   Chapter 16.9, Problem 1E

Chapter
Section
Textbook Problem

Verify that the Divergence Theorem is true for the vector field F on the region E.1. F(x, y, z) = 3x i + xy j + 2xz k, E is the cube bounded by the planes x =0, x = 1, y = 0, y = 1, z = 0, and z = 1

To determine

To Verify: The divergence theorem for the vector field F(x,y,z)=3xi+xyj+2xzk on the region E.

Explanation

Given data:

The vector field is F(x,y,z)=3xi+xyj+2xzk .

The region E is the cube bounded by the planes x=0,x=1,y=0,y=1,z=0 , and z=1 .

Formula used:

Write the formula of divergence theorem.

SFdS=EdivFdV (1)

Here,

E is the solid region.

Write the expression to find divergence of vector field F(x,y,z)=Pi+Qj+Rk .

divF=xP+yQ+zR (2)

Write the expression for SFdS in terms of normal vector to the plane as follows.

SFdS=SFndS (3)

Here,

n is the normal vector to the plane of the surface S.

Calculation of divF :

Substitute 3x for P , xy for Q , and 2xz for R in equation (2),

divF=x(3x)+y(xy)+z(2xz)=3+x+2x=3x+3

Calculation of flux EdivFdV :

Substitute (3x+3) for divF in the expression EdivFdV as follows.

EdivFdV=E(3x+3)dV

Apply the limits of x, y, and z, and rewrite the expression as follows.

EdivFdV=010101(3x+3)dxdydz

Simplify the expression as follows.

EdivFdV=01(1)dz01(1)dy01(3x+3)dx=[z]01[y]01[3(x22)+3x]01={[32(1)2+3(1)][32(0)2+3(0)]}=92

EdivFdV=92 (4)

Calculation of SFdS :

As the cube is bounded by the planes x=0,x=1,y=0,y=1,z=0 , and z=1 compute the expression for flux SFdS as follows.

Calculate the flux for each plane separately and add all the individual results to find the value of SFdS .

Write the expression to find the value of SFdS as follows.

SFdS=S1FdS+S2FdS+S3FdS+S4FdS+S5FdS+S6FdS (5)

For the plane x=0 :

Consider the surface for the plane x=0 is S1 .

The normal vector for the plane x=0 is n=i .

Substitute 0 for x in the vector field F=3xi+xyj+2xzk ,

F=3(0)i+(0)yj+2(0)zk=0i+0j+0k

Substitute S1 for S , (i) for n and (0i+0j+0k) for F in equation (3),

S1FdS=S1(0i+0j+0k)(i)dS=S1(0)dS=0

For the plane x=1 :

Consider the surface for the plane x=1 is S2 .

The normal vector for the plane x=1 is n=i .

Substitute 1 for x in the vector field F=3xi+xyj+2xzk .

F=3(1)i+(1)yj+2(1)zk=3i+yj+2zk

Substitute S2 for S , i for n and (3i+yj+2zk) for F in equation (3),

S2FdS=S2(3i+yj+2zk)(i)dS=S2(3)dS=3

For the plane y=0 :

Consider the surface for the plane y=0 is S3

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