Explanation Determine the region that is enclosed between the circle x 2 + y 2 = 4 , the plane z = x + 1 and the x y − plane, to determine the flux. Given : Let W be the region in Figure 19 bounded by the cylinder x 2 + y 2 = 4 , the plane z = x + 1 and the x y − plane. Formula used : ∬ S F . d S = ∭ W d i v ( F ) d V = ∭ W 2 d V Calculation : We compute the divergence of F ( x , y , z ) = 〈 z , x , y + z 2 〉 : d i v ( F ) = ∂ ∂ x ( z ) + ∂ ∂ z ( x ) + ∂ ∂ z ( y + z 2 ) = 2 z By Divergence Theorem we have ∬ S F . d S = ∭ W d i v ( F ) d V = ∭ W 2 d V To compute the triple integral, we identify that the region D is enclosed by the circle x 2 + y 2 = 4 and the line 0 = x + 1 = > x = − 1 Therefore, ∬ S F . d S = ∭ W 2 d V = ∬ D ∫ 0 x + 1 2 z d z d x d y = ∬ D ( x + 1 ) 2 d x d y We compute the double integral as the difference of two integrals: the integral over the disk D 2 of radius 2 and the integral over the part D 1 of the disk which can be written as ∬ S F ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18.3, Problem 23E

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The flux of $F (x, y, z) = 〈 z, x, y + z^{2} 〉$ through the boundary of $W$ using the Divergence Theorem.

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Chapter 18.3, Problem 22E

Chapter 18.3, Problem 24E

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The flux of F ( x , y , z ) = 〈 z , x , y + z 2 〉 through the boundary of W using the Divergence Theorem.

Solution Summary: The author calculates the flux of F through the boundary of W using the Divergence Theorem.

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To calculate:

The flux of $F (x, y, z) = 〈 z, x, y + z^{2} 〉$ through the boundary of $W$ using the Divergence Theorem.

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Chapter 18 Solutions

The flux of F ( x , y , z ) = 〈 z , x , y + z 2 〉 through the boundary of W using the Divergence Theorem.

Solution Summary: The author calculates the flux of F through the boundary of W using the Divergence Theorem.

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To calculate: The flux of F(x,y,z)=〈z,x,y+z2〉 through the boundary of W using the Divergence Theorem.

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Chapter 18 Solutions

To calculate:

The flux of $F (x, y, z) = 〈 z, x, y + z^{2} 〉$ through the boundary of $W$ using the Divergence Theorem.