Explanation Let the vector field be F → = y i ^ + z 2 j ^ − 2 z k ^ and S is the part of the surface cut from the cylinder z = 3 − x 2 by y = 0 , y = 1 , and z = 0 . So, f ( x , y ) = z = 3 − x 2 And, f x = − 2 x and f y = 0 . So, the value of the integral ∬ S F → ⋅ N → d S is, ∬ S F → ⋅ N → d S = ∬ D F → ( x , y , f ( x , y ) ) ⋅ 〈 − f x , − f y , 1 〉 d A = ∫ x = − 3 3 ∫ y = 0 1 〈 y , ( 3 − x 2 ) 2 , 2 x 2 − 6 〉 ⋅ 〈 − ( − 2 x ) , − ( 0 )

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ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 39PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=yi^+z2j^−2zk^ and S is the part of the surface cut from the cylinder z=3−x2 by y=0,y=1 , and z=0 .

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Chapter 13.5, Problem 38PS

Chapter 13.5, Problem 40PS

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7. Sketch the vector field F = (x − y)î + yĵ. Identify where F₁ vanishes. Identify where F2 vanishes. Plot a few flow lines. 8. Repeat the previous problem for F = yî + (x − y)î.

Incorrect. Use a computer or calculator with Euler's method to approximate the flow line through (1, 2) for the vector field v = y² i + 1.1x² j using 5 steps with At = 0.1. Find the exact values of x1,.…. ,X5 and yı, , y5 and then fill in the blanks rounding your numbers to •.... three decimal places. X1 = i 1.1 ,Yi = i 2.51 X2 i 1.2 ,y2 3.273 X3 = i 1.3 ,Y3 = i 4.50283 X4 = i 1.4 , Y4 i 6.7162 X5 = i 1.5 , Y5 = i 11.4427 eTextbook and Media Hint I| ||

Incorrect. Use a computer or calculator with Euler's method to approximate the flow line through (1, 2) for the vector field v = y² i +1.1x² j using 5 steps with At = 0.1. Find the exact values of x1, ... , x5 and y1, ... , y5 and then fill in the blanks rounding your numbers to three decimal places. X1 = !Yı = i X2 i 1.2 , y2 3.273 X3 = i 1.3 Y3 = i 4.50283 X4 i 1.4 6.7162 X5 = i 1.5 Y5 = i 11.4427 eTextbook and Media Assistance Used Hint Assistance Used The vector field is given by v = y i + 1.1x² j , that is, the flow line (x (t), y (t)) satisfies x' (t) = y² y' (t) = 1.1x².

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = y i ^ + z 2 j ^ − 2 z k ^ and S is the part of the surface cut from the cylinder z = 3 − x 2 by y = 0 , y = 1 , and z = 0 .

Solution Summary: The author calculates the value of the integral displaystyle 'underset' Siint's stackrel to a and S is the surface cut from the cylinder

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=yi^+z2j^−2zk^ and S is the part of the surface cut from the cylinder z=3−x2 by y=0,y=1 , and z=0 .

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Chapter 13 Solutions