Explanation Given: F ( x , y , z ) = ( y , x , z ) and region x 2 + y 2 + z 2 ≤ 4 . Theorem used: Divergence theorem, ∬ S F . d S = ∭ R d i v ( F ) d V Calculation: Now calculating ∬ S F . d S , for this we use parametrization of sphere surface Surface S : ϕ ( θ , ϕ ) = ( 2 cos θ sin ϕ , 2 sin θ sin ϕ , 2 cos ϕ ) , 0 ≤ θ ≤ 2 π , 0 ≤ ϕ ≤ π n = 4 sin ϕ 〈 cos θ sin ϕ , sin θ sin ϕ , cos ϕ 〉 F ( ϕ ( θ , ϕ ) ) . n = 2 〈 sin θ sin ϕ , cos θ sin ϕ , cos ϕ 〉 .4 sin ϕ 〈 cos θ sin ϕ , sin θ sin ϕ , cos ϕ 〉 = 8 ( sin θ cos θ sin 3 ϕ + cos θ sin θ sin 3 ϕ + cos 2 ϕ sin ϕ ) = 8 ( sin 2 θ sin 3 ϕ + cos 2 ϕ sin ϕ ) ⇨ ∬ S F . d S = ∫ 0 π ∫ 0 2 π F ( ϕ ( θ , ϕ ) ) ...

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Chapter 18.3, Problem 2E

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The verification of divergence theorem for the vector field $F (x, y, z) = (y, x, z)$ and region $x^{2} + y^{2} + z^{2} \leq 4$ .

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Chapter 18.3, Problem 1E

Chapter 18.3, Problem 3E

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Chapter 18 Solutions

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The verification of divergence theorem for the vector field F ( x , y , z ) = ( y , x , z ) and region x 2 + y 2 + z 2 ≤ 4 .

Solution Summary: The author explains the verification of the divergence theorem for the vector field F (x, y, and z) and region x 2, 2, 3, and 4.

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The verification of divergence theorem for the vector field $F (x, y, z) = (y, x, z)$ and region $x^{2} + y^{2} + z^{2} \leq 4$ .

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Chapter 18 Solutions

The verification of divergence theorem for the vector field F ( x , y , z ) = ( y , x , z ) and region x 2 + y 2 + z 2 ≤ 4 .

Solution Summary: The author explains the verification of the divergence theorem for the vector field F (x, y, and z) and region x 2, 2, 3, and 4.

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The verification of divergence theorem for the vector field F(x,y,z)=(y,x,z) and region x2+y2+z2≤4.

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Chapter 18 Solutions

The verification of divergence theorem for the vector field $F (x, y, z) = (y, x, z)$ and region $x^{2} + y^{2} + z^{2} \leq 4$ .