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∬SF(x,y,z)·ndS,where Sis the portion of the surface

x=z3that lies between the planes y=13and y=42and whereF(x,y,z)=cos(xy),1y2z2+1,x+y

Short Answer

Expert verified

No, the integral ∬SF(x,y,z)·ndS, cannot be evaluated by means of Divergence Theorem.

Step by step solution

01

Introduction

Consider the vector field below:

F(x,y,z)=cos(xy),1y2z2+1,x+y

The goal is to determine whether the integral ∬SF(x,y,z)·ndS,can be calculated using the Divergence Theorem, where the surfacerole="math" localid="1650777168836" x=z3is the piece of the surface Sthat falls between the planes y=13andy=42

Divergence According to the theorem,

"Let Wbe a bounded region in R3with a smooth or piecewise-smooth closed oriented surface as its boundary SIf on an open region containing W, a vector field F(x,y,z)is defined, then

∬SF(x,y,z)·ndS=∭WdivF(x,y,z)dV."........(1)

where ndenotes the outwardly unit normal vector

02

Explanation

The surface integral is converted to a triple integral using the Divergence Theorem. In R3, the triple integral is calculated over the bounded region W. A region Wcan only be found inside closed surfaces. A closed surface encloses a certain area. As a result, the Divergence Theorem does not apply to non-closed surfaces.

The surface Sis the portion of the surface x=z3that is not closed and lies between the planes y=13and y=42. As a result, it does not enclose any area.

No, the integral ∬SF(x,y,z)·ndS, cannot be evaluated by means of Divergence Theorem.

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