Explanation Given: The part of the plane: x + y + z = 0 The cylinder: x 2 + y 2 = 1 And, the function: f ( x , y , z ) = z 2 Calculation: Consider the part of the plane: x + y + z = 0 And the function: f ( x , y , z ) = z 2 Here, z = − x − y g ( x , y ) = − ( x + y ) The plane x + y + z = 0 is contained in the cylinder x 2 + y 2 = 1 . The cylinder has the circular cross-section and the surface is the circular disc parameterized by: G ( r , t ) = ( r cos t , r sin t , − r ( cos t + sin t ) ) , 0 ≤ r ≤ 1 , 0 ≤ t ≤ 2 π And, the tangent vectors are: T r = 〈 cos t , sin t , − ( cos t + sin t ) 〉 , T t = 〈 − r sin t , r cos t , r ( sin t − cos t ) 〉 Now, the Normal vector is: N = T r × T t = 〈 cos t , sin t , − ( cos t + sin t ) 〉 × 〈 − r sin t , r cos t , r ( sin t − cos t ) 〉 = i j k cos t sin t − ( cos t + sin t ) − r sin t r cos t r ( sin t − cos t ) = [ r sin t ( sin t − cos t ) + r cos t ( cos t + sin t ) ] i − [ r cos t ( sin t − cos t ) − r sin t ( cos t + sin t ) ] j + [ r cos 2 t + r sin 2 t ] k = r ( sin 2 t − sin t cos t + cos 2 t + sin t cos t ) i − r ( sin t cos t − cos 2 t − sin t cos t − sin 2 t ) j + r ( sin 2 t + cos 2 t

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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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 22E

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To calculate:

$\iint_{S} f (x, y, z) d S$ for the given surface and function.

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Chapter 17.4, Problem 21E

Chapter 17.4, Problem 23E

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Chapter 17 Solutions

Calculus - Standalone book

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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∬ S f ( x , y , z ) d S for the given surface and function.

Solution Summary: The author calculates S f ( x, y and z ) d S for the given surface and function. The cylinder has a circular cross-section and the surface is the circular disc.

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To calculate:

$\iint_{S} f (x, y, z) d S$ for the given surface and function.

Want to see the full answer?

Chapter 17 Solutions

∬ S f ( x , y , z ) d S for the given surface and function.

Solution Summary: The author calculates S f ( x, y and z ) d S for the given surface and function. The cylinder has a circular cross-section and the surface is the circular disc.

Concept explainers

Videos

To calculate: ∬Sf(x,y,z)dS for the given surface and function.

Want to see the full answer?

Chapter 17 Solutions

To calculate:

$\iint_{S} f (x, y, z) d S$ for the given surface and function.