Explanation Given: A r e a ( S ) = 2 π ∫ c d y 1 + g ' ( y ) 2 d y --------- ( 1 ) The cone described by x 2 + y 2 = z 2 for 0 ≤ z ≤ d . Formula used: g ( y ) [for c ≤ y ≤ d ]is the generating curve in z y − p l a n e for the surface of revolution S about z − a x i s A r e a ( S ) = 2 π ∫ c d y 1 + [ g ' ( y ) ] 2 d y

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 17.4, Problem 44E

To determine

The surface area of upper half of the cone described by $x^{2} + y^{2} = z^{2}$ for $0 \leq z \leq d$ using surface revolution technique.

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Chapter 17.4, Problem 43E

Chapter 17.4, Problem 45E

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

Ch. 17.1 - Prob. 7E Ch. 17.1 - Prob. 8E Ch. 17.1 - Prob. 9E Ch. 17.1 - Prob. 10E Ch. 17.1 - Prob. 11E Ch. 17.1 - Prob. 12E Ch. 17.1 - Prob. 13E Ch. 17.1 - Prob. 14E Ch. 17.1 - Prob. 15E Ch. 17.1 - Prob. 16E Ch. 17.1 - Prob. 17E Ch. 17.1 - Prob. 18E Ch. 17.1 - Prob. 19E Ch. 17.1 - Prob. 20E Ch. 17.1 - Prob. 21E Ch. 17.1 - Prob. 22E Ch. 17.1 - Prob. 23E Ch. 17.1 - Prob. 24E Ch. 17.1 - Prob. 25E Ch. 17.1 - Prob. 26E Ch. 17.1 - Prob. 27E Ch. 17.1 - Prob. 28E Ch. 17.1 - Prob. 29E Ch. 17.1 - Prob. 30E Ch. 17.1 - Prob. 31E Ch. 17.1 - Prob. 32E Ch. 17.1 - Prob. 33E Ch. 17.1 - Prob. 34E Ch. 17.1 - Prob. 35E Ch. 17.1 - Prob. 36E Ch. 17.1 - Prob. 37E Ch. 17.1 - Prob. 38E Ch. 17.1 - Prob. 39E Ch. 17.1 - Prob. 40E Ch. 17.1 - Prob. 41E Ch. 17.1 - Prob. 42E Ch. 17.1 - Prob. 43E Ch. 17.1 - Prob. 44E Ch. 17.1 - Prob. 45E Ch. 17.1 - Prob. 46E Ch. 17.1 - Prob. 47E Ch. 17.1 - Prob. 48E Ch. 17.1 - Prob. 49E Ch. 17.1 - Prob. 50E Ch. 17.1 - Prob. 51E Ch. 17.1 - Prob. 52E Ch. 17.1 - Prob. 53E Ch. 17.1 - Prob. 54E Ch. 17.1 - Prob. 55E Ch. 17.1 - Prob. 56E Ch. 17.1 - Prob. 57E Ch. 17.2 - Prob. 1PQ Ch. 17.2 - Prob. 2PQ Ch. 17.2 - Prob. 3PQ Ch. 17.2 - Prob. 4PQ Ch. 17.2 - Prob. 1E Ch. 17.2 - Prob. 2E Ch. 17.2 - Prob. 3E Ch. 17.2 - Prob. 4E Ch. 17.2 - Prob. 5E Ch. 17.2 - Prob. 6E Ch. 17.2 - Prob. 7E Ch. 17.2 - Prob. 8E Ch. 17.2 - Prob. 9E Ch. 17.2 - Prob. 10E Ch. 17.2 - Prob. 11E Ch. 17.2 - Prob. 12E Ch. 17.2 - Prob. 13E Ch. 17.2 - Prob. 14E Ch. 17.2 - Prob. 15E Ch. 17.2 - Prob. 