Explanation Use parametrization of the cone to determine the surface area. Given : The equation of the cone x 2 + y 2 = z 2 Formula used : Area () = ∬ D ‖ n ‖ d u d v Calculation The surface area of the cone can be parametrized as φ ( u , v ) = ( u cos v , u sin v , u ) The normal vector is given by n = 〈 −...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 17.4, Problem 36E

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The surface area of the part of the cone $x^{2} + y^{2} = z^{2}$ between the planes $z = 2$ and $z = 5$

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Chapter 17.4, Problem 35E

Chapter 17.4, Problem 37E

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Chapter 17 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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The surface area of the part of the cone x 2 + y 2 = z 2 between the planes z = 2 and z = 5

Solution Summary: The author explains how the cone's surface area can be parametrized by the equation x 2 + y 2 = z 2 between the planes.

Concept explainers

Videos

To calculate:

The surface area of the part of the cone $x^{2} + y^{2} = z^{2}$ between the planes $z = 2$ and $z = 5$

Want to see the full answer?

Chapter 17 Solutions

The surface area of the part of the cone x 2 + y 2 = z 2 between the planes z = 2 and z = 5

Solution Summary: The author explains how the cone's surface area can be parametrized by the equation x 2 + y 2 = z 2 between the planes.

Concept explainers

Videos

To calculate: The surface area of the part of the cone x2+y2=z2 between the planes z = 2 and z = 5

Want to see the full answer?

Chapter 17 Solutions

To calculate:

The surface area of the part of the cone $x^{2} + y^{2} = z^{2}$ between the planes $z = 2$ and $z = 5$