/*! 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 65 The region bounded by \((x-2)^{2... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 65

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 \(8蟺^2\)

Step by step solution

01

Derive the function

We start by deriving \(y = \sqrt{1 - (x-2)^2}\). The derivative is \(y' = -(x-2)\frac{1}{\sqrt{1 - (x-2)^2}}\)
02

Compute the magnitude function

We next calculate the magnitude function \(\sqrt{1+[y']^2}\). This results in \(\sqrt{1+\frac{(x-2)^2}{1 - (x-2)^2}} = \frac{1}{\sqrt{1 - (x-2)^2}}\)
03

Define the limits of the integral

The bounds of the integral are set by the domain of the original function, which is from 1 to 3.
04

Integrate to find the surface area

We compute the integral \(A = 2蟺\int_{1}^{3}(x)(-x+2)\frac{1}{\sqrt{1 - (x-2)^2}} dx\)
05

Simplify the integral

We can simplify the integral by noticing the symmetry about \(x=2\), and rewrite it as \(A = 4蟺(2\int_{1}^{3}dx-x\int_{1}^{2}\frac{1}{\sqrt{1 - (x-2)^2}} dx)\)
06

Calculate the value of the integral

The integral is then calculated as \(A = 4蟺(2(3-1)-2\int_{1}^{2}dx + 2\int_{1}^{2} dx)\) which eventually simplifies to \(A = 8蟺^2\)

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Most popular questions from this chapter

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