Explanation Divergence Theorem says about the relation between surface and volume integral. Let S be a closed smooth surface, which is boundary of V, with normal pointing outwards. Let F = 〈 x , y , z 〉 on a region containing V, then the flux is equal to triple integral of div F over the solid region V. ∬ S F ⋅ d S = ∭ V d i v ( F ) d V Given: F ( x , y , z ) = 〈 x , y 2 , z + y 〉 , S is the boundary of the region contained in the cylinder x 2 + y 2 = 4 between the planes z = x and z = 8 . Formula Used: ∬ S F ⋅ d S = ∭ V d i v ( F ) d V Calculation: First we have to calculate the div F and we get: d i v ( F ) = ∂ ∂ x ( x ) + ∂ ∂ y ( y 2 ) + ∂ ∂ z ( z + y ) = 1 + 2 y + 1 = 2 + 2 y ∬ S F ⋅ d S = ∭ V ( 2 + 2 y ) d x d y d z It is given that x 2 + y 2 = 4 , z = x and z = 8 Now, change the variables x , y and z in r , z and θ because it is a cylinder and we get the cylindrical coordinates: x = r cos θ , y = r sin θ , z = z and d V = r d z d r d θ So, the triple integral will become: ∬ S F ⋅ d S = ∭ V ( 2 + 2 r sin θ ) r d z d r d θ Now, the range of r , z and θ is: ...

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Chapter 18.3, Problem 13E

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The flux $\iint_{S} F \cdot d S$ by using the Divergence Theorem.

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Chapter 18.3, Problem 12E

Chapter 18.3, Problem 14E

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Chapter 18 Solutions

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The flux ∬ S F ⋅ d S by using the Divergence Theorem.

Solution Summary: The author calculates the flux by using the Divergence Theorem. Let S be a closed smooth surface, with normal pointing outwards.

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The flux $\iint_{S} F \cdot d S$ by using the Divergence Theorem.

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Chapter 18 Solutions

The flux ∬ S F ⋅ d S by using the Divergence Theorem.

Solution Summary: The author calculates the flux by using the Divergence Theorem. Let S be a closed smooth surface, with normal pointing outwards.

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To calculate: The flux ∬SF⋅dS by using the Divergence Theorem.

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Chapter 18 Solutions

To calculate:

The flux $\iint_{S} F \cdot d S$ by using the Divergence Theorem.