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 +

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.4, 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 17.4, Problem 44E

Chapter 17.4, Problem 46E

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Chapter 17 Solutions

CALCULUS (CLOTH)

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E

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(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

Solution Summary: The author explains the torus obtained by rotating a circle in the upper half of x y p l

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(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 17 Solutions

(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

Solution Summary: The author explains the torus obtained by rotating a circle in the upper half of x y p l

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(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 17 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 $A r e a (T) = 4 π^{2} a b$