Explanation Given : The surface is over the cylinder x 2 + y 2 = 4 , 1 ≤ z ≤ 5 outward pointing normal. Calculation : Step 1: The surface is parametrized by: Φ ( θ , z ) = ( 2 c o s θ , 2 s i n θ , z ) where the parameter domain is D : 0 ≤ θ ≤ 2 π , 1 ≤ z ≤ 5 For this parametrization, it holds that: T θ × T z = 〈 2 cos θ , 2 sin θ , 0 〉 Step 2: This vector is horizontal and points out of the cylinder. Therefore, n = 〈 2 cos θ , 2 sin θ , 0 〉 Step 3: The dot product F • n is computed as follows: F ( Φ ( θ , z ) ) • n ( θ , z ) = 〈

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.5, Problem 12E

To determine

To calculate: The value of $\int \int_{S} F . d s$ for $F = 〈 x, y, e^{z} 〉$ ,over the cylinder $x^{2} + y^{2} = 4, 1 \leq z \leq 5$ outward pointing normal.

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Chapter 17.5, Problem 11E

Chapter 17.5, Problem 13E

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Chapter 17 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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The value of ∫ ∫ S F . d s for F = 〈 x , y , e z 〉 ,over the cylinder x 2 + y 2 = 4 , 1 ≤ z ≤ 5 outward pointing normal.

Solution Summary: The author explains how the surface is parametrized by: (, z ) = ( 2 c o

Videos

To calculate: The value of $\int \int_{S} F . d s$ for $F = 〈 x, y, e^{z} 〉$ ,over the cylinder $x^{2} + y^{2} = 4, 1 \leq z \leq 5$ outward pointing normal.

Want to see the full answer?

Chapter 17 Solutions

The value of ∫ ∫ S F . d s for F = 〈 x , y , e z 〉 ,over the cylinder x 2 + y 2 = 4 , 1 ≤ z ≤ 5 outward pointing normal.

Solution Summary: The author explains how the surface is parametrized by: (, z ) = ( 2 c o

Videos

To calculate: The value of ∫∫SF.ds for F=〈x,y,ez〉,over the cylinder x2+y2=4,1≤z≤5 outward pointing normal.

Want to see the full answer?

Chapter 17 Solutions

To calculate: The value of $\int \int_{S} F . d s$ for $F = 〈 x, y, e^{z} 〉$ ,over the cylinder $x^{2} + y^{2} = 4, 1 \leq z \leq 5$ outward pointing normal.