Explanation Given: T is the torus obtained by rotating the circle in the y z − p l a n e of radius a centered at ( 0 , b , 0 ) about the z − a x i s . Assume b > a > 0 A r e a ( S ) = 2 π ∫ c d y 1 + g ' ( y ) 2 d y --------- ( 1 ) The torus obtained by rotating a circle of radius a . Formula used: Area of a semicircle in the upper half of x y − p l a n e having center at ( 0 , 0 ) and radius r is y = r 2 − x 2 Calculation: (a) Area of a semicircle is given by g ( y ) = a 2 − ( b − y ) 2 g ' ( y ) = 2 π 2 ( b − y ) 2 a 2 − ( b − y ) 2 = ( b − y ) a 2 − ( b − y ) 2 1 + [ g ' ( y ) ] 2 = 1 + ( b − y ) 2 a 2 − ( b − y ) 2 ⇒ 1 + [ g ' ( y ) ] 2 = a a 2 − ( b − y ) 2 Here, c = ( b − a ) , d = ( b + a ) b + a Thus, A r e a ( T ) = [ 2 π ∫ b − a b + a y 1 +

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Question

Chapter 16.4, Problem 45E

To determine

(a) Proof of equation $A r e a (T) = 4 π \int_{b - a}^{b + a} \frac{a y}{\sqrt{a^{2} - {(b - y)}^{2}}} d y$ using the equation $(1)$

(b) Proof of $A r e a (T) = 4 π^{2} a b$

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Chapter 16.4, Problem 44E

Chapter 16.4, Problem 46E

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Chapter 16 Solutions

Calculus: Early Transcendentals

Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E

Ch. 16.1 - Prob. 7E Ch. 16.1 - Prob. 8E Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E Ch. 16.1 - Prob. 11E Ch. 16.1 - Prob. 12E Ch. 16.1 - Prob. 13E Ch. 16.1 - Prob. 14E Ch. 16.1 - Prob. 15E Ch. 16.1 - Prob. 16E Ch. 16.1 - Prob. 17E Ch. 16.1 - Prob. 18E Ch. 16.1 - Prob. 19E Ch. 16.1 - Prob. 20E Ch. 16.1 - Prob. 21E Ch. 16.1 - Prob. 22E Ch. 16.1 - Prob. 23E Ch. 16.1 - Prob. 24E Ch. 16.1 - Prob. 25E Ch. 16.1 - Prob. 26E Ch. 16.1 - Prob. 27E Ch. 16.1 - Prob. 28E Ch. 16.1 - Prob. 29E Ch. 16.1 - Prob. 30E Ch. 16.1 - Prob. 31E Ch. 16.1 - Prob. 32E Ch. 16.1 - Prob. 33E Ch. 16.1 - Prob. 34E Ch. 16.1 - Prob. 35E Ch. 16.1 - Prob. 36E Ch. 16.1 - Prob. 37E Ch. 16.1 - Prob. 38E Ch. 16.1 - Prob. 39E Ch. 16.1 - Prob. 40E Ch. 16.1 - Prob. 41E Ch. 16.1 - Prob. 42E Ch. 16.1 - Prob. 43E Ch. 16.1 - Prob. 44E Ch. 16.1 - Prob. 45E Ch. 16.1 - Prob. 46E Ch. 16.1 - Prob. 47E Ch. 16.1 - Prob. 48E Ch. 16.1 - Prob. 49E Ch. 16.1 - Prob. 50E Ch. 16.1 - Prob. 51E Ch. 16.1 - Prob. 52E Ch. 16.1 - Prob. 53E Ch. 16.1 - Prob. 54E Ch. 16.1 - 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Knowledge Booster

(a) Proof of equation A r e a ( T ) = 4 π ∫ b − a b + a a y a 2 − ( b − y ) 2 d y using the equation ( 1 ) (b) Proof of A r e a ( T ) = 4 π 2 a b

(a) Proof of equation Area(T)=4π∫b−ab+aaya2−(b−y)2dy using the equation (1) (b) Proof of Area(T)=4π2ab

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Chapter 16 Solutions

(a) Proof of equation $A r e a (T) = 4 π \int_{b - a}^{b + a} \frac{a y}{\sqrt{a^{2} - {(b - y)}^{2}}} d y$ using the equation $(1)$

(b) Proof of $A r e a (T) = 4 π^{2} a b$