16E Ch. 17.2 - Prob. 17E Ch. 17.2 - Prob. 18E Ch. 17.2 - Prob. 19E Ch. 17.2 - Prob. 20E Ch. 17.2 - Prob. 21E Ch. 17.2 - Prob. 22E Ch. 17.2 - Prob. 23E Ch. 17.2 - Prob. 24E Ch. 17.2 - Prob. 25E Ch. 17.2 - Prob. 26E Ch. 17.2 - Prob. 27E Ch. 17.2 - Prob. 28E Ch. 17.2 - Prob. 29E Ch. 17.2 - Prob. 30E Ch. 17.2 - Prob. 31E Ch. 17.2 - Prob. 32E Ch. 17.2 - Prob. 33E Ch. 17.2 - Prob. 34E Ch. 17.2 - Prob. 35E Ch. 17.2 - Prob. 36E Ch. 17.2 - Prob. 37E Ch. 17.2 - Prob. 38E Ch. 17.2 - Prob. 39E Ch. 17.2 - Prob. 40E Ch. 17.2 - Prob. 41E Ch. 17.2 - Prob. 42E Ch. 17.2 - Prob. 43E Ch. 17.2 - Prob. 44E Ch. 17.2 - Prob. 45E Ch. 17.2 - Prob. 46E Ch. 17.2 - Prob. 47E Ch. 17.2 - Prob. 48E Ch. 17.2 - Prob. 49E Ch. 17.2 - Prob. 50E Ch. 17.2 - Prob. 51E Ch. 17.2 - Prob. 52E Ch. 17.2 - Prob. 53E Ch. 17.2 - Prob. 54E Ch. 17.2 - Prob. 55E Ch. 17.2 - Prob. 56E Ch. 17.2 - Prob. 57E Ch. 17.2 - Prob. 58E Ch. 17.2 - Prob. 59E Ch. 17.2 - Prob. 60E Ch. 17.2 - Prob. 61E Ch. 17.2 - Prob. 62E Ch. 17.2 - Prob. 63E Ch. 17.2 - Prob. 64E Ch. 17.2 - Prob. 65E Ch. 17.2 - Prob. 66E Ch. 17.2 - Prob. 67E Ch. 17.2 - Prob. 68E Ch. 17.2 - Prob. 69E Ch. 17.2 - Prob. 70E Ch. 17.2 - Prob. 71E Ch. 17.2 - Prob. 72E Ch. 17.2 - Prob. 73E Ch. 17.2 - Prob. 74E Ch. 17.2 - Prob. 75E Ch. 17.3 - Prob. 1PQ Ch. 17.3 - Prob. 2PQ Ch. 17.3 - Prob. 3PQ Ch. 17.3 - Prob. 4PQ Ch. 17.3 - Prob. 1E Ch. 17.3 - Prob. 2E Ch. 17.3 - Prob. 3E Ch. 17.3 - Prob. 4E Ch. 17.3 - Prob. 5E Ch. 17.3 - Prob. 6E Ch. 17.3 - Prob. 7E Ch. 17.3 - Prob. 8E Ch. 17.3 - Prob. 9E Ch. 17.3 - Prob. 10E Ch. 17.3 - Prob. 11E Ch. 17.3 - Prob. 12E Ch. 17.3 - Prob. 13E Ch. 17.3 - Prob. 14E Ch. 17.3 - Prob. 15E Ch. 17.3 - Prob. 16E Ch. 17.3 - Prob. 17E Ch. 17.3 - Prob. 18E Ch. 17.3 - Prob. 19E Ch. 17.3 - Prob. 20E Ch. 17.3 - Prob. 21E Ch. 17.3 - Prob. 22E Ch. 17.3 - Prob. 23E Ch. 17.3 - Prob. 24E Ch. 17.3 - Prob. 25E Ch. 17.3 - Prob. 26E Ch. 17.3 - Prob. 27E Ch. 17.3 - Prob. 28E Ch. 17.3 - Prob. 29E Ch. 17.3 - Prob. 30E Ch. 17.3 - Prob. 31E Ch. 17.3 - Prob. 32E Ch. 17.3 - Prob. 33E Ch. 17.3 - Prob. 34E Ch. 17.3 - Prob. 35E Ch. 17.4 - Prob. 1PQ Ch. 17.4 - Prob. 2PQ Ch. 17.4 - Prob. 3PQ Ch. 17.4 - Prob. 4PQ Ch. 17.4 - Prob. 5PQ Ch. 17.4 - Prob. 6PQ Ch. 17.4 - Prob. 1E Ch. 17.4 - Prob. 2E Ch. 17.4 - Prob. 3E Ch. 17.4 - Prob. 4E Ch. 17.4 - Prob. 5E Ch. 17.4 - Prob. 6E Ch. 17.4 - Prob. 7E Ch. 17.4 - Prob. 8E Ch. 17.4 - Prob. 9E Ch. 17.4 - Prob. 10E Ch. 17.4 - Prob. 11E Ch. 17.4 - Prob. 12E Ch. 17.4 - Prob. 13E Ch. 17.4 - Prob. 14E Ch. 17.4 - Prob. 15E Ch. 17.4 - Prob. 16E Ch. 17.4 - Prob. 17E Ch. 17.4 - Prob. 18E Ch. 17.4 - Prob. 19E Ch. 17.4 - Prob. 20E Ch. 17.4 - Prob. 21E Ch. 17.4 - Prob. 22E Ch. 17.4 - Prob. 23E Ch. 17.4 - Prob. 24E Ch. 17.4 - Prob. 25E Ch. 17.4 - Prob. 26E Ch. 17.4 - Prob. 27E Ch. 17.4 - Prob. 28E Ch. 17.4 - Prob. 29E Ch. 17.4 - Prob. 30E Ch. 17.4 - Prob. 31E Ch. 17.4 - Prob. 32E Ch. 17.4 - Prob. 33E Ch. 17.4 - Prob. 34E Ch. 17.4 - Prob. 35E Ch. 17.4 - Prob. 36E Ch. 17.4 - Prob. 37E Ch. 17.4 - Prob. 38E Ch. 17.4 - Prob. 39E Ch. 17.4 - Prob. 40E Ch. 17.4 - Prob. 41E Ch. 17.4 - Prob. 42E Ch. 17.4 - Prob. 43E Ch. 17.4 - Prob. 44E Ch. 17.4 - Prob. 45E Ch. 17.4 - Prob. 46E Ch. 17.4 - Prob. 47E Ch. 17.4 - Prob. 48E Ch. 17.4 - Prob. 49E Ch. 17.4 - Prob. 50E Ch. 17.4 - Prob. 51E Ch. 17.5 - Prob. 1PQ Ch. 17.5 - Prob. 2PQ Ch. 17.5 - Prob. 3PQ Ch. 17.5 - Prob. 4PQ Ch. 17.5 - Prob. 5PQ Ch. 17.5 - Prob. 6PQ Ch. 17.5 - Prob. 7PQ Ch. 17.5 - Prob. 1E Ch. 17.5 - Prob. 2E Ch. 17.5 - Prob. 3E Ch. 17.5 - Prob. 4E Ch. 17.5 - Prob. 5E Ch. 17.5 - Prob. 6E Ch. 17.5 - Prob. 7E Ch. 17.5 - Prob. 8E Ch. 17.5 - Prob. 9E Ch. 17.5 - Prob. 10E Ch. 17.5 - Prob. 11E Ch. 17.5 - Prob. 12E Ch. 17.5 - Prob. 13E Ch. 17.5 - Prob. 14E Ch. 17.5 - Prob. 15E Ch. 17.5 - Prob. 16E Ch. 17.5 - Prob. 17E Ch. 17.5 - Prob. 18E Ch. 17.5 - Prob. 19E Ch. 17.5 - Prob. 20E Ch. 17.5 - Prob. 21E Ch. 17.5 - Prob. 22E Ch. 17.5 - Prob. 23E Ch. 17.5 - Prob. 24E Ch. 17.5 - Prob. 25E Ch. 17.5 - Prob. 26E Ch. 17.5 - Prob. 27E Ch. 17.5 - Prob. 28E Ch. 17.5 - Prob. 29E Ch. 17.5 - Prob. 30E Ch. 17.5 - Prob. 31E Ch. 17.5 - Prob. 32E Ch. 17.5 - Prob. 33E Ch. 17.5 - Prob. 34E Ch. 17.5 - Prob. 35E Ch. 17.5 - Prob. 36E Ch. 17.5 - Prob. 37E Ch. 17.5 - Prob. 38E Ch. 17 - Prob. 1CRE Ch. 17 - Prob. 2CRE Ch. 17 - Prob. 3CRE Ch. 17 - Prob. 4CRE Ch. 17 - Prob. 5CRE Ch. 17 - Prob. 6CRE Ch. 17 - Prob. 7CRE Ch. 17 - Prob. 8CRE Ch. 17 - Prob. 9CRE Ch. 