Explanation Given : The surface is over the cone z 2 = x 2 + y 2 , x 2 + y 2 = 4 , z ≥ 0 downward pointing normal Calculation : Step 1: The surface is parametrized by: Φ ( θ , t ) = ( t c o s θ , t s i n θ , t ) where the parameter domain is D : { ( θ , t ) : 0 ≤ θ ≤ 2 π , 0 ≤ t ≤ 2 } For this parametrization, it holds that: T θ × T t = 〈 t cos θ , t sin θ , − t 〉 Step 2: Since the normal is pointing downward, the z-coordinate must be negative. Therefore, n = 〈 t cos θ , t sin θ , − t 〉 Step 3: We express F in terms of the parameters and then compute the dot product F • n . We obtain: F ( Φ ( θ , t ) ) = 〈 x y , y , 0 〉 = 〈 t cos θ t sin θ , t sin θ , 0 〉 = 〈 t 2 cos θ sin θ , t sin θ , 0 〉 F ( Φ ( θ , t ) ) • n ( θ , t ) = t sin θ 〈 t cos θ , 1 , 0 〉 . t 〈 cos θ , sin θ , − 1 〉 = t 2 sin θ ( t cos 2 θ + sin θ ) = t 3 cos 2 θ sin θ + t 2 sin 2 θ Step 4: The surface integral is equal to the following double integral : ∫ ∫ S F ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 17.5, Problem 14E

To determine

To calculate: The value of $\int \int_{S} F . d s$ for $F = 〈 x y, y, 0 〉$ ,over the cone $z^{2} = x^{2} + y^{2}, x^{2} + y^{2} = 4, z \geq 0$ downward pointing normal.

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Chapter 17.5, Problem 13E

Chapter 17.5, Problem 15E

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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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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 - 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The value of ∫ ∫ S F . d s for F = 〈 x y , y , 0 〉 ,over the cone z 2 = x 2 + y 2 , x 2 + y 2 = 4 , z ≥ 0 downward pointing normal.

Solution Summary: The author calculates the value of S F. d s for x, y, and z 0 downward pointing normal.

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To calculate: The value of $\int \int_{S} F . d s$ for $F = 〈 x y, y, 0 〉$ ,over the cone $z^{2} = x^{2} + y^{2}, x^{2} + y^{2} = 4, z \geq 0$ downward pointing normal.

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Chapter 17 Solutions

The value of ∫ ∫ S F . d s for F = 〈 x y , y , 0 〉 ,over the cone z 2 = x 2 + y 2 , x 2 + y 2 = 4 , z ≥ 0 downward pointing normal.

Solution Summary: The author calculates the value of S F. d s for x, y, and z 0 downward pointing normal.

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To calculate: The value of ∫∫SF.ds for F=〈xy,y,0〉,over the cone z2=x2+y2,x2+y2=4,z≥0 downward pointing normal.

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Chapter 17 Solutions

To calculate: The value of $\int \int_{S} F . d s$ for $F = 〈 x y, y, 0 〉$ ,over the cone $z^{2} = x^{2} + y^{2}, x^{2} + y^{2} = 4, z \geq 0$ downward pointing normal.