Explanation Theorem used: Divergence theorem, ∬ S F . d S = ∭ R d i v ( F ) d V Calculation: The region of box can be converted to 6 surfaces S 1 , S 2 , S 3 , S 4 , S 5 a n d S 6 Surface S 1 : ϕ 1 ( x , z ) = ( x , 0 , z ) , 0 ≤ x ≤ 4 , 0 ≤ z ≤ 3 and n = 〈 0 , − 1 , 0 〉 Surface S 2 : ϕ 2 ( y , z ) = ( 0 , y , z ) , 0 ≤ y ≤ 2 , 0 ≤ z ≤ 3 and n = 〈 − 1 , 0 , 0 〉 Surface S 3 : ϕ 3 ( x , z ) = ( x , 2 , z ) , 0 ≤ x ≤ 4 , 0 ≤ z ≤ 3 and n = 〈 0 , 1 , 0 〉 Surface S 4 : ϕ 4 ( y , z ) = ( 0 , y , z ) , 0 ≤ y ≤ 2 , 0 ≤ z ≤ 3 and n = 〈 1 , 0 , 0 〉 Surface S 5 : ϕ 5 ( x , y ) = ( x , y , 0 ) , 0 ≤ x ≤ 4 , 0 ≤ y ≤ 2 and n = 〈 0 , 0 , − 1 〉 Surface S 6 : ϕ 6 ( x , y ) = ( x , y , 3 ) , 0 ≤ x ≤ 4 , 0 ≤ z ≤ 3 and n = 〈 0 , 0 , 1 〉 Now calculating ∬ S F . d S = ∬ S 1 F . d S + ∬ S 2 F . d S + ∬ S 3 F . d S + ∬ S 4 F . d S + ∬ S 5 F . d S + ∬ S 6 F . d S ∬ S 3 F . d S = ∫ 0 3 ∫ 0 4 F ( ϕ 1 ( x , z ) ) . 〈 0 , − 1 , 0 〉 d x d z = ∫ 0 3 ∫ 0 4 〈 z , x , 0 〉 . 〈 0 , − 1 , 0 〉 d x d z = ∫ 0 3 ∫ 0 4 − x d x d z = 3 ( − x 2 2 ) | 0 4 = − 24 ∬ S 2 F . d S = ∫ 0 3 ∫ 0 2 F ( ϕ 2 ( y , z ) ) . 〈 − 1 , 0 , 0 〉 d y d z = ∫ 0 3 ∫ 0 2 〈 z , 0 , y 〉 . 〈 − 1 , 0 , 0 〉 d y d z = ∫ 0 3 ∫ 0 2 − z d y d z = 2 ( − z 2 2 ) | 0 3 = − 9 ∬ S 3 F . d S = ∫ 0 3 ∫ 0 4 F ( ϕ 3 ( x , z ) ) ...

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Author: FRANZOSA

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Chapter 18.3, Problem 1E

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The verification of divergence theorem for the vector field $F (x, y, z) = (z, x, y)$ and region of the box $[0, 4] \times [0, 2] \times [0, 3]$ .

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Chapter 18.3, Problem 5PQ

Chapter 18.3, Problem 2E

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The verification of divergence theorem for the vector field F ( x , y , z ) = ( z , x , y ) and region of the box [ 0 , 4 ] × [ 0 , 2 ] × [ 0 , 3 ] .

Solution Summary: The author explains the verification of the divergence theorem for the vector field F (x, y

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The verification of divergence theorem for the vector field $F (x, y, z) = (z, x, y)$ and region of the box $[0, 4] \times [0, 2] \times [0, 3]$ .

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Chapter 18 Solutions

The verification of divergence theorem for the vector field F ( x , y , z ) = ( z , x , y ) and region of the box [ 0 , 4 ] × [ 0 , 2 ] × [ 0 , 3 ] .

Solution Summary: The author explains the verification of the divergence theorem for the vector field F (x, y

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The verification of divergence theorem for the vector field F(x,y,z)=(z,x,y) and region of the box [0,4]×[0,2]×[0,3].

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Chapter 18 Solutions

The verification of divergence theorem for the vector field $F (x, y, z) = (z, x, y)$ and region of the box $[0, 4] \times [0, 2] \times [0, 3]$ .