17 - Prob. 10CRE Ch. 17 - Prob. 11CRE Ch. 17 - Prob. 12CRE Ch. 17 - Prob. 13CRE Ch. 17 - Prob. 14CRE Ch. 17 - Prob. 15CRE Ch. 17 - Prob. 16CRE Ch. 17 - Prob. 17CRE Ch. 17 - Prob. 18CRE Ch. 17 - Prob. 19CRE Ch. 17 - Prob. 20CRE Ch. 17 - Prob. 21CRE Ch. 17 - Prob. 22CRE Ch. 17 - Prob. 23CRE Ch. 17 - Prob. 24CRE Ch. 17 - Prob. 25CRE Ch. 17 - Prob. 26CRE Ch. 17 - Prob. 27CRE Ch. 17 - Prob. 28CRE Ch. 17 - Prob. 29CRE Ch. 17 - Prob. 30CRE Ch. 17 - Prob. 31CRE Ch. 17 - Prob. 32CRE Ch. 17 - Prob. 33CRE Ch. 17 - Prob. 34CRE Ch. 17 - Prob. 35CRE Ch. 17 - Prob. 36CRE Ch. 17 - Prob. 37CRE Ch. 17 - Prob. 38CRE Ch. 17 - Prob. 39CRE Ch. 17 - Prob. 40CRE Ch. 17 - Prob. 41CRE Ch. 17 - Prob. 42CRE Ch. 17 - Prob. 43CRE Ch. 17 - Prob. 44CRE Ch. 17 - Prob. 45CRE Ch. 17 - Prob. 46CRE Ch. 17 - Prob. 47CRE Ch. 17 - Prob. 48CRE Ch. 17 - Prob. 49CRE Ch. 17 - Prob. 50CRE Ch. 17 - Prob. 51CRE Ch. 17 - Prob. 52CRE Ch. 17 - Prob. 53CRE Ch. 17 - Prob. 54CRE Ch. 17 - Prob. 55CRE Ch. 17 - Prob. 56CRE Ch. 17 - Prob. 57CRE Ch. 17 - Prob. 58CRE Ch. 17 - Prob. 59CRE Ch. 17 - Prob. 60CRE Ch. 17 - Prob. 61CRE

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The surface area of upper half of the cone described by x 2 + y 2 = z 2 for 0 ≤ z ≤ d using surface revolution technique.

Solution Summary: The author calculates the surface area of upper half of the cone using surface revolution technique.

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The surface area of upper half of the cone described by $x^{2} + y^{2} = z^{2}$ for $0 \leq z \leq d$ using surface revolution technique.

Want to see the full answer?

Chapter 17 Solutions

The surface area of upper half of the cone described by x 2 + y 2 = z 2 for 0 ≤ z ≤ d using surface revolution technique.

Solution Summary: The author calculates the surface area of upper half of the cone using surface revolution technique.

Videos

The surface area of upper half of the cone described by x2+y2=z2 for 0≤z≤d using surface revolution technique.

Want to see the full answer?

Chapter 17 Solutions

The surface area of upper half of the cone described by $x^{2} + y^{2} = z^{2}$ for $0 \leq z \leq d$ using surface revolution technique.