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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Verify that the Divergence Theorem is true for the vector field F on the region E.

4. F(x, y, z) = ⟨x2, -y, z, E is the solid cylinder y2 + z2 ≤ 9, 0 ≤ x ≤ 2

To determine

To Verify: The divergence theorem for the vector field F(x,y,z)=x2,y,z on the region E.

Explanation

Given data:

The vector field is F(x,y,z)=x2,y,z .

The region E is the solid cylinder y2+z29 and 0x2 .

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 as follows.

SFdS=SFndS (3)

Here,

n is the unit outward normal of the surface S.

Calculation of divF :

Substitute x2 for P , (y) for Q , and z for R in equation (2),

divF=x(x2)+y(y)+z(z)=2x1+1=2x

Calculation of flux EdivFdV :

Substitute 2x for divF in the expression EdivFdV as follows.

EdivFdV=E(2x)dV

Apply the limits of the region E and rewrite the expression as follows.

EdivFdV=y2+z29[022xdx]dA=y2+z29(4)dA=4A(D)

Here,

A(D) is the area of the circle with radius 3.

As the area of the circle with radius r is πr2 , the area with radius 3 is π(3)2 , which is 9π .

Substitute 9π for A(D) in the expression EdivFdV=4A(D) .

EdivFdV=4(9π)

EdivFdV=36π (4)

Consider the surface S1 is the front of the cylinder in the plane x=2 , surface S2 is the back of the cylinder in the yz-plane, and S3 is the lateral surface of the cylinder.

Calculate flux on each surface and add the individual results to find the value of SFdS .

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

SFdS=S1FdS+S2FdS+S3FdS (5)

Calculation of S1FdS :

The surface S1 is the disk in the plane x=2 , y2+z29 .

The unit normal vector to the surface S1 is written as follows.

n=1,0,0

Substitute 2 for x in the vector field F=x2,y,z to find the vector field in the surface S1 .

F=22,y,z=4,y,z

Substitute S1 for S , 1,0,0 for n and 4,y,z for F in equation (3),

S1FdS=S14,y,z1,0,0dS=S1(4)dS=4(Surface area of S1)=4[π(3)2]

S1FdS=36π

Calculation of S2FdS :

The surface S2 is the disk in the plane x=0 , y2+z29 .

The unit normal vector to the surface S2 is written as follows.

n=1,0,0

Substitute 0 for x in the vector field F=x2,y,z to find the vector field in the surface S2 .

F=02,y,z=0,y,z

Substitute S2 for S , 1,0,0 for n and 0,y,z for F in equation (3),

S2FdS=S20,y,z1,0,0dS=S2(0)dS=0

The parametric equations of the surface S3 are written as follows

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Chapter 16 Solutions

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Sect-16.1 P-11ESect-16.1 P-12ESect-16.1 P-13ESect-16.1 P-14ESect-16.1 P-15ESect-16.1 P-16ESect-16.1 P-17ESect-16.1 P-18ESect-16.1 P-21ESect-16.1 P-22ESect-16.1 P-23ESect-16.1 P-24ESect-16.1 P-25ESect-16.1 P-26ESect-16.1 P-29ESect-16.1 P-30ESect-16.1 P-31ESect-16.1 P-32ESect-16.1 P-33ESect-16.1 P-34ESect-16.1 P-35ESect-16.1 P-36ESect-16.2 P-1ESect-16.2 P-2ESect-16.2 P-3ESect-16.2 P-4ESect-16.2 P-5ESect-16.2 P-6ESect-16.2 P-7ESect-16.2 P-8ESect-16.2 P-9ESect-16.2 P-10ESect-16.2 P-11ESect-16.2 P-12ESect-16.2 P-13ESect-16.2 P-14ESect-16.2 P-15ESect-16.2 P-16ESect-16.2 P-17ESect-16.2 P-18ESect-16.2 P-19ESect-16.2 P-20ESect-16.2 P-21ESect-16.2 P-22ESect-16.2 P-23ESect-16.2 P-24ESect-16.2 P-25ESect-16.2 P-26ESect-16.2 P-31ESect-16.2 P-32ESect-16.2 P-33ESect-16.2 P-34ESect-16.2 P-35ESect-16.2 P-36ESect-16.2 P-37ESect-16.2 P-38ESect-16.2 P-39ESect-16.2 P-40ESect-16.2 P-41ESect-16.2 P-42ESect-16.2 P-43ESect-16.2 P-44ESect-16.2 P-45ESect-16.2 P-46ESect-16.2 P-47ESect-16.2 P-48ESect-16.2 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