/*! 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 41 Identify the quadric surface. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 41

Identify the quadric surface. $$ x^{2}+\frac{y^{2}}{4}+z^{2}=1 $$

Short Answer

Expert verified
The quadric surface represented by the given equation \(x^{2}+\frac{y^{2}}{4}+z^{2}=1\) is an ellipsoid.

Step by step solution

01

Analyze the given equation

The provided equation is \(x^{2}+\frac{y^{2}}{4}+z^{2}=1\). As a quadric surface, we know this value must represent a three-dimensional figure. All terms involving variables (x, y and z) are squared in this equation, thus we should look at those types of quadric surfaces.
02

Compare the equation to the standard forms of quadric surfaces

Considering different types of quadric surfaces and their standard forms: \n 1. Ellipsoid: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1\) \n Here, a, b, and c are all positive constants, and a, b, and c are all non-zero. \n 2. Hyperboloid: \(x^{2}/a^{2}卤y^{2}/b^{2}-z^{2}/c^{2}=1\) \n 3. Paraboloid: \(x^{2}/a^{2}卤y^{2}/b^{2}=z/c\) \n \n Our given equation \(x^{2}+\frac{y^{2}}{4}+z^{2}=1\) matches the form of the ellipsoid, with \(a^{2}=1\), \(b^{2}=4\), and \(c^{2}=1\). Thus, this equation describes an ellipsoid.
03

Confirm the description of the surface

The final step in the solution involves confirming that the given equation indeed describes an ellipsoid. For this equation, a, b, and c (whenever exist) satisfy a > 0, b > 0, and c > 0, and thus describe an ellipsoid. The values for a, b, and c impact the 'stretch' of the ellipsoid along the x, y, and z axes respectively. As the equation \(x^{2}+\frac{y^{2}}{4}+z^{2}=1\) satisfies this condition, this confirms that the given equation describes an ellipsoid.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The population density (in people per square mile) for a coastal town can be modeled by \(f(x, y)=\frac{120,000}{(2+x+y)^{3}}\) where \(x\) and \(y\) are measured in miles. What is the population inside the rectangular area defined by the vertices \((0,0)\), \((2,0),(0,2)\), and \((2,2) ?\)

Evaluate the partial integral. $$ \int_{x}^{x^{2}} \frac{y}{x} d y $$

Evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{2}(x+y) d y d x $$

A company sells two products whose demand functions are given by \(x_{1}=500-3 p_{1}\) and \(x_{2}=750-2.4 p_{2}\) So, the total revenue is given by \(R=x_{1} p_{1}+x_{2} p_{2}\) Estimate the average revenue if the price \(p_{1}\) varies between \(\$ 50\) and \(\$ 75\) and the price \(p_{2}\) varies between \(\$ 100\) and \(\$ 150\).

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{1}^{2} \int_{2}^{4} d x d y $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.