/*! 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 75 Surface Area The region bounded ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 75

Surface Area The region bounded by \((x-2)^{2}+y^{2}=1\) is revolved about the \(y\) -axis to form a torus. Find the surface area of the torus.

Short Answer

Expert verified
The surface area of the torus is obtained by evaluating the integral \(A = 2\pi \int_{-1}^1 (\sqrt{1-y^2}+2) \sqrt{1+ \[(-y/\sqrt{1-y^2})\]^2} \, dy\)

Step by step solution

01

Understand the Problem

The equation \((x-2)^{2}+y^{2}=1\) represents a circle with center at (2,0) and radius 1. We're revolving this circle around the y-axis to form a torus.
02

Identify the Revolution

The region is revolved about the y-axis. In this setup, the radius from the axis of revolution to the curve is 'x' and the height element is \(dy\). The circle equation can be expressed in terms of y: \(x= \sqrt{1-y^2}+2\). Thus the radius is \(x= \sqrt{1-y^2}+2\).
03

Calculate Surface Area

The general flat surface area formula is \(A = 2\pi xy \, ds\), where ds is the infinitesimal arc length, x is the radial distance from the axis of revolution (in this case the x position), and y is the height (our infinitesimal distance).\nThe arc length can be expressed in terms of differentials as \(ds = \sqrt{1+(dx/dy)^2} \, dy\). Differentiating our above 'x' equation in terms of 'y' gives us \(dx/dy = \frac{-y}{\sqrt{1-y^2}}\). Inserting 'x' and 'dx/dy' into the formula for 'A', translating 'dx/dy' into 'ds' and taking the limit as \(dy->0\) will give the integral formula for the surface area.
04

Integrating for the Surface Area

The integral for the surface area is \(A = 2\pi \int_{-1}^1 (\sqrt{1-y^2}+2) \sqrt{1+ \[(-y/\sqrt{1-y^2})\]^2} \, dy\). Solving this integral will give the final surface area of the torus.

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 Gamma Function The Gamma Function \(\Gamma(n)\) is defined in terms of the integral of the function given by \(f(x)=x^{n-1} e^{-x}, \quad n>0 .\) Show that for any fixed value of \(n,\) the limit of \(f(x)\) as \(x\) approaches infinity is zero.

Making an Integral Improper For each integral, find a nonnegative real number \(b\) that makes the integral improper. Explain your reasoning. $$ \begin{array}{ll}{\text { (a) } \int_{0}^{b} \frac{1}{x^{2}-9} d x} & {\text { (b) } \int_{0}^{b} \frac{1}{\sqrt{4-x}} d x} \\ {\text { (c) } \int_{0}^{b} \frac{x}{x^{2}-7 x+12} d x} & {\text { (d) } \int_{b}^{10} \ln x d x} \\\ {\text { (e) } \int_{0}^{b} \tan 2 x d x} & {\text { (f) } \int_{0}^{b} \frac{\cos x}{1-\sin x} d x}\end{array} $$

Evaluating an Improper Integral In Exercises \(33-48\) determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{8} \frac{3}{\sqrt{8-x}} d x $$

Indeterminate Forms Show that the indeterminate forms \(0^{0}, \infty^{0},\) and \(1^{\infty}\) do not always have a value of 1 by evaluating each limit. (a) \(\lim _{x \rightarrow 0^{+}} x^{\ln 2 /(1+\ln x)}\) (b) \(\lim _{x \rightarrow \infty} x^{\ln 2 /(1+\ln x)}\) (c) \(\lim _{x \rightarrow 0}(x+1)^{(\ln 2) / x}\)

Volume Find the volume of the solid generated by revolving the region bounded by the graph of \(f\) about the \(x\) -axis. $$f(x)=\left\\{\begin{array}{ll}{x \ln x,} & {0
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

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.