Explanation Let the vector field be F → = x i ^ 2 + y 2 j ^ + z 2 k ^ and S is the part of cone z = x 2 + y 2 below the plane z = 2 . We will use the divergence theorem, so div F ^ = ∂ ∂ x ( x 2 ) + ∂ ∂ y ( y 2 ) + ∂ ∂ z ( z 2 ) = 2 x + 2 y + 2 z = 2 ( x + y + z ) Thus, ∬ S F → ⋅ N → d S = ∭ R div F ^ d V = ∫ − 2 2 ∫ − 4 − x 2 4 − x 2 ∫ 0 2 ( 2 ( x + y + z ) ) d z d y d x = ∫ − 2 2 ∫ − 4 − x 2

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

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

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 40PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^2+y2j^+z2k^ and S is the part of cone z=x2+y2 below the plane z=2 .

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Chapter 13.5, Problem 39PS

Chapter 13.5, Problem 41PS

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1. For each curve, construct dr. • Let Ci be r? + y² = 16 • Let C2 be y = 5 – 2x • Let C3 be r = 2 – y? 2. Create a plot of each vector field below. F = y& +5 ŷ • G = -4x x – 4y ŷ • H= =&+rŷ 3. Compute fe, F- dĩ 4. Compute fe G · dr 5. Сompute fo Н. dr

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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Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = x i ^ 2 + y 2 j ^ + z 2 k ^ and S is the part of cone z = x 2 + y 2 below the plane z = 2 .

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^2+y2j^+z2k^ and S is the part of cone z=x2+y2 below the plane z=2 .

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Chapter 13 Solutions