell's equations relating the electric field \(\mathbf{E}\) and mag- netic field \(\mathbf{H}\) as they vary vith time in a region containing no charge and no current can be stated as follows: $$\begin{array}{ll}{\operatorname{div} \mathbf{E}=0} & {\operatorname{div} \mathbf{H}=0} \\ {\operatorname{curl} \mathbf{E}=-\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}} & {\operatorname{curl} \mathbf{H}=\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}}\end{array}$$ where \(c\) is the speed of light. Use these equations to prove the following: $$\begin{array}{l}{\quad\left(\text { a) } \nabla \times(\nabla \times \mathbf{E})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}\right.} \\ {\text { (b) } \nabla \times(\nabla \times \mathbf{H})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}} \\\ {\text { (c) } \nabla^{2} \mathbf{E}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} \quad[\text {Hint} \text { : Use Exercise } 27 .]} \\\ {\text { (d) } \nabla^{2} \mathbf{H}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}}\end{array}$$

\(1-4\) Verify that the Divergence Theorem is true for the vector field \(F\) on the region \(E .\) $$\mathbf{F}(x, y, z)=\langle z, y, x\rangle$$ \(E\) is the solid ball \(x^{2}+y^{2}+z^{2} \leqslant 16\)
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